Solution manual For Business Analytics Applied Modelling and Prediction 1e James Abdey

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 1: Decision-making under uncertainty

ii

Chapter 1 Decision-making under uncertainty 1.1

Exercises

1.1.1

Test your understanding

1. What is meant by ‘uncertainty’ ?

Solution: Uncertainty refers to a lack of knowledge, information or predictability about a particular situation, event or outcome. It signifies a state of doubt or ambiguity where the future or the true nature of something is uncertain. Uncertainty can arise from various factors, such as incomplete data, conflicting evidence, complexity, or the presence of unpredictable variables. It often poses challenges in decision-making, planning, and risk assessment as it hampers our ability to accurately foresee or quantify potential outcomes. Managing uncertainty involves acknowledging its presence, assessing probabilities, considering alternative scenarios, and adopting flexible approaches to adapt to changing circumstances.

2. What is the difference between ‘dependent’ and ‘independent’ variables?

Solution: A dependent variable is the variable being measured or observed in an experiment or study. It is the outcome or result in which we are interested in understanding, explaining or predicting. Examples of dependent variables in business analytics include: • behavioural responses, such as customer awareness, brand perception, knowledge, intention to purchase and preference.

• performance metrics, such as sales revenues, production costs, profits, market share, cash-flow and share price.

Independent variables are those that impact the dependent variable – think of it in terms of ‘cause and effect’. Examples of independent variables in business analytics include: • marketing mix variables, such as product, price, placement and promotion (the 4 Ps), which are controllable, i.e. are under the control of the decision-maker.

• situational factors, such as demand, competition, economic climate, political climate, exchange rates, supply chain, regulations and technology, which are uncontrollable, i.e. are not under the control of the decision-maker.

The relationship between the dependent and independent variables is often explored through statistical analysis to determine whether changes in the independent variable cause changes in the dependent variable. This helps establish cause-and-effect relationships and provides insights into the phenomenon under investigation. Understanding the distinction between dependent and

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1. Decision-making under uncertainty

independent variables is fundamental for designing experiments, conducting research, and drawing meaningful conclusions from data.

3. Identify examples of real-world dependent variables for any business problem of your choice.

Solution: Consider a hotel. Management may be interested in the following dependent variables.

• Customer satisfaction score – this could be measured using surveys or feedback forms where customers rate their overall satisfaction with the hotel, its services, staff, amenities, cleanliness, etc. The satisfaction score can be a numerical rating or a categorical measure (for example, ranging from ‘very satisfied’ to ‘ very dissatisfied’).

• Revenue per customer – this variable represents the average amount of money a customer spends during their stay. Higher revenue per customer can indicate customer satisfaction, as satisfied guests are more likely to use additional services, such as room upgrades, spa treatments, or dining options.

• Customer retention rate – this metric indicates the percentage of customers who continue to stay at the hotel over a specific period of time. It reflects customer satisfaction and loyalty, as satisfied customers are more likely to return in the future!

4. For each dependent variable identified in Question 3, what are the likely independent variables?

Briefly justify your choice.

Solution: Customer satisfaction score would likely depend on the following: • Service quality.

• Price/value for money.

• Food and beverage quality.

• Room cleanliness.

• Location.

• Facilities.

Revenue per customer would likely depend on the following: • Room type booked.

• Length of stay.

• Season of the booking (peak vs. off-peak).

• Ancillary services used.

• In-room dining.

• Purpose of visit (business vs. leisure).

Customer retention rate would likely depend on the following: • Service quality (note also for customer satisfaction score!).

• Price/value for money (note also for customer satisfaction score!).

• Loyalty programs (accrue points for free nights, perhaps).

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1.1. Exercises

• Special packages (think conferences, weddings etc.).

• Brand image.

• Partnership with other hotel brands.

Note the above are not necessarily collectively exhaustive lists!

5. What is meant by the ‘four Ps’ ?

Solution: The ‘four Ps’ are known as the marketing mix variables: • Product – what the business is selling (which could be a tangible good, or intangible service).

• Price – what price to sell at and whether to offer a discount.

• Placement – the provision of customer access and convenience.

• Promotion – the choice of marketing communications.

6. Explain the DIKW pyramid.

Solution: The pyramid models the conversion of (i) data, into (ii) information, from which we gain (iii) knowledge, leading to our enlightenment with (iv) wisdom. The pyramid illustrates these as hierarchical relationships.

7. What is a model?

Solution: A model is a deliberate simplification of reality.

A good model retains the most important features of reality and ignores less important details.

Immediately we see that we face a trade-off. The benefit of a model is that we simplify the complex real world. The cost of a model is that by simplifying reality we engineer a departure from reality.

Broadly speaking, we would be happy if the benefit exceeded the cost, i.e. if the simplicity made it easier for us to understand and analyse the real world while incurring only a minimal departure from reality.

8. For a city other than London with a metro system, look at the official map of the network. Critique the map, noting any possible improvements you would recommend are made.

Solution: Possible improvements may relate to: • geographic accuracy • clarity of presentation • ease of navigation.

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1. Decision-making under uncertainty

9. A friend of yours is a contestant on the Monty Hall game show. Your friend tells you ‘No matter what, I will not switch door’. Would you support your friend’s decision (to maintain your friendship!) or would you advise otherwise?

Solution: At the risk of jeopardising your friendship, you should (politely) point out that probabilistically it is optimal to switch door. Of course, you need to balance this with the possibility that the original door chosen is the correct door, in which case you may never be forgiven if your friend followed your advice!

10. Suppose the producers of the Monty Hall show changed the format of the game. Instead of there being three doors, there are now n > 3 doors. The contestant chooses a door. Monty (who continues to know where the car is) opens n − 2 doors, leaving the contestant’s original door and one other door unopened. (a) In terms of ‘to switch, or not to switch?’ what would you recommend is the optimal strategy and why? (b) Do you support the producers in increasing the number of doors to n > 3? Justify your view.

Solution: (a) With n > 3 doors, the probability that the correct door is chosen at the start by the contestant is 1/n. Hence the probability that the correct door is one of the other n − 1 doors is (n − 1)/n, and clearly: n−1 1 < . n n Once Monty opens n − 2 doors (excluding the contestant’s original choice, of course), the probability of winning by switching is (n − 1)/n since this amount of probability is ‘inherited’ by the one door Monty chooses not to open. Decision-making under uncertainty requires us to ‘play to the probabilities’, hence it is optimal to switch.

Chapter 4 solves the Monty Hall problem more formally using Bayes’ theorem. (b) If contestants are savvy enough to understand the probabilistic argument for switching door each time, then we would expect all such contestants to switch. This would mean that (in percentage terms) the win rate of contestants (and hence the loss rate of the show’s producers) in the long run would be: n−1 × 100%. n As n increases, this worsens the situation for the show’s producers (more prizes are won, which costs money!) so increasing the number of doors would seem to be a bad idea. Also, television shows are meant to be for entertainment, so with n = 3 doors this maximises the extent of unpredictability about whether the contestant would win, which in turn would likely be optimal for audience viewing figures. Suspense is more exciting!

1.1.2

Case study corner

For a company or industry of your choice, formulate a possible quantitative business problem which could be analysed. In particular, think about the following.

What is the specific business question to be addressed?

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1.1. Exercises

What are the business objectives of the analysis?

What are the dependent and independent variables?

Are there any challenges in collecting relevant data to analyse the problem?

Are there any uncertainties involved with the business problem? If so, what are they?

Solution: Formulating a problem suitable for quantitative analysis begins with posing a question. Suppose the company’s question is: ‘What price should we set for our new product?’

While we may have previously relied on gut instinct decisions, we now aim to approach this question using a more quantitative approach. To answer it, we must identify the factors which may influence the pricing decision. The challenge lies in the lack of knowledge regarding consumers’ willingness to pay.

However, we would likely have data on manufacturing costs and marketing expenses for our product (or at least reasonable estimates).

Market research could be conducted to examine the current market landscape and the pricing strategies adopted by competitors for similar products. Such research encompasses various approaches, each with its own set of assumptions and uncertainties. By analysing this market research, we could estimate what consumers might be willing to pay for a similar product and develop a pricing structure accordingly.

However, there are additional uncertainties that need to be considered. For instance: Will there still be demand for our product when it is launched?

Will consumers have the financial means to purchase it at that time? (If there was a recession, consumers may reduce their discretionary spending.)

Even if we price our product lower than our competitor(s), will consumers remain loyal to their familiar brand and refrain from purchasing our product?

To address these uncertainties, we could conduct surveys to collect data on brand loyalty, make predictions regarding consumers’ future willingness to pay, and assume that demand for our product will be strong at launch. By gathering relevant data related to these uncertainties and incorporating logical assumptions, we can create data-driven models to predict the outcomes of our proposed pricing structures.

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Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 2: Descriptive statistics

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Chapter 2 Descriptive statistics 2.1

Exercises

2.1.1

Test your understanding

1. Why do we tend to work with sample data rather than population data?

Solution: There are several reasons why sample data are preferred over population data.

• Collecting data from an entire population can be very time-consuming and expensive. Sampling allows data analysts to gather information from a smaller subset of the population, which is typically more feasible in practice.

• In many cases, accessing the entire population data is simply not possible. For example, for large populations (such as that of a country) details of the population may be unknown. As such, working with a representative sample becomes the only viable option.

• If a sample is well-designed and representative of the population, it can provide accurate and precise estimates of population parameters. Statistical techniques allow analysts to generalise their findings from the sample to the broader population with reasonable confidence.

• Analysing and interpreting sample datasets is often more manageable than dealing with much larger population datasets. It can simplify data analysis, making it easier to identify patterns, trends and relationships within the data.

2. What are the different levels of measurement of variables?

Solution: Qualitative (or categorical) variables are either nominal (no rank order) or ordinal (ranked in order). These categories represent an attribute such that there is a non-numeric value assigned to that attribute. Examples include market sector, socio-economic group and brand name.

Quantitative (or measurable) variables represent an attribute which takes a numeric value.

Examples include monetary amounts (revenues, costs, profits), market share and sales.

3. Provide some real-world business examples of discrete and continuous variables.

Solution: Some examples of discrete variables are the following: • Number of employees – the number of employees, or headcount, in a company is a discrete variable because it takes integer values only. You cannot have a fraction of an employee.

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2. Descriptive statistics

• Customer satisfaction rating – often measured on a ‘Likert scale’ (covered in Chapter 8) ranging from 1 to 5 representing distinct levels of satisfaction.

• Product defects – the number of defective items in a manufacturing process. Defects are whole, they cannot take fractional values.

Some examples of continuous variables are the following: • Sales revenue – the total amount of sales generated by a business is a continuous variable because it can take on any value within a range, including fractional values, such as £748,589.29.

• Duration of a worker’s shift – an amount of time measured in hours, minutes, seconds etc.

• Percentage discount – goods and services may be discounted in terms of percentage points, such as 12.50% off.

Note that some variables can be discrete or continuous depending on how they are measured or categorised. For example, if we measure time in hours, it becomes a discrete variable as it can only take on whole numbers. However, if we measure time in seconds or milliseconds, it becomes a continuous variable as it can have fractional values.

4. What does MECE stand for, and what do these terms (ME and CE) mean in practice?

Solution: MECE stands for mutually exclusive and collectively exhaustive. A MECE mentality allows us to partition problems. Mutually exclusive means at most one event can happen, while collectively exhaustive means at least one event can happen.

The twenty chapters of this textbook are mutually exclusive (each chapter is distinct) and collectively exhaustive (there are twenty chapters in total, excluding the appendices).

5. What is skewness, and what is the difference between positively skewed and negatively skewed?

Solution: Skewness measures the departure from symmetry, i.e. the extent of asymmetry of a distribution.

Positively skewed distributions have a longer right tail, with the majority of values concentrated on the left side. Negatively skewed distributions have a longer left tail, with the majority of values concentrated on the right side.

6. What is the purpose of measures of central tendency, and what are the main measures (along with their Excel functions)?

Solution: Measures of central tendency summarise and describe the centre or typical value of a dataset. They provide a single value which represents the central or average value around which the data tend to cluster. The main measures of central tendency are: • mean – the arithmetic average, which is calculated using the AVERAGE function in Excel.

• median – the middle value when the data are ordered from the smallest to the largest. The MEDIAN function in Excel can be used to calculate it.

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2.1. Exercises

• mode – the most frequently occurring value in the dataset. The MODE function in Excel can determine the mode(s) of a dataset (there could be more than one).

7. Why is the mean a good summary of the central tendency?

Solution: The sum of deviations from the mean is 0, i.e. we have: n X

(Xi − X̄) = 0.

i=1

The mean is ‘in the middle’ of the observations X1 , X2 , . . . , Xn , in the sense that positive and negative values of the deviations Xi − X̄ cancel out, when summed over all the observations.

8. What is an outlier, and how might you identify one?

Solution: An outlier is an extreme observation which departs greatly from ‘typical’ observations.

Unfortunately, there is no universally agreed definition of an outlier. You might define an outlier as any value more than two or three standard deviations from the mean, but this is only a rule of thumb.

9. What is the difference between a percentile and a quartile? Are they ever the same?

Solution: For any percentage p, the pth percentile is the value such that a percentage p of all values are less than it. The quartiles divide the data into four groups, each with (approximately) a quarter of all observations. Naturally, the first, second and third quartiles are the percentiles corresponding to p = 25, p = 50 and p = 75, respectively.

10. Why should we consider measures of dispersion in addition to measures of central tendency?

Solution: Measures of dispersion complement measures of central tendency by providing information about the spread, or variability, of data. They help us understand how much the data deviate from the central value, providing a more comprehensive picture of the dataset and enabling better analysis, comparison and decision-making.

11. Which is better, the (sample) variance or the (sample) standard deviation? Briefly justify your view and state what the respective Excel functions are.

Solution: The (sample) standard deviation is generally preferred over the (sample) variance. The standard deviation is in the same units as the original data, making it easier to interpret and compare. It provides a measure of dispersion that is directly relatable to the data, whereas the sample variance is in squared units, which can be less intuitive. The sample standard deviation is more sensitive to outliers compared to the sample variance. By taking the square root of the variance, the standard deviation emphasises the deviations from the mean, including the effect of outliers.

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2. Descriptive statistics

The respective Excel functions for calculating these measures are the following. Sample variance: VAR; population variance: VAR.P; sample standard deviation: STDEV; population standard deviation: STDEV.P. 12. Why do we divide by n − 1 when calculating the sample variance?

Solution: We divide by n − 1 to ensure that the sample variance, S 2 , is an unbiased estimator of σ 2 , the population variance. This means that using the divisor n − 1 ensures that the sample variance is equal to the population variance on average. Using the n divisor systematically underestimates σ 2 , resulting in a (negatively) biased estimator. We would rather estimate something on average correctly, than incorrectly, ceteris paribus. Dividing by n − 1 ensures this.

13. Name two types of quantile-based measures of dispersion. Which do you prefer, and why?

Solution: One is the range, calculated as X(n) − X(1) = maximum − minimum. The other is the interquartile range (IQR), where IQR = q75 − q25 = Q3 − Q1 .

The range is very sensitive to outliers, hence the IQR is generally preferred.

14. What information can be deduced from a boxplot?

Solution: A boxplot provides information about the distribution of a dataset, including the median, quartiles, range, and potential outliers. It helps identify the central tendency, spread, skewness, and presence of extreme values in the data.

15. For what purpose would we consider using side-by-side boxplots?

Solution: Side-by-side boxplots are used to compare and visualise the distributions of multiple groups or categories simultaneously. They help in identifying differences in central tendency, spread, skewness, and outliers between the groups, allowing effective comparative analyses and insights into the data.

16. In a population, would you expect the mean income to be greater than, less than, or equal to the median income?

Solution: Income is positively-skewed, hence the mean income will be greater than the median income.

17. For a set of n observations x1 , x2 , . . . , xn , with mean x̄, show that: n X (xi − x̄) = 0. i=1

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2.1. Exercises

Solution: We have: n X

(xi − x̄) =

i=1

n X

xi −

i=1

=

n X

n X

x̄

(expansion of brackets)

i=1

xi − nx̄

(x̄ is a constant added n times)

i=1

=

n X

n P

xi − n

i=1

=

n X i=1

xi −

xi

i=1

n X

(re-expressing x̄)

n xi

(cancel the ns)

i=1

= 0.

(et voilà!)

18. For a set of n observations x1 , x2 , . . . , xn , with sample mean x̄, explain why it is not sensible to use the following as a measure of dispersion: n 1 X (xi − x̄). n − 1 i=1

Solution: Since:

n X

(xi − x̄) = 0

i=1

then the proposed statistic is always zero so it is useless at measuring dispersion.

19. For a sample of observations x1 , x2 , . . . , xn on people’s income, in UK pounds, would you recommend the sample variance or the sample standard deviation as the preferred measure of dispersion when explaining these results in a newspaper? Briefly justify your choice.

Solution: Standard deviation would be preferable since the units of £ would be more meaningful than the units of variance, £2 .

20. The annual before-tax earnings for a random sample of a company’s employees, measured in thousands of UK pounds, produced the following summary statistics: 12 X

xi = 398 and

i=1

12 X

x2i = 14,418.

i=1

(a) Calculate the sample standard deviation.

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2. Descriptive statistics

(b) Explain, without calculations, what the effect would be on the mean and standard deviation if each employee was given an annual bonus of £1,000.

Solution: (a) Since n = 12, we have: 1 s2 = n−1 hence s =

√

n X

! x2i − nx̄2

i=1

1 = 11

398 14,418 − 12 × = 110.70 12

110.70 = £10,521 (noting the data are measured in £000s).

(b) If each employee received a fixed bonus of £1,000, then the average annual earnings would increase by £1,000, but there would be no effect on the standard deviation. This is because adding a constant shifts a (sample) distribution, so its central tendency shifts by the constant, but does not affect the dispersion of the distribution.

21. Explain, by using a definition, why the range is influenced by extreme values.

Solution: The range is a measure of dispersion defined as the difference between the maximum and minimum values in a dataset, i.e. x(n) − x(1) . It is influenced by extreme values because their presence can significantly affect the span between the highest and lowest values since the extreme values would typically be x(n) (the sample maximum) and/or x(1) (the sample minimum), thereby affecting the range.

22. Summary statistics of the performance of job applicants in a screening online aptitude test are as follows:

Question # 1 2 3 4 5 6 7 8

Number of attempts 467 465 466 224 392 273 328 330

Here are boxplots of the distributions:

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Mean mark 10.03 (out of 16) 9.48 (out of 12) 8.17 (out of 12) 9.37 (out of 20) 12.57 (out of 20) 10.51 (out of 20) 13.33 (out of 20) 12.51 (out of 20)

Standard deviation 2.93 3.18 2.21 4.88 5.41 6.81 6.10 3.68

2.1. Exercises

What can you deduce from this information? How might you use this information to inform the next stage of the job recruitment process?

Solution: Ultimately, the distribution of marks in an online aptitude test will depend on the difficulty of the test. The shape could easily be skewed to the left, with the job applicants who did not prepare comprising the long left tail. However, it could also be symmetric or even skewed to the right, with a few top-performing job applicants comprising the long right tail.

For the data provided, which note is population level data (we have data on all job applicants), it is meaningful to compare Questions 2 and 3 (since they are both marked out of 12), and also Questions 4 to 8 (each marked out of 20). Comments here are numerous, although worthy of comment is how the boxplots for Questions 4 and 7 clearly reflect the difference in mean scores (9.37 and 13.33, respectively), indicating job applicants’ relative difficulty with Question 4 compared with Question 7. In terms of dispersion, Question 8 has the smallest standard deviation (of the 20-mark questions) and the box clearly indicates how the central 50% of marks were in a narrow range, compared with the high-variation Question 6.

Note that such analysis of the descriptive statistics and boxplots is purely descriptive – we cannot conclude anything about the ‘why’ in terms of variability in performance. Ideally we would wish to review the questions in the online aptitude test and see if we could explain the distributions by question!

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Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 3: Data visualisation

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Chapter 3 Data visualisation 3.1

Exercises

3.1.1

Test your understanding

1. Is data visualisation an art or a science? Briefly justify your view.

Solution: Data visualisation can be considered both an art and a science, as it combines elements of creativity and design with principles of information representation and communication.

Data visualisation involves the use of aesthetics, colour, layout, and visual storytelling techniques to effectively convey information and engage the audience. Design choices, such as the selection of visual elements and the arrangement of data, require creativity and subjective decision-making, similar to the artistic process.

Data visualisation also follows scientific principles and best practices for effective communication and accurate representation of data. Scientific rigour and data-driven decision-making guide the selection of appropriate visualisation techniques.

Ultimately, successful data visualisation requires a careful balance between artistic creativity and scientific rigour. It involves understanding the data and the message it conveys, while applying design principles to create visually appealing and informative representations.

2. Should you always trust your first impressions when viewing data visualisations produced by others? If not, what might you wish to look at more closely?

Solution: No! First impressions may be misleading. One should examine the data source, methodology, and any potential biases. It is also important to assess if the visualisation accurately represents the data, uses appropriate scales, and provides sufficient context.

3. Over time, what will you see on the start page in Tableau?

Solution: Over time, your start page will fill with thumbnails of recent workbooks that you will have worked on.

4. Distinguish between a ‘worksheet’, ‘dashboard’ and ‘story’ in Tableau.

Solution: In Tableau, we have the following hierarchy, all contained within a Tableau ‘workbook’:

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3. Data visualisation

• Worksheet – where individual visualisations are created, starting from a blank canvas.

• Dashboard – a collection of visualisations, i.e. formed by combining one or more worksheets (think of a dashboard like a single PowerPoint slide).

• Story – a sequence of dashboards (think of a story like a full PowerPoint slide deck).

5. How many worksheets should appear in a single dashboard? (‘Well, James, it depends’ is a fair answer, but explain on what basis ‘it depends’ !)

Solution: Well, James, it depends! The number of worksheets in a single dashboard can vary depending on the complexity and purpose of the visualisation. However, you should exercise sound judgement and not crowd out a dashboard by attempting to include too many worksheets.

6. What is the purpose of a filter in Tableau?

Solution: Filters selectively restrict or limit the data displayed in a visualisation or dashboard based on specific criteria or conditions. They allow users to focus on a subset of data that is relevant to their analysis, enabling them to explore and analyse data from different perspectives and answer specific questions. By applying filters, users can dynamically control the data shown in visualisations, allowing for interactive data exploration and providing more meaningful insights. Filters can be based on dimensions (categorical data) or measures (quantitative data) and help refine and customise the view of data in Tableau.

7. What does the data source in Tableau show us?

Solution: The data source provides an overview of the underlying data structure, including the tables, columns, and relationships. It allows users to explore and understand the available data fields, their data types and metadata.

8. Can multiple raw data worksheets be linked in Tableau? If so, how?

Solution: Multiple raw data worksheets can be linked in Tableau using relationships.

9. How do ‘dimensions’ and ‘measures’ differ in Tableau?

Solution: A dimension is a variable that cannot be aggregated in some sense, i.e. adding up or taking an average of a dimension field is meaningless. A measure is a variable that can be aggregated (summed, averaged etc.).

10. What is a ‘blue pill’ ?

Solution: A blue pill is used for discrete data.

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3.1. Exercises

11. What is a ‘green pill’ ?

Solution: A green pill is used for continuous data.

12. What use is Show Me in Tableau?

Solution: ‘Show Me’ is a feature in Tableau which provides a visual recommendation of the most appropriate visualisation types based on the data fields selected. It helps users quickly explore different visualisation options and choose the most suitable one to represent their data effectively. ‘Show Me’ saves time by offering pre-configured visualisations which align with the data and desired analysis goals, allowing the creation of insightful and engaging visual representations with ease!

13. What are some of the data visualisation types in the Show Me gallery? Which is your favourite, and why?

Solution: The ‘Show Me’ gallery in Tableau offers various data visualisation types such as bar charts, histograms, line graphs, scatterplots, maps and treemaps.

14. Is it possible for worksheet titles to update automatically, such as when filters are applied? If so, how?

Solution: Yes, it is possible, and can be achieved by using dynamic or parameter-driven titles. By incorporating parameters or calculated fields that respond to filter selections, the worksheet titles can reflect the filtered context dynamically, ensuring they stay up to date with the applied filters.

Clever stuff.

15. If you wanted to create a map, how would you identify geographic fields in Tableau’s list of variables?

Solution: Geographic fields can be identified by the globe icon in the list of variables.

16. What is a ‘calculated field’ ?

Solution: A calculated field in Tableau is a user-defined field created using a formula or expression that operates on existing fields or values. It allows users to perform custom calculations, aggregations, transformations and logical operations on data. Calculated fields can be created to derive new insights or supplement existing data in visualisations and analysis.

17. Give an example of a parameter and how it could be used in a calculated field.

Solution: An example is the churn rate, Churn Rate, i.e. the rate of customer attrition, which could be used in calculating a sales forecast based on existing sales, Sales.

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3. Data visualisation

18. For the parameter suggested in Question 17, write the formula for the calculated field using Tableau syntax.

Solution: The calculated field would be: [Sales] ? (1-[Churn Rate]).

19. For the parameter used in Questions 17 and 18, what range of values would you suggest it takes, and what would be your choice of default value? Briefly justify your view.

Solution: Churn Rate cannot be negative. Its minimum value is 0% (no customer attrition) with, in principle, a maximum value of 100% (meaning we lose all of our customers – oh no!). In practice we might specify a feasible upper bound of perhaps 50%. A sensible default value could be the most recent churn rate observed, or perhaps the average churn rate for the particular industry. Churn rates will vary considerably – for example, banks tend to experience low churn rates, while mobile phone network providers tend to experience high churn rates.

20. Have you watched Tableau’s own screencasts yet? If no, please do. If yes, proceed to the case study task below!

Solution: Hopefully by now you have!

3.1.2

Case study corner

Source online any dataset of your choice (ideally with a large number of variables/fields). Experiment with creating some visualisations in Tableau, and produce a set of dashboards and an accompanying executive summary (a one-page document) summarising your findings. Are there any data-driven recommendations resulting from your data viz?

Solution: This is over to you. Remember that your dashboards should be: engaging informative self-explanatory relevant interactive.

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Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 4: Probability

ii

Chapter 4 Probability 4.1

Exercises

4.1.1

Test your understanding

1. What values can probabilities take?

Solution: Probabilities can take values between 0 and 1, inclusive. A probability of 0 indicates an impossible event, while a probability of 1 indicates a certain event. Values between 0 and 1 indicate varying degrees of likelihood, with events becoming more probable as probabilities get closer to 1.

2. How do probabilities relate to the likelihood of occurrence of an event?

Solution: Probabilities quantify the likelihood of occurrence of an event. A higher probability indicates a higher chance of the event happening, while a lower probability indicates a lower chance.

3. Distinguish between the following terms: ‘experiment’, ‘outcome’, ‘sample space’ and ‘event’.

Solution: An experiment is a (possibly controlled) procedure or process conducted to observe or study a specific phenomenon. When controlled, this involves manipulating variables and observing the outcomes. A classic example of a random experiment is rolling a die.

An outcome is the result of the experiment. It represents a specific realisation or occurrence within the sample space. For example, rolling a 6.

In turn, the sample space is the set of all possible outcomes which can result from an experiment. It includes all the mutually exclusive and collectively exhaustive outcomes which can occur. For conventional (six-sided) die, the sample space consists of the scores 1, 2, 3, 4, 5 and 6.

An event is a subset of the sample space, representing a specific collection of outcomes of interest from an experiment. For example, rolling an even score.

4. Probabilities can be determined (i) subjectively, (ii) by experimentation, and (iii) theoretically.

Briefly explain the difference between these three approaches, and provide a practical example when each might be used.

Solution: Subjective probabilities are determined based on personal beliefs, opinions or judgements. They rely on an individual’s subjective assessment of the likelihood of an event occurring. These probabilities

1

4. Probability

are not based on empirical data or formal calculations but are instead influenced by factors such as personal experiences, biases, and intuition. For example, a marketing manager might subjectively estimate the probability of success of a new product launch based on their experience, industry knowledge and intuition. This approach is useful when historical data or scientific evidence is limited, and decision-making relies heavily on expert judgement.

Experimental probabilities are determined through empirical observations and data collected from conducting experiments or observations. These probabilities are based on the observed frequencies of events occurring in a sample or series of experiments. For example, a retail store might conduct ‘A/B testing’ on two different pricing strategies for a product to observe which one generates more sales. By analysing the collected data, they can estimate the probability of each strategy resulting in a purchase. This approach allows businesses to test hypotheses and make data-driven decisions.

Theoretical probabilities are determined based on probability theory and statistical analysis. These probabilities are calculated using formulae, rules and assumptions. For example, a financial analyst may use theoretical probabilities derived from option pricing models to assess the likelihood of different price movements in the stock market. These probabilities help in risk management and hence in making informed investment decisions.

5. When elementary outcomes of an experiment are equally likely, how is the probability of an event, such as A, determined?

Solution: For an experiment with equally likely elementary outcomes: • let N be the total number of equally likely elementary outcomes • let n be the number of these elementary outcomes which are favourable to our event of interest, A, then: n P (A) = .

N 6. How is a factorial calculated?

Solution: The factorial of a non-negative integer n, denoted n!, is the number of ways which n different objects can be placed in order, where: n! = n × (n − 1) × (n − 2) × · · · × 3 × 2 × 1.

The exclamation mark, ‘!’, is what we call ‘factorial’.

In Excel, use the function =FACT(number).

7. Provide expressions for the number of permutations and the number of combinations when choosing r objects from n.

Solution: The number of different groups of r objects which can be chosen in order from n different objects is called the number of permutations of r from n, defined as: n

2

Pr =

n! . (n − r)!

4.1. Exercises

In Excel, use the function =PERMUT(number, number chosen).

The number of different groups of r objects which can be chosen from n different objects (i.e. order does not matter) is called the number of combinations of r from n, defined as:

n n! n Cr = = . r r! (n − r)!

In Excel, use the function =COMBIN(number, number chosen).

8. When would permutations be used? When would combinations be used?

Solution: Permutations would be used when considering the order or arrangement of items. For example, determining the number of ways to arrange a sales team lineup. Combinations would be used when order does not matter, such as selecting a committee from a pool of employees.

9. Explain the birthday paradox.

Solution: The birthday paradox states that in a group of only 23 people, the probability of two people having the same birthday is surprisingly high, at around 50% (a probability of 0.507 to be precise!). It is a paradox as the result is counterintuitive, arising from the fact that the number of possible pairs of individuals increases exponentially as the group size grows, making it more likely for at least one pair to share a birthday. Many happy returns.

10. What is the relative frequency approach to probability?

Solution: Suppose the event A associated with some experiment either does or does not occur. Also, suppose we conduct this experiment independently a total of F times, representing the overall frequency of these repeated experiments. Suppose the event A occurs f times across the F repeated experiments.

The ‘frequentist’ approach to probability would regard the probability of A as: P (A) =

f F

as F → ∞.

11. What is meant by ‘randomness’ ?

Solution: When we repeat an experiment under apparently identical conditions, we get different results.

Although the conditions may be as identical as we are able to control them to be, there will inevitably be a large number of uncontrollable (and frequently unknown) variables which we do not measure and which have a cumulative effect on the result of the sample or experiment. For example, weather conditions may affect the outcomes of ‘field’ experiments.

Therefore, the cumulative effect is to cause variation in our results. It is this variation which we term randomness and, although we never fully know the true generating mechanism for our data, we can take the random component into account via the concept of probability which is, of course, why probability plays such an important role in data analysis!

3

4. Probability

12. State the three axioms of probability.

Solution: 1. For any event A, P (A) ≥ 0.

2. For the sample space S, P (S) = 1.

3. If {Ai }, i = 1, 2, . . ., are mutually exclusive events, then the probability of their ‘union’ is the sum of their respective probabilities, i.e. we have: ! ∞ ∞ [ X P Ai = P (Ai ). i=1

i=1

When the number of events is finite, say n of them, the finite additivity version of axiom 3 is: ! n n [ X P Ai = P (Ai ). i=1

i=1

13. Distinguish between the symbols ‘∪’, ‘∩’, ‘c ’ and ‘|’.

Solution: ‘∪’ indicates the union of events. ‘∩’ indicates the intersection of events. ‘c ’ indicates the complement of an event. ‘|’ indicates conditioning.

14. State the general ‘additive law’ and two special cases of it.

Solution: Let A and B be any two events. The additive law states that: P (A ∪ B) = P (A) + P (B) − P (A ∩ B).

So P (A ∪ B) is the probability that at least one of A and B occurs, and P (A ∩ B) is the probability that both A and B occur.

Special case 1: If A and B are mutually exclusive events, i.e. they cannot occur simultaneously, then P (A ∩ B) = 0. Hence: P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = P (A) + P (B) − 0 = P (A) + P (B).

Special case 2: The probability of an event A not occurring, i.e. the complement, Ac , is: P (Ac ) = 1 − P (A).

15. What does it mean for two events to be independent?

Solution: Two events are independent if the occurrence or non-occurrence of one event does not affect the probability of the other event happening. They are unrelated and do not influence each other.

4

4.1. Exercises

16. State the multiplicative law.

Solution: The multiplicative law (for independent events): Events A and B are independent if: P (A ∩ B) = P (A) P (B).

17. How is the conditional probability P (A | B) calculated?

Solution: For any two events A and B, we define conditional probabilities as: P (A | B) =

P (A ∩ B) P (B)

and P (B | A) =

P (A ∩ B) P (A)

provided P (A) > 0 and P (B) > 0.

In words: ‘the probability of one event, given a second event, is equal to the probability of both, divided by the probability of the second (conditioning) event.’

18. What is the total probability formula?

Solution: The (simple form of the) total probability formula is: P (A) = P (A | B) P (B) + P (A | B c ) P (B c ).

In words: ‘the probability of an event is equal to its conditional probability on a second event times the probability of the second event, plus its probability conditional on the second event not occurring times the probability of that non-occurrence.’

19. State the simple and general forms of Bayes’ theorem.

Solution: The simple form of Bayes’ theorem is: P (A | B) =

P (B | A) P (A) .

P (B)

General form: For a general partition of the sample space S into B1 , B2 , . . . , Bn , and for some event A, then: P (A | Bk ) P (Bk ) P (Bk | A) = P . n P (A | Bi ) P (Bi ) i=1

20. The successful operation of three separate switches is needed to control a machine. If the probability of failure of each switch is 0.1 and the failure of any switch is independent of any other switch, what is the probability that the machine will break down?

Solution: The probability that the machine will not break down equals the probability that all three switches work. This equals 0.9 × 0.9 × 0.9 = 0.729, by independence. Hence the probability it will break down is 1 − 0.729 = 0.271.

5

4. Probability

21. Suppose that 8 in every 100 people (i.e. 8%) have a particular disease. A diagnostic test for the disease has 76% sensitivity, that is, if a person has the disease, the test will give a positive result with a probability of 0.76. The test has 65% specificity, that is, if a person does not have the disease, the test will give a negative result with a probability of 0.65. Calculate the probability that a person does not have the disease, given that the person has received a negative test result.

Solution: Let B denote the presence of the disease, and B c denote no disease. Let A denote a negative test result. We have that: P (A) = P (A | B) P (B) + P (A | B c ) P (B c ) = 0.24 × 0.08 + 0.65 × 0.92 = 0.6172.

Hence the probability that a person does not have the disease, B c , given that the person has received a negative test result, A, is: P (B c | A) =

0.65 × 0.92 P (A | B c ) P (B) = = 0.9689. c c P (A | B) P (B) + P (A | B ) P (B ) 0.6172

22. A bank classifies borrowers as being either ‘high risk’ or ‘low risk’. Twelve per cent of the bank’s loans are made to high-risk borrowers. Overall, 7% of the bank’s loans are in default, and 30% of loans in default were made to high-risk borrowers. What is the probability that a high-risk borrower will default?

Solution: Let H = high-risk borrower and let D = loan in default. We have P (H) = 0.12, P (D) = 0.07 and P (H | D) = 0.30. Hence the probability that a high-risk borrower will default is: P (D | H) =

0.30 × 0.07 P (H | D) P (D) = = 0.175.

P (H) 0.12

23. A box contains 15 light bulbs, of which two are defective. If a person selects six light bulbs at random, without replacement, what is the probability that both defective light bulbs will be selected?

Solution: The sample space consists of all (unordered) subsets of 6 out of the 15 light bulbs in the box. There

15 are 6 such subsets. The number of subsets which contain the two defective bulbs is the number of

13 subsets of size 4 out of the other 13 bulbs, 4 , so the probability we want is:

13 6×5 4

= 0.1429.

15 = 15 × 14 6 24. A firm produces electrical components using three non-overlapping shifts. It is known that 50% of the components are produced during shift 1, 20% during shift 2, and 30% during shift 3. A further look at product quality reveals that 6% of the components produced during shift 1 are defective.

The corresponding percentages for shifts 2 and 3 are 8% and 15%, respectively. (a) What percentage of all components are not defective? (b) Given that a defective component is found, what is the probability that it was produced during shift 3?

6

4.1. Exercises

Solution: We have the following, using obvious notation for shifts and defective components: P (S1 ) = 0.50, P (D | S1 ) = 0.06,

P (S2 ) = 0.20,

P (S3 ) = 0.30

P (D | S2 ) = 0.08 and P (D | S3 ) = 0.15.

(a) The percentage of all components that are not defective is, using the total probability formula: c

P (D ) =

3 X

P (Dc | Si ) P (Si ) = 0.94 × 0.50 + 0.92 × 0.2 + 0.85 × 0.3 = 0.909

i=1

i.e. 90.9%. (b) Given that a defective component is found, the probability that it was produced during shift 3 is: 0.15 × 0.3 P (S3 ∩ D) P (S3 | D) = = = 0.495.

P (D) 0.091 25. Assume that a calculator has a ‘random number’ key and that when pressed generates an integer between 0 and 99 inclusive at random, all numbers being equally likely and generated independently of one another. (a) Let X denote the random number generated when the key is pressed once. Write down the probability function of X. (b) If two numbers are generated, what is the probability that the first number exceeds the second number? (c) If five numbers are generated, what is the probability that at least one number occurs more than once?

Solution: (a) We have: ( 1/100 for x = 0, 1, 2, . . . , 99 p(x) = 0 otherwise. (b) Suppose P (first greater) = x, then by symmetry we have that P (second greater) = x. However, the probability that both are equal is (by counting): {0, 0}, {1, 1}, . . . , {99, 99} 100 = = 0.01.

10,000 10,000 Hence x + x + 0.01 = 1, so x = 0.495. (c) The probability that they are all different is: 1×

99 98 97 96 × × × .

100 100 100 100

Subtracting from 1 gives the required probability, i.e. 0.09655.

7

4. Probability

26. Inventor A repeats an experiment three times, while inventor B repeats it twice. The probability of success of each experiment is 0.6 and the outcomes of different experiments are independent of each other. (a) What is the probability that all three experiments of inventor A were successful and both experiments of inventor B failed? (b) What is the probability that inventor B performed more successful experiments than inventor A? (c) Given that inventor B performed more successful experiments than inventor A, what is the probability that both experiments of inventor B were successful? (d) Inventor B is now planning to perform the experiment for a third time. Given that she had more successes than inventor A in her previous two experiments and given that both her previous experiments were successful, what is the probability she will also be successful next time?

Solution: (a) The three experiments of inventor A will all be successful with probability (0.6)3 and both experiments of inventor B will fail with probability (0.4)2 . Hence the required probability is, due to independence: (0.6)3 × (0.4)2 = 0.03456. (b) Inventor B will have more successes than inventor A if she either has one success and inventor A has none with probability: 2 × 0.6 × 0.4 × (0.4)3 = 0.03072 or if she has 2 successes and inventor A has less than 2 which has probability: (0.6)2 × ((0.4)3 + 3 × 0.6 × (0.4)2 ) = 0.12672.

Hence the probability she has more successes than inventor A is: 0.12672 + 0.03072 = 0.15744. (c) Using the results in (b), the conditional probability that inventor B had two successes is: 0.12672 = 0.804878 0.15744 (d) The success of the next experiment is unaffected by what happened in the past, due to independence, so it is 0.6.

27. James is a salesman for a company and sells two products, X and Y. He visits three different customers each day. For each customer, the probability that James sells product X is 1/3 and the probability that he sells product Y is 1/4. The sale of product X is independent of the sale of product Y during any visit, and the results of the three visits are mutually independent.

Calculate the probability that James will: (a) sell both products, X and Y, on the first visit

8

4.1. Exercises

(b) sell only one product during the first visit (c) make no sales of product X during the day (d) make at least one sale of product Y during the day.

Solution: (a) Using the obvious notation X is ‘product X sold’ and Y is ‘product Y sold’ we have that P (X) = 1/3 and P (Y ) = 1/4. Due to independence: P (X ∩ Y ) = P (X) P (Y ) =

1 1 1 × = = 0.0833.

3 4 12

(b) We have: c

c

P (X ∩ Y ) + P (X ∩ Y ) =

1 3 × 3 4

+

2 1 × 3 4

=

5 = 0.4167.

12

(c) We have P (X c ) = 1 − P (X) = 1 − 1/3 = 2/3, hence: 3 8 2 = = 0.2963.

P (no X sales all day) = 3 27 (d) We have: 3 3 37 P (at least 1 Y sale) = 1 − P (no Y sales) = 1 − = 0.5781. = 4 64

9

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 5: Probability distributions

ii

Chapter 5 Probability distributions 5.1

Exercises

5.1.1

Test your understanding

1. In what sense is a variable ‘random’ ?

Solution: A variable is considered ‘random’ if its value is subject to uncertainty or unpredictability. The value of a random variable is not determined in advance and can vary based on chance, probability, or external factors.

2. What types of data would be represented with a discrete random variable? What types of data would be represented by a continuous random variable?

Solution: Discrete random variables represent data that can take on a countable or finite set of distinct values, such as the number of sales per day or the number of defective products in a batch.

Continuous random variables represent data that can take on any value within a range or interval, such as revenue, production output, or customer satisfaction scores.

3. If X denotes a random variable, what is the difference between ‘X’ and ‘x’ ?

Solution: X represents the random variable itself, which is a mathematical concept that describes all possible outcomes and their associated probabilities, while x represents a specific value or realisation of the random variable, denoting an observed or potential outcome.

4. What must be true about the possible outcomes of a discrete uniform distribution?

Solution: In a discrete uniform distribution, all possible outcomes must have equal probabilities. Each outcome has the same likelihood of occurring. With k possible outcomes, the probability of each outcome is 1/k.

5. What is the ‘relative frequency’ interpretation of a probability distribution?

Solution: A probability distribution has a natural relative frequency interpretation – if the experiment is repeated a very large number of times, then the probability of any particular value of the random

1

5. Probability

variable is equal to the limit of its relative frequency as the number of experiments becomes infinitely large.

6. If a company classified its customers into two types – ‘satisfied’ and ‘dissatisfied’ – what would be the appropriate probability distribution to model these customer types, and why?

Solution: A suitable probability distribution would be a Bernoulli distribution which models a single trial with two possible outcomes, where each customer is classified as either ‘satisfied’ or ‘dissatisfied’ with probabilities π and 1 − π, respectively.

7. In a game show each contestant has two chances out of three of winning a prize, independently of other contestants. If six contestants take part, determine the probability distribution of the number of winners. Find the mean and variance of the number of winners.

Solution: The number of winners, X, is binomial with n = 6 and π = 2/3. Using =BINOM.DIST(number s, trials, probability s, cumulative) we can obtain the binomial probabilities as follows: X=x P (X = x) x P (X = x) x2 P (X = x)

0 0.0014 0 0

1 0.0165 0.0165 0.0165

2 0.0823 0.1646 0.3292

3 0.2195 0.6585 1.9755

4 0.3292 1.3168 5.2672

5 0.2634 1.3170 6.5850

6 0.0878 0.5268 3.1608

Total 1 4 17.33

This yields a mean of: E(X) =

6 X

x p(x) = 0 + 0.0165 + · · · + 0.5268 = 4

x=0

alternatively E(X) = nπ = 6 × 2/3 = 4, and a variance of: 2

2

Var(X) = E(X ) − µ =

6 X

x2 p(x) − µ2 = (0 + 0.0165 + · · · + 3.1608) − 42 = 17.33 − 42 = 1.33

x=0

alternatively Var(X) = nπ(1 − π) = 6 × 2/3 × 1/3 = 1.33.

8. Components for assembly are delivered in batches of 100, and experience shows that 5% of each batch are defective. On arrival, five pieces are selected at random from each batch and tested. If two or more of the five are found to be faulty, the entire batch is rejected. What is the probability that a 5% defective batch will be accepted?

Solution: Since this question involves sampling without replacement, we should in principle use the ‘hypergeometric distribution’, or at least the ‘finite population correction factor’ as an approximation to it. However, we frequently want to save effort in these calculations, and here n = 5 (both the sample size and the upper limit of the values of interest) and N = 100, so that π = n/N = 5/100 = 0.05 is small. Moreover, N π = n = 5 is also ‘small’.

2

5.1. Exercises

Therefore, we can use a binomial approximation to these preferable, but more complicated, methods, with n = 5 and π = 0.05. We find that: P (X = 0) = (0.95)5 = 0.7738 and P (X = 1) = 5 × (0.05)1 × (0.95)4 = 0.2036.

Hence P (batch accepted) = 0.7738 + 0.2036 = 0.9774.

9. One in ten of the new cars leaving a factory has minor faults of one kind or another. (a) Assuming that a batch of ten cars delivered to a dealer represents a random sample of the output, what is the probability that: i. at least 1 will be faulty? ii. more than 3 will be faulty?

On receiving a delivery of ten new cars from the manufacturer, the dealer checks out four of these, chosen at random, before they are delivered to customers. (b) If, in fact, two of the cars are faulty, what is the probability that both faults will be discovered?

Solution: (a) Using the binomial distribution with n = 10 and π = 0.1 we have the following: i. P (at least 1 faulty) = 1 − P (X = 0) = 1 − (0.9)10 = 0.6513. ii. We have, using =BINOM.DIST(number s, trials, probability s, cumulative): P (more than 3 faulty) = 1 − P (X = 0) − P (X = 1) − P (X = 2) − P (X = 3) = 1 − 0.3487 − 0.3874 − 0.1937 − 0.0574 = 0.0128. (b) We can consider this as involving a population of 10 cars. A sample of 4 cars is taken without replacement. There are 2 ‘defectives’ in the population – what is the probability that both are in the sample?

There are a total of 10 = 210 ways of choosing 4 cars from 10 cars. If the two defectives are 4 included in the sample, there are 8 candidates for the remaining 2 places, so there are 82 = 28 samples which contain the faulty cars. Hence the required probability is 28/210 = 0.1333.

10. Over a period of time the number of break-ins per month in a given district has been observed to follow a Poisson distribution with mean 2. (a) For a given month, find the probability that the number of break-ins is: i. fewer than 2 ii. more than 4 iii. at least 1, but no more than 3. (b) What is the probability that there will be fewer than ten break-ins in a 6-month period?

3

5. Probability

Solution: (a) Using =POISSON.DIST(x, mean, cumulative) we obtain the first few entries in the table of the probability function below. They all follow from the Poisson probability function: P (X = x) =

e−λ λx x!

with λ = 2.

X=x P (X = x)

0 0.1353

1 0.2707

2 0.2707

3 0.1804

4 0.0902

5 ...

i. We have: P (X < 2) = P (X = 0) + P (X = 1) = 0.1353 + 0.2707 = 0.4060. ii. We have: P (X > 4) = 1 − P (X = 0) − P (X = 1) − P (X = 2) − P (X = 3) − P (X = 4) = 1 − 0.1353 − 0.2707 − 0.2707 − 0.1804 − 0.0902 = 0.0527. iii. We have: P (1 ≤ X ≤ 3) = P (X = 1) + P (X = 2) + P (X = 3) = 0.2707 + 0.2707 + 0.1804 = 0.7218. (b) If there are an average of 2 break-ins per month, there will be an average of 12 break-ins in a 6-month period. Therefore, the number of break-ins will have a Poisson distribution with λ = 12. We need to calculate P (X < 10). This is 0.2424, returned by =POISSON.DIST(9,12,1).

11. Arrivals at a supermarket are assumed to follow a Poisson distribution with a rate parameter of 120 per hour. If T is the time to arrival of the next customer (in minutes), calculate: P (T > 1.0 | T > 0.5).

Solution: The time to the next customer is more than t if there are no arrivals in the interval from 0 to t, which means that we need to use λ = t × 2. Now the conditional probability formula yields: P (T > 1.0 | T > 0.5) =

4

P ({T > 1.0} ∩ {T > 0.5}) P (T > 0.5)

5.1. Exercises

and, as in other instances, the two events {T > 1.0} and {T > 0.5} collapse to a single event, {T > 1.0}. Hence: P (T > 1.0) P (T > 1.0 | T > 0.5) = .

P (T > 0.5) Hence: P (T > 1.0) = Hence:

e−2 20 = e−2 0!

and P (T > 0.5) =

e−1 10 = e−1 .

0!

e−2 P (T > 1.0 | T > 0.5) = −1 = e−1 = 0.3679. e

12. Two per cent of items of a particular type produced by a company are known to be defective. If a random sample of 100 items is selected for inspection, calculate the probability of getting no defectives by using: (a) the binomial distribution (b) the Poisson distribution.

Solution: (a) Using the binomial with n = 100 and π = 0.02, P (X = 0) = (0.98)100 = 0.1326. (b) Using the Poisson with λ = 100 × 0.02 = 2, P (X = 0) = exp(−2) = 0.1353.

Note that the answers are almost equal. This is because n is large and π is small.

13. Suppose X ∼ N (10, 4). (a) Find: i. P (X > 13.4) ii. P (8 < X < 9). (b) Find the value a such that P (10 − a < X < 10 + a) = 0.95. (c) Find the value b such that P (10 − b < X < 10 + b) = 0.99. (d) How far above the mean of the standard normal distribution must we go such that only 1% of the probability remains in the right-hand tail? (e) How far below the mean of the standard normal distribution must we go such that only 5% of the probability remains in the left-hand tail?

Solution: Since√X ∼ N (10, 4), we use the transformation Z = (X − µ)/σ with the values µ = 10 and σ = 4 = 2. (a)

i. We have P (X > 13.4) = P (Z > (13.4 − 10)/2) = P (Z > 1.7) = 0.0446. We could use =1-NORM.DIST(13.4,10,2,1) or =1-NORM.S.DIST(1.7,1).

5

5. Probability

ii. We have:

P (8 < X < 9) = P

8 − 10 9 − 10 <Z< 2 2

= P (−1 < Z < −0.5) = P (Z < −0.5) − P (Z < −1) = 0.1498.

We could use =NORM.DIST(9,10,2,1)-NORM.DIST(8,10,2,1) or =NORM.S.DIST(-0.5,1)-NORM.S.DIST(-1,1). (b) We want to find the value a such that P (10 − a < X < 10 + a) = 0.95, that is:

(10 − a) − 10 (10 + a) − 10 0.95 = P <Z< 2 2 a

a =P − <Z< 2 2

a a −P Z <− =1−P Z > 2 2

a =1−2×P Z > .

2

This is the same as 2 × P (Z > a/2) = 0.05, i.e. P (Z > a/2) = 0.025. Hence a/2 = 1.96, using =NORM.S.INV(0.975), and so a = 3.92. (c) We want to find the value b such that P (10 − b < X < 10 + b) = 0.99. Similar reasoning shows that P (Z > b/2) = 0.005. Hence, using =NORM.S.INV(0.995), b/2 = 2.58, so that b = 5.16. (d) We want k such that P (Z > k) = 0.01. Hence k = 2.33, using =NORM.S.INV(0.99). (e) We want x such that P (Z < x) = 0.05. This means that x < 0 and P (Z > |x|) = 0.05, so |x| = 1.65, using =NORM.S.INV(0.95), and hence x = −1.65.

14. Your company requires a special type of light bulb which is available from only two suppliers.

Supplier A’s light bulbs have a mean lifetime of 2000 hours with a standard deviation of 180 hours.

Supplier B’s light bulbs have a mean lifetime of 1850 hours with a standard deviation of 100 hours.

The distribution of the lifetimes of each type of light bulb is normal. Your company requires that the lifetime of a light bulb be not less than 1500 hours. All other things being equal, which type of bulb should you buy, and why?

Solution: Let A and B be the random variables representing the lifetimes (in hours) of light bulbs from supplier A and supplier B, respectively. We are told that: A ∼ N (2,000, (180)2 ) and B ∼ N (1,850, (100)2 ).

Since the relevant criterion is that light bulbs last at least 1,500 hours, the company should choose

6

5.1. Exercises

the supplier whose light bulbs have a greater probability of doing so. We find that:

1,500 − 2,000 P (A > 1,500) = P Z > 180 = P (Z > −2.78) = 0.9973 using =1-NORM.S.DIST(-2.78,1), and:

1,500 − 1,850 P (B > 1,500) = P Z > 100 = P (Z > −3.50) = 0.9998 using =1-NORM.S.DIST(-3.5,1). Therefore, the company should buy light bulbs from supplier B, since they have a greater probability of lasting the required time.

Note it is good practice to define notation and any units of measurement and to state the distributions of the random variables. Note also that here it is not essential to compute the probability values in order to determine what the company should do, since −2.78 > −3.5 implies that P (Z > −2.78) < P (Z > −3.5).

15. The life, in hours, of a light bulb is normally distributed with a mean of 200 hours. If a consumer requires at least 90% of the light bulbs to have lives exceeding 150 hours, what is the largest value that the standard deviation can have?

Solution: Let X be the random variable representing the lifetime of a light bulb (in hours), so that for some value σ we have X ∼ N (200, σ 2 ). We want P (X > 150) = 0.9, such that:

50 150 − 200 =P Z>− = 0.9.

P (X > 150) = P Z > σ σ Note that this is the same as P (Z > 50/σ) = 1 − 0.9 = 0.1, so 50/σ = 1.28, using =NORM.S.INV(0.9), giving σ = 39.06.

16. A company manufactures rods whose diameters are normally distributed with a mean of 5 mm and a standard deviation of 0.05 mm. It also drills holes to receive the rods and the diameters of these holes are normally distributed with a mean of 5.2 mm and a standard deviation of 0.07 mm. The rods are allocated to the holes at random. What proportion of rods will fit into the holes?

Solution: Let X and Y , respectively, denote the random variables for the diameters of the rods and of the holes (in millimetres), so that: X ∼ N (5, (0.05)2 ) and Y ∼ N (5.2, (0.07)2 ).

Therefore, P (rod fits hole) = P (X < Y ) = P (Y − X > 0). Now note that: Y − X ∼ N (5.2 − 5, (0.05)2 + (0.07)2 )

⇒

Y − X ∼ N (0.2, 0.0074).

7

5. Probability

So:

0 − 0.2 P (rod fits hole) = P Z > √ 0.0074 = P (Z > −2.33) = 0.9901 using =1-NORM.S.DIST(-2.33,1).

17. An investor has the choice of two out of four investments: X1 , X2 , X3 and X4 . The profits (in thousands of pounds per annum) from these may be assumed to be independently distributed and: profit from X1 ∼ N (2, 1),

profit from X2 ∼ N (3, 3),

profit from X3 ∼ N (1, 0.25),

profit from X4 ∼ N (2.5, 4).

Which pair of investments should the investor choose in order to maximise the probability of making a total profit of at least £2000? What is this maximum probability?

Solution: Let A1 , A2 , A3 and A4 be the profits from the investments X1 , X2 , X3 and X4 , respectively. There are 4 C2 = 6 possible pairs, and we simply have to work out the probabilities for each. From the information in the question, we see that:

2−5 = 0.9332 A1 + A2 ∼ N (5, 4) ⇒ P (A1 + A2 > 2) = P Z > √ 4

2−3 A1 + A3 ∼ N (3, 1.25) ⇒ P (A1 + A3 > 2) = P Z > √ = 0.8133 1.25

2 − 4.5 A1 + A4 ∼ N (4.5, 5) ⇒ P (A1 + A4 > 2) = P Z > √ = 0.8686 5

2−4 A2 + A3 ∼ N (4, 3.25) ⇒ P (A2 + A3 > 2) = P Z > √ = 0.8665 3.25

2 − 5.5 A2 + A4 ∼ N (5.5, 7) ⇒ P (A2 + A4 > 2) = P Z > √ = 0.9066 7

2 − 3.5 = 0.7673.

A3 + A4 ∼ N (3.5, 4.25) ⇒ P (A3 + A4 > 2) = P Z > √ 4.25 Therefore, the investor should choose X1 and X2 , for which the maximum probability is 0.9332.

18. An advertising agency claims that 40% of all television viewers watch a particular programme. In a random sample of 500 viewers, what is the probability that fewer than 170 will be watching the programme if the agency’s claim is correct?

Solution: Let X be the number of television viewers in a random sample of 500, such that if the advertising agency’s claim is true, we would have X ∼ Bin(500, 0.4). Given the sample size, we want to use the

8

5.1. Exercises

normal approximation to X, with µ = 500 × 0.4 = 200 and σ 2 = 500 × 0.4 × (1 − 0.4) = 120.

Therefore, we can assume that X ∼ N (200, 120).

We might use a continuity correction, but in this case (chiefly because the sample size is so much larger than 30) it makes little difference. So we want:

170 − 200 P (X < 170) = P Z < √ 120 = P (Z < −2.74) = 0.0031 using =NORM.S.DIST(-2.74,1).

19. The number of newspapers sold daily at a kiosk is normally distributed with a mean of 350 and a standard deviation of 30. (a) Find the probability that fewer than 300 newspapers are sold on Monday. (b) Find the probability that fewer newspapers are sold on Tuesday than on Monday. (c) Find the probability that fewer than 1700 newspapers are sold in a (5-day) week. What assumption have you made in order to answer this? (d) How many newspapers should the newsagent stock each day such that the probability of running out on any particular day is 10%?

Solution: Let X be the number of newspapers sold, hence X ∼ N (350, 900). (a) P (X < 300) = P (Z < (300 − 350)/30) = P (Z < −1.67) = 0.0475, using =NORM.S.DIST(-1.67,1). (b) If XM and XT are the random variables for newspaper sales on Mondays and on Tuesdays, respectively, the mean of XM − XT is zero, and we want to find P (XM − XT > 0). Since we can clearly suppose that the distribution of XM − XT is left-right symmetric about its mean, 0, P (XM − XT > 0) = 0.5. (c) If T is the number of newspapers sold in a five-day week then we have T ∼ N (1,750, 4,500) (since it is the sum of five copies of X, which may be assumed to be independent). Therefore:

1,700 − 1,750 √ = P (Z < −0.75) = 0.2266 P (T < 1,700) = P Z < 4,500 using =NORM.S.DIST(-0.75,1). Note that since we have assumed that daily newspaper sales are independent, which we were not told explicitly in the question, we need to (i.) think if it is reasonable – in the given question – to assume independence, and then (ii.) say we have assumed it. (d) Let s be the required stock, then we require P (X > s) = 0.1. Hence, using =NORM.S.INV(0.9):

s − 350 P Z> = 0.1 30 ⇒ ⇒

s − 350 ≥ 1.28 30 s ≥ 350 + 1.28 × 30 = 388.4.

9

5. Probability

Rounding up, the required stock is 389.

20. Explain the purpose of a continuity correction.

Solution: The purpose of a continuity correction is to improve the accuracy of approximations when using continuous probability distributions (such as the normal distribution) to estimate discrete probabilities (such as binomial probabilities) in statistical calculations.

21. An electrical device contains 10 components connected in series. The device fails if any one of the components fails. For each component the probability that it survives a year of use without failing is π, and the failures of different components can be regarded as independent events. (a) What is the probability that the device fails in a year of use? (b) How large must π be for the probability of failure in (a) to be less than 0.1?

Solution: (a) It is often easier to evaluate the probability of the complement of the event specified. Here, we calculate: P (device does not fail) = P (every component works) = π 10 and hence P (device fails) = 1 − π 10 .

It is always a good idea to do a quick ‘reality check’ of your answer. If you calculated, say, the probability to be 10(1 − π), this must be wrong because for some values of π you would have a probability greater than 1! √ (b) We require 1 − π 10 < 0.1, which is true if π > 10 0.9 ≈ 0.9895.

22. The possible annual percentage return of the stocks of ABC and DEF share a common probability distribution, given below.

Return of ABC Return of DEF Probability

−7.3 38.9 0.05

−1.9 25.6 0.20

6.4 16.3 0.25

13.7 5.6 0.25

15.9 −2.5 0.20

34.1 −5.3 0.05

(a) Determine the expected annual return of ABC and DEF. (b) Determine the standard deviation of the annual return of ABC and DEF. (c) If you could only invest in one of these two stocks, which would you choose and why? (d) What can be concluded about the relationship of the annual returns of these two stocks?

Would this impact your investment decision-making?

Solution: (a) The expected annual return of ABC, E(RABC ), is: E(RABC ) = (−7.3 × 0.05) + (−1.9 × 0.20) + · · · + (34.1 × 0.05) = 9.17%.

Similarly, for DEF the expected annual return is: E(RDEF ) = (38.9 × 0.05) + (25.6 × 0.20) + · · · + (−5.3 × 0.05) = 11.78%.

10

5.1. Exercises

(b) For ABC, we have: 2 ) = ((−7.3)2 × 0.05) + ((−1.9)2 × 0.20) + · · · + ((34.1)2 × 0.05) = 169.25 E(RABC

hence the standard deviation is: q p 2 E(RABC ) − (E(RABC ))2 = 169.25 − (9.17)2 = 9.23%.

Similarly, for DEF, we have: 2 ) = ((38.9)2 × 0.05) + ((25.6)2 × 0.20) + · · · + ((−5.3)2 × 0.05) = 283.65 E(RDEF

hence the standard deviation is: q p 2 E(RDEF ) − (E(RDEF ))2 = 283.65 − (11.78)2 = 12.04%. (c) DEF has a higher expected annual return. However, DEF is also more volatile with a higher standard deviation. Each stock also has a 25% chance of a negative return. Therefore, you could select DEF based on the greater expected return or ABC based on the lower volatility. (d) These two stocks are negatively correlated with one another. Therefore, if you want a more balanced portfolio, you may want to invest in both stocks for diversification purposes.

23. At a charity event, the organisers sell 100 tickets for a raffle. At the end of the event, one of the tickets is selected at random and the person with that number wins a prize. Person A buys ticket number 22. Person B buys tickets numbered 1–5. What is the probability of each of them winning the prize?

Solution: Let X denote the number on the winning ticket. Since all values between 1 and 100 are equally likely, X has a discrete ‘uniform’ distribution such that: P (‘A wins’) = P (X = 22) = p(22) = and: P (‘B wins’) = P (X ≤ 5) = F (5) =

1 = 0.01 100

5 = 0.05.

100

24. Toward the end of the financial year, James is considering whether to accept an offer to buy his stock option now, rather than wait until the normal exercise time. If he sells now, his profit will be £120,000. If he waits until the exercise time, his profit will be £200,000, provided that there is no crisis in the markets before that time; if there is a crisis, the option will be worthless and he would expect a net loss of £50,000. What action should he take to maximise his expected profit if the probability of crisis is: (a) 0.5? (b) 0.1?

11

5. Probability

(c) For what probability of a crisis would James be indifferent between the two courses of action if he wishes to maximise his expected profit?

Solution: Let π = probability of crisis, then: S = E(profit given James sells) = £120,000 and: W = E(profit given James waits) = £200,000(1 − π) + (−£50,000)π. (a) If π = 0.5, then S = £120,000 and W = £75,000, so S > W , hence James should sell now. (b) If π = 0.1, then S = £120,000 and W = £175,000, so S < W , hence James should wait until the exercise time.

To be indifferent, we require S = W , i.e. we have: £200,000 − £250,000 π = £120,000 so π = 8/25 = 0.32.

25. The waiting time X (in minutes) of a customer at a taxi rank has the following probability density function: ( 0.25e−0.25x for x ≥ 0 f (x) = 0 otherwise.

This is the exponential distribution with parameter λ = 0.25. For this distribution, calculate the probability that the waiting time is: (a) 5 minutes (b) at most 4 minutes (c) at least 8 minutes (d) at most 0.5 standard deviations from its expected value.

Solution: This is the probability density function of an exponential distribution withpλ = 0.25. The expected value is E(X) = 1/λ = 1/0.25 = 4 and the standard deviation is sd(X) = Var(X) = 1/λ = 4.

Using these, we get the following. (a) P (X = 5) = 0. The probability that a continuous distribution has any particular value exactly is always 0. (b) P (X ≤ 4) = 0.6321, using =EXPON.DIST(4,0.25,1). (c) P (X ≥ 8) = 0.1353, using =1-EXPON.DIST(8,0.25,1).

12

5.1. Exercises

(d) We have: P (E(X) − 0.5 × sd(X) ≤ X ≤ E(X) + 0.5 × sd(X)) = P (4 − 2 ≤ X ≤ 4 + 2) = P (2 ≤ X ≤ 6) = 0.3834 using =EXPON.DIST(6,0.25,1)-EXPON.DIST(2,0.25,1).

26. In a survey of the population of employed workers in the UK, respondents are asked how many hours they usually work per week in their main job. The average of the responses is 38.74 hours, with a standard deviation of 15.72 hours. In answering the following question, suppose that working hours in the population follow a normal distribution with these values of the mean and standard deviation. (a) What percentage of people in the population work between 35 and 45 hours per week? (b) How many hours do the middle 50% of people in the population work? (c) What is the number of hours such that only 10% of people work that long or longer per week?

Solution: Use the random variable X ∼ N (38.74, (15.72)2 ). (a) We have:

P (35 < X < 45) = P

45 − 38.74 35 − 38.74 <Z< 15.72 15.72

= P (−0.24 < Z < 0.40) = 0.2502 (i.e. about 25%) using =NORM.S.DIST(0.4,1)-NORM.S.DIST(-0.24,1). (b) Let 38.74 ± c be the required interval, where c > 0, so: P (38.74 − c ≤ X ≤ 38.74 + c) = 0.50.

It is convenient to write this as P (38.74 ≤ X ≤ 38.74 + c) = 0.25, by symmetry. This also means that P (X ≤ 38.74 + c) = 0.75, since P (X < 38.74) = 0.5, because the normal distribution of X is symmetric around its mean (and median and mode) of 38.74.

Standardise by way of Z = (X − 38.74)/15.72 to give: c P Z≤ = 0.75 15.72

i.e. P (Z > c/15.72) = 0.25. P (Z > 0.67) ≈ 0.25, using =NORM.S.INV(0.75).

Therefore, we have c/15.72 = 0.67 and c = 10.53. Hence the required interval is 38.74 ± 10.53 or (28.21, 49.27).

13

5. Probability

(c) Here we need k such that P (X > k) = 0.10. Standardising:

k − 38.74 P Z> = 0.10.

15.72 We have that P (Z > 1.28) ≈ 0.10, using =NORM.S.INV(0.9), so here: k − 38.74 = 1.28.

15.72 Hence k = 1.28 × 15.72 + 38.74 = 58.87 hours.

27. Each morning telephone calls arrive at a reception desk according to a Poisson process with on average three calls every 5 minutes. Let T denote the waiting time, in minutes, until the first call is received after 09:00. (a) What is the distribution of T ? (b) Calculate P (T > 4).

Solution: (a) The waiting time until the first call will follow an exponential distribution with λ = 0.6. Hence: ( 0.6e−0.6t for t ≥ 0 f (t) = 0 otherwise. (b) The required probability can simply be calculated using =1-EXPON.DIST(4,0.6,1), returning 0.0907.

28. The times taken, in minutes, for three commuters to travel to work can be regarded as independent and normally distributed random variables with means and standard deviations as given in the following table: Commuter A B C

Mean 28 50 35

Standard deviation 4 6 5

Departure time 08:30 08:10 08:20

In order to arrive at work for a 09:00 start, the commuters leave home at the times shown in the right-hand column of the table. Let A, B and C be the random variables denoting the journey times, in minutes, of commuters A, B and C, respectively. (a) Find the probabilities of the following events: i. Commuter A is late for work. ii. Commuter A arrives at work before Commuter B. iii. At least one commuter is late for work. iv. All three commuters are late for work, given that at least one commuter is late for work. v. Commuter C is late for work, given that exactly one commuter is late for work.

14

5.1. Exercises

(b) At what time should Commuter C leave home if there is to be a 2.5% chance of her being late for work?

Solution: Letting A, B and C, respectively, denote the random variables representing A’s, B’s and C’s journey times (in minutes), we have: A ∼ N (28, 16),

(a)

B ∼ N (50, 36) and C ∼ N (35, 25).

i. We have P (A late):

P (A > 30) = P

30 − µA Z> σA

30 − 28 =P Z> = P (Z > 0.5) = 0.3085.

4

Although not needed here, but useful for later calculations, we also have:

40 − µC 40 − 35 P (C late) = P (C > 40) = P Z > =P Z> σC 5 = P (Z > 1) = 0.1587.

Note P (B late) = 0.5, since B’s available journey time is 50 minutes (= µB ). Since the normal distribution is symmetric about the mean, B will be late 50% of the time. No calculations are needed here! ii. Consider arrival times in minutes after 09:00. Hence A’s and B’s arrival times are, respectively: A + 30 ∼ N (58, 16) and B + 10 ∼ N (60, 36).

‘A arrives at work before B’ means A + 30 < B + 10, that is A + 20 < B, alternatively B − A > 20. (Of course, we could write this as A − B < −20, but better to avoid negative values whenever possible.) Hence: B − A ∼ N (50 − 28, 36 + 16) = N (22, 52) = N (22, (7.2111)2 ).

So:

20 − 22 P (A arrives before B) = P (B − A > 20) = P Z > 7.2111 = P (Z > −0.28) = 0.6103.

Note that the above has made use of the following facts: • B − A is of the form αA + βB with α = −1 and β = 1.

• Since A and B are normal random variables, any linear combination αA + βB is also a normal random variable (provided the coefficients are both non-zero).

15

5. Probability

• E(αA + βB) = α E(A) + β E(B).

• Since A and B are independent: Var(αA + βB) = α2 Var(A) + β 2 Var(B). iii. Noting independence: P (at least 1 late) = 1 − P (all early) = 1 − P (A early ∩ B early ∩ C early) = 1 − (P (A early) × P (B early) × P (C early)) = 1 − ((1 − 0.3085) × (1 − 0.5) × (1 − 0.1587)) = 0.7091.

For the next two parts, it is essential to identify the whole sample space, i.e. the relevant outcomes, for the situations under discussion, and then to determine the events, i.e. the collection of outcomes, of interest in each part.

This means that you need to think of – and it may help to write out – some representation of the sample space(s) involved. If we let ‘E’ denote ‘early’ and ‘L’ denote ‘late’, then one way to do this is via the table below. Note there are eight possible outcomes overall, of which seven are of interest when considering the event ‘at least one late’ and three are of interest when considering the event ‘exactly one late’.

A B C

‘at least one late’ ↓ ↓ ↓ ↓ ↓ ↓ E E E E L L L E E L L E E L E L E L E L E ↑ ↑ ↑ ‘exactly one late’

↓ L L L

It is also possible to use a Venn diagram approach, but because we have three random variables (rather than two) in (v.) it is somewhat harder to work out in detail how the cases should be represented. iv. From Bayes’ formula, use P (A | B) P (B) = P (A ∩ B) = P (B | A) P (A). So: P (all late | ≥ 1 late) P (≥ 1 late) = P (all late ∩ ≥ 1 late) = P (≥ 1 late | all late) P (all late).

Now, P (≥ 1 late | all late) = 1, P (≥ 1 late) = 0.7091 and, using independence, P (all late) = 0.3085 × 0.5 × 0.1587 = 0.024479. Hence: P (all late | ≥ 1 late) =

16

1 × 0.024479 = 0.0345.

0.7091

5.1. Exercises

v. Again, Bayes’ formula yields: P (C late | exactly 1 late) P (exactly 1 late) = P (exactly 1 late ∩ C late) = P (only C late).

Under the independence of the distributions:    0.3085 × (1 − 0.5) × (1 − 0.1587)  +(1 − 0.3085) × 0.5 × (1 − 0.1587) P (exactly 1 late) = = 0.4755.   +(1 − 0.3085) × (1 − 0.5) × 0.1587 So: P (C late | exactly 1 late) = =

P (A early ∩ B early ∩ C late) P (exactly 1 late) (1 − 0.3085) × (1 − 0.5) × 0.1587 0.4755

= 0.1154. (b) This part reverses what we have done earlier – instead of finding probabilities given numerical values, here we are given a probability and want to find the corresponding numerical value.

We want to find C’s journey time of x minutes such that: P (C > x) = 0.025

⇒

x − 35 = 1.96.

5

Hence this gives a journey time of x = 35 + (5 × 1.96) = 44.8 ≈ 45 minutes. So C’s departure time should be 08:15.

General remarks: • The question asks for a clock time, not a duration!

• Different rounding conventions at various stages during the calculation can give slightly different values.

• It is best not to round values when using them in a subsequent calculation, especially for subtraction and/or division. This is because final results can be very sensitive to rounding errors.

17

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 6: Decision tree analysis and game theory

ii

Chapter 6 Decision tree analysis and game theory 6.1

Exercises

6.1.1

Test your understanding

1. What are the elements of a decision tree?

Solution: A decision tree consists of: • decision nodes, denoted by , which represent points where the decision-maker decides which path to take • chance nodes, denoted by

, which represent points where there is resolution of uncertainty

• branches which connect the nodes • probabilities which indicate the likelihood of each branch stemming from a chance node • payoffs at the end of terminal branches which indicate the monetary outcome.

2. How is an expected monetary value calculated?

Solution: An expected monetary value (EMV) is calculated as a probability-weighted average, i.e. by multiplying the possible monetary outcomes of a decision by their respective probabilities and summing them up. For n outcomes, the formula for calculating the EMV is: EMV =

n X

(outcome i × probability of outcome i).

i=1

Each outcome represents a potential monetary gain or loss, and each probability represents the likelihood of that outcome occurring. By calculating the EMV, one can assess the average expected monetary outcome of a decision or project, incorporating both the probabilities and the associated values.

3. Why should a risk-neutral decision-maker base their decision on a comparison of expected monetary values?

Solution: A risk-neutral decision-maker does not consider the level of risk or uncertainty associated with the outcomes. Hence the EMV provides a straightforward assessment of the expected monetary outcome, disregarding risk. Hence a risk-neutral decision-maker should base their decision on a comparison of expected monetary values.

1

6. Decision tree analysis and game theory

4. Explain why decision trees with multiple decision nodes should be solved using backward induction.

Solution: Backward induction allows for the systematic evaluation of the optimal decision at each stage.

Starting with the final decision node, the decision-maker considers the expected values associated with the different options and selects the one with the highest expected monetary value (profit maximisation or loss minimisation). This process is repeated backwards, evaluating decision nodes at each preceding stage until reaching the initial decision node. Backward induction ensures that decisions are made in a way which maximises the expected value at each step, leading to an optimal overall decision strategy. While in practice not every decision node may ultimately be reached, it is important to contingency plan to prepare for every possible scenario.

5. Distinguish between ‘risk-averse’, ‘risk-neutral’ and ‘risk-loving’ decision-makers.

Solution: A decision-maker is risk-averse if they strictly prefer the certain outcome of £x over a risky alternative with an EMV also of £x.

A decision-maker is risk-neutral if they are indifferent between a certain outcome and an uncertain outcome with the same EMV. A decision-maker is risk-loving if they strictly prefer a risky alternative with an EMV of £x over the certain outcome of £x.

6. Do you consider yourself to be risk-averse, risk-neutral or risk-loving? Why do you think this?

Solution: I suspect you are all three(!) with your risk appetite varying depending on what is at stake.

Risk-aversion will likely increase as the magnitude of the payoffs increase. Of course, you may be a thrill-seeker or adrenaline junkie and actively seek out risky opportunities at all times!

7. What is meant by the term ‘certainty equivalent’ ?

Solution: The certainty equivalent (CE) of a risky alternative is defined as the amount of money that would make the decision-maker indifferent between receiving this amount with certainty and the risky alternative.

8. Define the term ‘risk premium’ and explain how an individual’s risk attitude can be determined from the risk premium.

Solution: The risk premium of a risky alternative is defined as: risk premium = EMV − CE and represents how much money the decision-maker is willing to pay to receive a certain payoff of x rather than face the risky alternative with an expected payoff of x.

2

6.1. Exercises

Risk attitudes can be determined as follows: EMV > CE

⇒

risk-averse, with positive risk premium

EMV = CE

⇒

risk-neutral, with zero risk premium

EMV < CE

⇒

risk-loving, with negative risk premium.

For risk-neutral decision-makers, we simply choose the maximum EMV option.

9. What is the difference between ‘perfect information’ and ‘imperfect information’ ?

Solution: Perfect information turns decision-making under uncertainty into decision-making under certainty.

In a decision tree, chance nodes appear before decision nodes, such that the decision-maker has full knowledge of the ‘state of the world’ by the time they must make a decision.

Imperfect information is the receipt of an imperfect signal (such as an expert opinion) which reduces – but does not eliminate – the uncertainty in a decision tree. Bayes’ theorem updates the probabilistic beliefs in light of the new information allowing the decision-maker to make a better-informed decision.

10. How is the expected value of perfect information calculated?

Solution: 1. Calculate the EMV in the usual way from the problem of decision-making under uncertainty.

2. Determine the optimal decision under each possible state of the world and its associated payoff.

3. Calculate the expected payoff with perfect information as a probability-weighted average.

4. Under risk-neutrality, then: EVPI = E(payoff | perfect information) − EMV.

11. How is the expected value of imperfect information calculated?

Solution: 1. Calculate the EMV in the usual way from the problem of decision-making under uncertainty.

2. Determine the optimal decision under each possible predicted state of the world.

3. Calculate the expected payoff with imperfect information as a probability-weighted average.

4. Under risk-neutrality, then: EVII = E(payoff | imperfect information) − EMV.

3

6. Decision tree analysis and game theory

12. Suppose an expert predicts (imperfectly) whether a prospective borrower would default on a loan by classifying the prospective borrower as either a ‘good credit risk’ (resulting in loan approval) or a ‘bad credit risk’ (resulting in a loan rejection). In qualitative/practical terms, distinguish between a: (a) true positive (b) false positive (c) true negative (d) false negative.

Solution: (a) True positive – the expert predicts that the borrower is a bad credit risk, and the borrower actually defaults on the loan. (b) False positive – the expert predicts that the borrower is a bad credit risk, but the borrower does not default on the loan. (c) True negative – the expert predicts that the borrower is a good credit risk, and the borrower does not default on the loan. (d) False negative – the expert predicts that the borrower is a good credit risk, but the borrower actually defaults on the loan.

13. What is meant by ‘strategic interaction’ and how does this differ from unilateral decision-making?

Solution: Strategic interaction refers to a situation where the actions and decisions of one party are influenced by the anticipated actions and decisions of other parties. It involves considering the impact of others’ choices and adjusting one’s own strategy accordingly.

Unilateral decision-making involves making decisions without considering or being influenced by the actions of other parties involved in the situation. It is a decision-making process where only one party’s choices and objectives are taken into account.

14. Outline the components of a simultaneous-move game.

Solution: A simultaneous-move game (sometimes called a ‘normal form game’) has three components: 1. A set of players (we will consider two players only; Player 1 and Player 2).

2. Each player’s set of strategies, Si for i = 1, 2, i.e. the possible decisions they can make.

3. The payoffs for each player (typically these can be thought of as monetary payoffs, or at least a monetary equivalent).

15. Assume player 1 has payoff profiles u1 ({Left, Up}) = u1 ({Right, Down}) = 4 and u1 ({Left, Down}) = u1 ({Right, Up}) = 2 and assume the payoff profiles of player 2 are u2 ({Left, Up}) = u2 ({Right, Down}) = 2 and u2 ({Left, Down}) = u2 ({Right, Up}) = 3.

Construct the payoff matrix in normal form.

4

6.1. Exercises

Solution: To construct the payoff matrix in normal form, we need to list the payoffs for each combination of strategies chosen by players 1 and 2. In this case, we have two players and each player has two possible strategies: ‘Left’ and ‘Right’ for player 1 and ‘Up’ and ‘Down’ for player 2.

Hence the payoff matrix in normal form, where the values in each cell represent the payoffs for the respective combination of strategies chosen by the players (with player 1’s payoff given first), is: Player 1 Left Right Player 2 Up (4, 2) (2, 3) Down (2, 3) (4, 2) 16. What is a dominant strategy?

Solution: Assume a player has a particular strategy which always yields a greater payoff regardless of what the other player does. We call such a strategy a dominant strategy. When a player possesses a dominant strategy, then it is rational to always play this. If both players have a dominant strategy, then they each play their respective dominant strategies.

Strategy s∗i of player i is a dominant strategy if: πi (s∗i , s−i ) > πi (si , s−i ) for all si different from s∗i and for all s−i .

This means that s∗i is superior than any other strategy of player i, regardless of the strategies of other players.

17. How would you explain the Prisoner’s dilemma to a friend?

Solution: The Prisoner’s dilemma is a scenario where two individuals are arrested and offered separate deals.

If one stays silent and the other confesses, the confessor gets a reduced sentence while the silent one gets a harsher penalty. However, if both confess, they both receive moderate sentences. It shows how self-interest can lead to suboptimal outcomes when cooperation would yield better results.

18. What is a dominated strategy?

Solution: A dominated strategy results in a strictly worse payoff relative to any other strategy, regardless of the strategies of other players. Formally, strategy si for player i is a dominated strategy if another strategy, say s0i , results in a higher payoff no matter what other players do. Hence: πi (si ) < πi (s0i ).

19. Explain the process of iterative elimination of strictly dominated strategies.

Solution: Iterative elimination of strictly dominated strategies involves repeatedly eliminating strategies which are always worse than other available options. Starting with the initial set of strategies, any

5

6. Decision tree analysis and game theory

strategy that is strictly dominated (that is, which always yields lower payoffs regardless of the opponent’s choices) is removed. This elimination process continues iteratively until no strictly dominated strategies remain.

20. Distinguish between a ‘pure strategy’ and a ‘mixed strategy’.

Solution: A pure strategy is when a player chooses a single action that is a member of their strategy profile.

For example, in the Prisoner’s dilemma, the pure strategies for each prisoner are ‘Confess’ and ‘Silent’.

A mixed strategy assigns a probability to each pure strategy. For example, a prisoner could choose ‘Confess’ with probability π and ‘Silent’ with probability 1 − π. Formally, a mixed strategy si is a probability distribution over the set of (pure) strategies.

21. What is a Nash equilibrium?

Solution: A Nash equilibrium (in pure strategies) is a strategy profile (s∗1 , s∗2 , . . . , s∗n ) if each s∗i is a mutual best response. This means that s∗i is a best response to s∗−i (the strategies of all players other than player i). As a best response, it is impossible for player i to be better off (in terms of their payoff) if they deviate from s∗i . Formally: πi (s∗i , s∗−i ) ≥ πi (si , s∗−i ) for all strategies si in Si .

22. In qualitative terms, briefly explain the Cournot competition model.

Solution: The Cournot competition model is used to analyse oligopolistic markets. Firms simultaneously choose their output levels, assuming that each firm believes its competitors’ outputs are fixed. The firms act as quantity setters rather than price takers. The model demonstrates how firms interact strategically, considering the impact of their output decisions on market prices and profits, leading to a non-cooperative equilibrium.

23. Is collusion desirable? (Ignore any legal implications!)

Solution: Desirability is, of course, subjective! From a strategic perspective, collusion among competitors can lead to increased profits and stability in the short term. It may allow firms to coordinate their actions, reduce competition, and collectively exert market power. However, collusion can be detrimental to overall market efficiency, consumer welfare, and innovation. It can limit choices, lead to higher prices, and stifle competition. Ethical considerations and long-term implications need to be carefully evaluated when assessing the desirability of collusion!

6

6.1. Exercises

24. In qualitative terms, briefly explain the Bertrand competition model.

Solution: The Bertrand competition model is used to analyse price competition between firms. Firms simultaneously set their prices, assuming that consumers will always choose the lowest-priced product. The model assumes perfect substitutes and assumes that firms compete only on price. It shows that in a competitive market, firms will undercut each other’s prices until the price reaches the level of marginal cost, resulting in zero economic profits in equilibrium.

25. In qualitative terms, briefly explain the Stackelberg leadership model.

Solution: The Stackelberg leadership model is used to analyse strategic interaction between firms in an oligopolistic market. One firm, the leader, sets its quantity or price first, and then the follower firms adjust their strategies accordingly. The leader’s actions influence the follower’s decisions, and the leader typically has a first-mover advantage in terms of market power or cost structure. It highlights the importance of strategic decision sequencing and the ability of the leader to anticipate and shape the behaviour of other firms in the market.

26. A (risk-loving?) individual decides to play Russian roulette. The revolver to be used has a cylinder with room for a maximum of six bullets. The cylinder is spun, and the revolver placed against the player’s head. The possible outcomes are either ‘survive with net worth’ or ‘die’. (a) James decides to play. He places five bullets in the revolver’s cylinder. James has the option of paying an amount of money, m1 , to remove one bullet (meaning there would only be four bullets). Construct James’ decision tree. (b) James decides to play. He places one bullet in the revolver’s cylinder. James has the option of paying an amount of money, m2 , to remove the bullet (meaning there would be no bullets in the cylinder). Construct James’ decision tree. (c) Explain in qualitative terms how a solution could be reached for the decision trees in (a) and (b).

Solution: To construct the decision trees for James’ choices in (a) and (b), we will represent the possible outcomes as branches of the tree. Since James has the option of paying an amount of money to change the setup of the revolver, the decision tree will include decision nodes where James chooses whether to pay or not. The probability of surviving will depend on his decision, since this affects the number of bullets in the revolver’s cylinder. (a) James decides to play with five bullets in the cylinder and has the option to pay an amount m1 to remove one bullet. In this tree, the first node represents James’ decision to pay m1 or not.

The probabilities on the branches indicate the likelihood of the corresponding outcome. If he pays m1 , one bullet is removed, and the probability of surviving becomes 1/3 (= 2/6), while the probability of death is 2/3 (= 4/6). If he doesn’t pay, the probability of surviving is 1/6, while the probability of death is 5/6.

7

6. Decision tree analysis and game theory

(b) James decides to play with one bullet in the cylinder and has the option to pay an amount m2 to remove the bullet. In this tree, the first node represents James’ decision to pay m2 or not. If he pays m2 , the bullet is removed, and the probability of surviving becomes 1 (he survives with certainty). If he doesn’t pay, the probability of surviving is 5/6.

(c) To reach a solution for the decision trees in parts (a) and (b), James needs to consider his risk attitudes and the potential outcomes. He will calculate the expected value for each decision branch and choose the option which maximises his expected payoff (potential problem: can you put a monetary value on death?). For a risk-loving individual, he might be willing to take higher risks for an added adrenaline rush from greater danger. Therefore, he might choose to play without paying m1 in part (a). However, in part (b), he might choose to pay m2 to eliminate the risk of dying altogether (for the present moment in time at least, assuming James is not immortal). In practice, the actual decisions will also depend on the specific values of m1 and m2 , James’ risk appetite and whether James has a death wish (this James does not!).

27. James, who is risk-neutral, is deciding whether to launch a new product to market. He also has an option to pilot test the new product (and observe whether the pilot test is successful) before committing to the full launch. Suppose that for both the pilot test and the full launch he judges commercial success and failure to be equally likely. Initially he has wealth of £100,000.

8

6.1. Exercises

• If the pilot test is successful, James makes £10,000, but if the pilot test is unsuccessful he loses £5,000.

• If the full product launch is successful (regardless of the outcome of any pilot test), James makes £1,000,000, but if the full product launch is unsuccessful he loses £500,000.

Draw the decision tree and determine James’ optimal decisions.

Solution: The decision tree is:

Consider decision node 2. James’ expected monetary value from chance node B is: EMV(launch) = £1,010,000 × 0.5 − £490,000 × 0.5 = £260,000 which exceeds the (certain) payoff of £10,000 from ‘don’t launch’, hence the optimal decision at decision node 2 is ‘launch’.

Consider decision node 3. James’ expected monetary value from chance node C is: EMV(launch) = £995,000 × 0.5 − £505,000 × 0.5 = £245,000 which exceeds the (certain) payoff of −£5000 from ‘don’t launch’, hence the optimal decision at decision node 3 is ‘launch’.

Consider decision node 1. James’ expected monetary value from chance node A is: EMV(pilot test) = £260,000 × 0.5 + £245,000 × 0.5 = £252,500

9

6. Decision tree analysis and game theory

and from chance node D is: EMV(no pilot test) = £1,000,000 × 0.5 − £500,000 × 0.5 = £250,000 hence the optimal decision at decision node 1 is ‘pilot test’ with an expected monetary value of £252,500. Since this is positive, it is worth proceeding. Had this expected payoff been negative, James would be better off doing nothing, preserving his initial wealth of £100,000.

Note that since the outcome of the pilot test does not affect the probabilities of the launch being successful and unsuccessful, we could have simply observed that the expected monetary value of only the pilot test is: EMV(pilot test) = £10,000 × 0.5 − £5000 × 0.5 = £2500 which is positive so it makes sense to proceed with the pilot test regardless. (In practice we might expect the outcome of the pilot test to signal how likely a successful launch would be, but we are not given such conditional probabilities in the question.)

28. A risk-neutral media executive is deciding whether to offer a new ‘freemium’ subscription service. If successful, the company will make £20 million. If unsuccessful, the company will make a loss of £5 million. Doing nothing costs nothing. (a) It is (subjectively) estimated that there is a 25% chance of success. What should the executive do? (b) For what probability of success, π, would the executive be indifferent in terms of their decision-making? (c) The executive could undertake test marketing which would inform them with certainty about whether the freemium subscription service will be a success. Calculate the expected value of perfect information, and hence determine the maximum willingness to pay for the test marketing. (d) Unfortunately, the test marketing is discovered not to be perfect after all. Suppose there is an 85% chance of a true positive (the test marketing correctly indicating success), and a 95% chance of a true negative (the test marketing correctly indicating failure). Calculate the expected value of imperfect information, and hence determine the maximum willingness to pay for the test marketing.

Solution: (a) The decision tree is:

10

6.1. Exercises

The expected monetary value from offering the freemium service is: EMV(offer freemium) = £20m × 0.25 − £5m × 0.75 = £1.25m which exceeds the (certain) payoff of £0 from ‘do nothing’, hence the optimal decision for the (risk-neutral) media executive is ‘offer freemium’. (b) Let π be the probability of success. To be indifferent, the expected monetary value from offering the freemium service must be £0. Hence: EMV(offer freemium) = £20m × π − £5m × (1 − π) = £0

⇒

π = 0.20.

(c) The decision tree is:

A certain prediction of success would result in the decision to ‘offer freemium’ at decision node 1 (with a £20m payoff), while a certain prediction of failure would result in the decision to ‘do nothing’ at decision node 2 (with a £0 payoff, which is better than a loss of £5m!). Hence: E(payoff | perfect information) = £20m × 0.25 + £0 × 0.75 = £5m and so the expected value of perfect information is: EVPI = E(payoff | perfect information) − EMV = £5m − £1.25m = £3.75m which is therefore the maximum willingness to pay for the (perfect) test marketing. (d) From the information provided, we have that: P (predict success | actual success) = 0.85 and P (predict failure | actual success) = 0.15 and also: P (predict success | actual failure) = 0.05 and P (predict failure | actual failure) = 0.95.

The prior probabilities of success and failure are 0.25 and 0.75, respectively. Using the total probability formula we have: P (predicts success) = P (predicts success | actual success) P (actual success) + P (predicts success | actual failure) P (actual failure) = 0.85 × 0.25 + 0.05 × 0.75 = 0.25

11

6. Decision tree analysis and game theory

and hence P (predicts failure) = 1 − 0.25 = 0.75. Note that while these are the same as the prior probabilities of success and failure, this is purely a coincidence in this case!

Now we seek the conditional probability of actual success, given the test marketing’s prediction of success. Applying Bayes’ theorem, we have: P (actual success | predicts success) = =

P (predicts success | actual success) P (actual success) P (predicts success) 0.85 × 0.25 0.25

= 0.85 and hence P (actual failure | predicts success) = 1 − 0.85 = 0.15. Note the coincidence of the two 0.25 probabilities cancelling, which will not be the case in general.

Similarly: P (actual success | predicts failure) = =

P (predicts failure | actual success) P (actual success) P (predicts failure) 0.15 × 0.75 0.75

= 0.15 and hence P (actual failure | predicts failure) = 1 − 0.15 = 0.85. Again note the coincidence of the two 0.75 probabilities cancelling, which will not be the case in general.

Notice that the probability of actual success changes in response to the test marketing’s prediction. The prior probability of actual success increases from 0.25 to 0.85 in response to a prediction of success (a correct positive signal; a true positive) and decreases from 0.25 to 0.05 in response to a prediction of failure (an incorrect negative signal; a false negative). Similarly, the prior probability of failure decreases from 0.75 to 0.15 in response to a prediction of success (an incorrect positive signal; a false positive) and increases from 0.75 to 0.95 in response to a prediction of failure (a correct negative signal; a true negative).

The decision tree with imperfect information is shown below.

12

6.1. Exercises

At chance node B, the EMV from offering freemium is: £20m × 0.85 − £5m × 0.15 = £16.25m hence at decision node 1 it is optimal to offer freemium since £0 < £16.25m.

At chance node C, the EMV from offering freemium is: £20m × 0.05 − £5m × 0.95 = −£3.75m. hence at decision node 2 it is optimal to do nothing since £0 > −£3.75m.

We now compute the EVII. At chance node A we have that: E(payoff | imperfect information) = £16.25m × 0.25 + £0 × 0.75 = £4.0625m.

Therefore: EVII = E(payoff | imperfect information) − EMV = £4.0625m − £1.25m = £2.8125m which is therefore the maximum willingness to pay for the (imperfect) test marketing. To note this is a lower value than the EVPI calculated in part (c), as it must be since the quality of imperfect information is inferior to the quality of perfect information.

29. Consider the following normal form of a game.

Player 2

Up Down

Player 1 Left Right 1, 1 0, 2 2, 0 π1 , π2

(a) Suppose that π1 = π2 . For what values of π1 do players 1 and 2 both have a dominant strategy?

What are these dominant strategies? (b) Suppose that π1 = π2 . To make this game a Prisoner’s dilemma, what would be the required values of π1 ? (c) Suppose that π1 > 0 and π2 < 0. Check whether there are any pure-strategy Nash equilibria. (d) Do any values of π1 and π2 exist such that there is no pure-strategy Nash equilibrium?

Solution: (a) Suppose π1 = π2 = π. To find the dominant strategies for both players, we compare the payoffs for each strategy.

Consider Player 1’s payoffs: • If Player 2 plays ‘Up’, Player 1’s payoffs are 1 and 0 for ‘Left’ and ‘Right’, respectively.

• If Player 2 plays ‘Down’, Player 1’s payoffs are 2 and π for ‘Left’ and ‘Right’, respectively.

For Player 1 to have a dominant strategy, this strategy must always result in a higher payoff.

Since 1 > 0, Player 1 can only have a dominant strategy of ‘Left’ which must mean that 2 > π.

13

6. Decision tree analysis and game theory

Consider Player 2’s payoffs: • If Player 1 plays ‘Left’, Player 2’s payoffs are 1 and 0 for ‘Up’ and ‘Down’, respectively.

• If Player 1 plays ‘Right’, Player 2’s payoffs are 2 and π for ‘Up’ and ‘Down’, respectively.

For Player 2 to have a dominant strategy, this strategy must always result in a higher payoff.

Since 1 > 0, Player 2 can only have a dominant strategy of ‘Up’ which must mean that 2 > π.

Hence both players have dominant strategies for π1 = π2 < 2. (b) To turn the game into a Prisoner’s dilemma, we need to ensure that betraying is a dominant strategy for both players, hence π1 = π2 < 2. However, the players need to be worse off in the cell (Left, Up) than had they been in the cell (Right, Down). Hence we also need π1 = π2 > 1.

Therefore, the required values are all those in the inteval (1, 2), i.e. 1 < π1 = π2 < 2. For example, suppose π1 = π2 = 1.5, then:

Player 2

Up Down

Player 1 Left Right *1, 1* 0, 2* *2, 0 1.5, 1.5

The pure-strategy Nash equilibrium is (Left, Up), but both players would be better off with (Right, Down). (c) Suppose π1 > 0 and π2 < 0. Since π2 < 0, then Player 2 has a dominant strategy of ‘Up’. Hence there is one pure-strategy Nash equilibrium of (Left, Up).

Player 2

Up Down

Player 1 Left Right *1, 1* 0, 2* **2, 0 **> 0, < 0

** as it depends on the value of π1 . (d) There are no values of π1 and π2 for which there is no pure-strategy Nash equilibrium because (Left, Up) is always a pure-strategy Nash equilibrium since when Player 1 plays Left, Player 2 always plays Up (0 < 1), and when Player 2 plays Up, Player 1 always plays Left (1 > 0).

30. Suppose that WFH (Work From Home) and WTF (Work Through Flexibility) are the only two firms facilitating remote working technology. If WFH and WTF each decide to enter the market, a price war is triggered causing both firms to be loss-making. In the event that only one of WFH and WTF enters, both firms are profitable due to the non-entrant being a supplier to the entrant of some of its technology. The payoffs, in billions of pounds, are as follows, shown as (WFH, WTF).

WFH

14

Enter market Stay out

WTF Enter market Stay out −5, −5 6, 1 1, 6 0, 0

6.1. Exercises

Assume the two firms make their decisions simultaneously. (a) Identify any pure-strategy Nash equilibria. (b) Identify the mixed-strategy Nash equilibrium. (c) Suppose that, before deciding whether to enter, WTF signs a binding agreement with a remote recruitment agency. Business news media widely report the details of the agreement so it is common knowledge. The agreement promises the remote recruitment agency compensation of £7 billion from WTF if WTF chooses not to enter the market. Find any pure-strategy Nash equilibria of the game in this case.

Solution: (a) If WFH plays ‘Enter market’, WTF will choose ‘Stay out’ (since −5 < 1), and if WFH plays ‘Stay out’, WTF will choose ‘Enter market’ (since 6 > 0).

By symmetry, if WTF plays ‘Enter market’, WFH will choose ‘Stay out’ (since −5 < 1), and if WTF plays ‘Stay out’, WFH will choose ‘Enter market’ (since 6 > 0).

Hence there are two pure-strategy Nash equilibria: (Enter market, Stay out) and (Stay out, Enter market).

WFH

Enter market Stay out

WTF Enter market Stay out −5, −5 *6, 1* *1, 6* 0, 0

(b) A mixed-strategy Nash equilibrium involves both players randomising their strategies based on certain probabilities to maximise their expected payoffs. To find the mixed-strategy Nash equilibrium, we need to consider each player’s willingness to randomise their actions.

Let p be the probability that WFH enters the market and q be the probability that WTF enters the market. Since both players are indifferent between their options when they enter, we can set up the following equations for their expected payoffs: E(πWFH (Enter market)) = −5q + 6(1 − q) and E(πWFH (Stay out)) = q + 0(1 − q) = q We now solve for q such that: E(πWFH (Enter market)) = E(πWFH (Stay out))

⇐⇒

−5q + 6(1 − q) = q

which gives q = 1/2. By symmetry, p = 1/2. Hence the mixed strategy Nash equilibrium can be written as:

1 1 1 1 , , , .

2 2 2 2 (c) With the binding agreement, the payoff matrix changes:

WFH

Enter market Stay out

WTF Enter market Stay out −5, −5* *6, −6 *1, 6* 0, −7

15

6. Decision tree analysis and game theory

If WFH plays ‘Enter market’, WTF will choose ‘Enter market’ (since −5 > −6), and if WFH plays ‘Stay out’, WTF will choose ‘Enter market’ (since 6 > −7). Hence ‘Enter market’ is now a dominant strategy for WTF due to the binding agreement.

If WTF plays ‘Enter market’, WFH will choose ‘Stay out’ (since −5 < 1), and if WTF plays ‘Stay out’, WFH will choose ‘Enter market’ (since 6 > 0).

Hence there is one pure-strategy Nash equilibrium: (Stay out, Enter market). The binding agreement signals to WFH that WTF is better off entering the market regardless, which means WFH chooses to stay out.

31. Suppose N competitor firms decide simultaneously whether or not they wish to enter a particular product market. Suppose that the market can only support a maximum of Q firms, where Q < N .

If at most Q of the N firms choose enter the market, suppose each entrant earns £1 million. If more than Q firms enter, each entrant earns nothing. Each non-entering firm earns £0.5 million.

Determine whether there are any pure-strategy Nash equilibria.

Solution: Consider a single firm. If it believes that exactly Q or fewer firms will enter the market, then it should definitely enter since it will earn £1 million. However, if it believes that more than Q firms will enter, then it should not enter, as it will earn £0.5 million if it stays out.

Suppose Q firms enter the market. Each will earn £1 million, and there is no incentive for any firm to deviate from entering. However, if fewer than Q firms enter, any firm that stayed out will earn £0.5 million, and there will be an incentive for a firm to deviate and enter the market to earn £1 million.

If more than Q firms enter, all entrants will earn nothing, and there will be an incentive for any firm that entered to deviate and stay out to earn £0.5 million.

There are no pure-strategy Nash equilibria in this game. For each firm, its best response depends on the choices made by the other firms. If the firm believes that exactly Q firms will enter, it will enter; otherwise, it will stay out. However, this reasoning depends on the firm’s beliefs about the actions of the other firms, and these beliefs may not be consistent with what actually happens.

There may be mixed-strategy Nash equilibria where firms randomise their choices according to a probability distribution. For example, if a firm believes that exactly Q firms will enter with some probability π and that fewer than Q firms will enter with probability 1 − π, it may decide to randomise between entering and staying out with specific probabilities to maximise its expected payoff.

32. An industry has two identical firms, producing output of q1 and q2 , respectively. Hence industry output is Q = q1 + q2 . Total production costs for firm i are 2 + qi , for i = 1, 2. Suppose the market demand curve is Q = 10 − P . (a) Determine the Cournot–Nash equilibrium quantities for each firm, and the resulting market price, P . (b) Could these two firms engage in collusion? Justify your answer. (c) Suppose firm 1 moved first. Using the Stackelberg leadership model how would this affect the quantities for each firm, and the resultant market price, P ?

16

6.1. Exercises

Solution: (a) Each firm’s profit is given by its revenue less its total production costs. Revenue is the product of the market price and the quantity sold. The market price is determined by the total quantity supplied in the market (i.e. Q) and the market demand curve.

From the market demand curve, we have that Q = 10 − P . Substitute Q = q1 + q2 into the demand equation to find the market price: q1 + q2 = 10 − P

⇒

P = 10 − q1 − q2

Firm 1’s profit is given by: Π1 (q1 , q2 ) = P q1 − (2 + q1 ) = (10 − q1 − q2 )q1 − (2 + q1 ) = 9q1 − q12 − q1 q2 − 2 and similarly firm 2’s profit is given by: Π2 (q1 , q2 ) = P q2 − (2 + q2 ) = (10 − q1 − q2 )q2 − (2 + q2 ) = 9q2 − q22 − q1 q2 − 2 subject to the constraints 0 ≤ qi ≤ 10 − P for i = 1, 2 (since each firm’s production cannot exceed the market demand).

To find each firm’s best response, we maximise their profits with respect to their quantity decisions. For firm 1: max Π1 q1

⇒

∂Π1 = 9 − 2q1 − q2 = 0 ∂q1

⇒

q1∗ =

9 − q2 .

2

⇒

∂Π2 = 9 − 2q2 − q1 = 0 ∂q2

⇒

q2∗ =

9 − q1 .

2

By symmetry, for firm 2: max Π2 q2

Therefore, we have that q1∗ = q2∗ = q (i.e. we can drop the subscript) such that: q∗ =

9 − q∗ 2

⇒

q ∗ = 3.

Hence the industry supply is: Q∗ = q ∗ + q ∗ = 3 + 3 = 6.

The market equilibrium price is then: P ∗ = 10 − Q∗ = 10 − 6 = 4. (b) Collusion occurs when firms cooperate to maximise their joint profits, usually by agreeing on a specific output level or price. In a Cournot duopoly like this, firms may collude by coordinating their production levels to achieve higher joint profits. Whether these firms can collude successfully depends on various factors, such as the level of trust between the firms, the possibility of detecting deviations from the agreed-upon strategy, and the presence of antitrust regulations.

17

6. Decision tree analysis and game theory

(c) In the Stackelberg leadership model, the leader makes its production decision first. The other firm is the follower and observes the leader’s decision before making its own production choice.

The leader’s decision will affect the market price, and the follower will take this into account when making its decision.

If firm 1 is the leader (and hence firm 2 is the follower), the leader’s best response is still to maximise its profit based on market demand, but it takes into account the follower’s reaction function. The follower’s reaction function is how firm 2 responds to firm 1’s production decision, which is simply the Cournot solution of: q2∗ =

9 − q1 .

2

Aware of this, the leader maximises their profit function: Π1 (q1 , q2 ) = P q1 − (2 + q1 ) = (10 − q1 − q2 )q1 − (2 + q1 )

9 − q1 q1 − (2 + q1 ) = 10 − q1 − 2 = 4.5q1 − 0.5q12 − 2.

Maximising, we have: max Π1 q1

⇒

∂Π1 = 4.5 − q1 = 0 ∂q1

⇒

q1∗ = 4.5.

We now substitute this into the reaction function of firm 2 to determine their best response: q2∗ =

9 − 4.5 9 − q1∗ = 2 2

⇒

q2∗ = 2.25 =

q1∗ 2

meaning that the follower only produces half of the output of the leader. Hence the leader has larger market share resulting in a clear first-mover advantage.

The resultant market price is: P ∗ = 10 − Q∗ = 10 − q1∗ − q2∗ = 10 − 4.5 − 2.25 = 3.25.

18

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 7: Sampling and sampling distributions

ii

Chapter 7 Sampling and sampling distributions 7.1

Exercises

7.1.1

Test your understanding

1. What are the main considerations in sampling design?

Solution: Sampling design involves several basic questions.

• Should a sample even be taken (over a census)?

• If we should sample, what process should be followed?

• What kind of sample should be drawn?

• How large should the sample size, n, be?

2. Define each of the following terms: ‘target population’, ‘members’, ‘sampling unit’, ‘census’, ‘sample’ and ‘sampling error’.

Solution: Target population – the collection of all population members, sharing some common set of characteristics, which comprise the universe for the purpose of the business analytics problem about which inferences are sought.

Members – ‘objects’ in which we are interested; for example, customers, ex-customers, or potential customers.

Sampling unit – a member (such as a customer) or a unit which contains the member (such as a household).

Census – a complete enumeration of the members of a population of study ‘objects’; again, this could be customers, ex-customers, potential customers etc.

Sample – a subgroup of the population members selected for participation in the study.

Sampling error (or estimation error) – the difference between an estimate of a population parameter and its true value. The less representative a sample is, the greater the risk of a large sampling error.

3. What factors determine whether a census or a sample is preferred?

Solution: The factors are: • Population size.

1

7. Sampling and sampling distributions

• Time horizon.

• Budget.

• Homogeneity of the population.

• Sampling errors.

• Non-sampling errors.

• Type of measurement.

4. What are the characteristics of non-random sampling techniques, and when would these be used in practice?

Solution: Non-random sampling techniques are characterised by the fact that some units in the population have an unknown probability or a zero probability of selection. These are used when a sampling frame is not available and/or when only exploratory research is required.

5. What are the characteristics of random sampling techniques, and when would these be used in practice?

Solution: Random sampling techniques mean every population member has a known, non-zero probability of being selected. Probability sampling makes it possible to estimate the margins of sampling error, and hence all statistical techniques (such as confidence intervals and hypothesis tests) can be applied. These are used when conclusive research is required.

6. Outline the features of a convenience sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A convenience sample is a non-random sampling technique where individuals or elements are selected based on their easy availability and accessibility to the researcher.

Participants are chosen based on their convenient proximity or availability. The sample is easily accessible and requires minimal effort to recruit. Convenience sampling can be quick and cost-efficient compared to other sampling methods. However, the sample may not be representative of the entire population due to potential biases and lack of random selection.

A practical example is conducting a survey about customer satisfaction in a shopping mall.

Researchers might approach customers exiting the mall and ask for their feedback, as these individuals are readily available and accessible at the location. While the sample may not represent the entire population of mall visitors, it can still provide insights into the satisfaction levels of those who happened to be present at that specific time and place.

7. Outline the features of a judgemental sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A judgemental sample is a non-probability sampling method where the researcher relies on their expertise to select specific individuals or elements for the study based on their judgement and

2

7.1. Exercises

knowledge of the population. This approach is subjective and hence may not represent the entire population. An example would be selecting key industry experts for a qualitative market research study on emerging technology trends.

8. Outline the features of a quota sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A quota sample is a non-probability sampling method where the researcher divides the population into subgroups (quotas) based on specific and relevant characteristics, after which individuals are selected until the quota for each subgroup is met. It allows for control over the sample composition but lacks randomisation. An example would be conducting a customer satisfaction survey by setting quotas for age groups, gender, and income levels to ensure representation of diverse customer segments in the analysis.

9. Outline the features of a snowball sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A snowball sample is a non-probability sampling method where participants refer others to the study, forming a chain-like (or snowball, hence the name!) recruitment process. It is useful when the target population is difficult to access, such as hidden or marginalised groups. An example is researching niche industries or underground markets, where personal referrals can help identify and connect with potential participants.

10. Outline the features of a simple random sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A simple random sample is a probability sampling method where each member of the population has an equal and known chance of being selected. It ensures unbiased representation and allows generalisability to the whole population. For example, when conducting customer satisfaction surveys for a large retail chain, a simple random sample would be appropriate to ensure fair selection of customers from all store locations and demographics.

11. Outline the features of a systematic sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A systematic sample is a probability sampling method where researchers select every ith element from a list of the population, known as the sampling frame, where the sampling interval i = bN/ne.

It provides a representative sample and is more efficient than simple random sampling for large populations. For example, when analysing employee performance in a company, selecting every 10th employee from the HR database would be practical and time-saving. Whenever the order of the population members in the sampling frame is related to the characteristic of interest, systematic sampling will increase the representativeness of the sample. However, whenever the order of the population members produces a cyclical pattern, systematic sampling may decrease the representativeness of the sample.

3

7. Sampling and sampling distributions

12. Outline the features of a stratified sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A stratified sample is a probability sampling method where the population is divided into mutually exclusive and collectively exhaustive subgroups (strata) based on relevant characteristics. Simple random samples are then drawn from each stratum, usually in proportion to its relative size (proportionate stratified sampling) or proportional to the relative size of the stratum and the standard deviation (disproportionate stratified sampling). It ensures representativeness of all strata in the sample. For example, when conducting a market survey for a smartphone, stratifying the sample by age groups (18–25, 26–35, 36–45 etc.) would ensure diverse user perspectives across age ranges.

13. Outline the features of a cluster sample, and provide a practical example when you would consider this the most appropriate sampling technique.

Solution: A cluster sample is a probability sampling method where the population is divided into clusters (such as geographic regions or office branches), and a random sample of clusters is selected, either using simple random sampling or with probability proportional to size. If all members within the chosen clusters are included in the study then this is called one-stage cluster sampling. If a random sample of cluster members is chosen then this is called ‘two-stage cluster sampling’ and is suitable for large cluster sizes. A business may use a cluster sample when conducting employee satisfaction surveys by selecting random office branches for assessment.

14. In what sense are the techniques of quota sampling and stratified sampling similar? In what sense are they different?

Solution: Quota sampling and stratified sampling are both methods used to ensure adequate representation of different groups within a population. Both involve dividing the population into subgroups. However, they differ in how the samples are selected. In quota sampling, researchers choose participants non-randomly based on their judgement, while in stratified sampling samples are randomly selected from each subgroup (often in proportion to their size), ensuring a probability-based approach.

15. What is a pseudo-random number generator?

Solution: In order to perform simple random sampling we require a random number generator to randomly select (i.e. probabilistically select) which members are to be in the sample. In practice, we would appeal to a pseudo-random number generator, which generates numbers algorithmically, but for practical purposes the generated numbers have the appearance of being random. As the generated numbers are determined by an initial value called a seed, they are not truly random as they are entirely determined by the seed value.

16. Which Excel function could be used to select a simple random sample? Why is this function appropriate?

4

7.1. Exercises

Solution: In Excel RANDBETWEEN(bottom, top) returns a (pseudo-)random integer value where • bottom is the smallest possible integer value returned • top is the largest possible integer value returned such that all integer values {bottom, bottom+1, ..., top-1, top} are equally likely to be returned.

17. Why is a representative sample desirable, and to what extent do different sampling techniques achieve representative samples?

Solution: A representative sample is desirable because it accurately reflects the characteristics of the larger population it represents. It allows analysts to draw valid and reliable conclusions about the population based on the sample data. Different sampling techniques achieve representative samples to varying extents. Probability sampling methods, like simple random sampling and stratified sampling, have a higher likelihood of achieving representativeness as they allow each member of the population to have an equal or known chance of being selected. Non-probability sampling methods, such as convenience sampling and quota sampling, may introduce bias and are less likely to produce truly representative samples, as they rely on subjective criteria for selection.

18. When conducting a survey, does the choice of contact method matter?

Solution: Yes, it can! An interesting example is that of the systematically different results from telephone and online polling ahead of the UK’s EU membership referendum, held in June 2016.

More generally, the choice of contact method can have an impact on the response rate, response quality, and overall representativeness of the sample. There are also cost and logistics issues to consider.

In some cases, using multiple contact methods (mixed-mode surveys) can be beneficial to reach a broader and more diverse sample.

19. Explain what a sampling distribution is.

Solution: A sampling distribution is a theoretical distribution that shows all the possible values of a statistic (such as the sample mean or sample proportion) obtained from multiple random samples drawn from the same population.

20. Distinguish between a (point) estimator and a (point) estimate.

Solution: A (point) estimator is a statistic, that is, a known function of data, used to calculate an estimate of a population parameter. A (point) estimate is the specific numerical value obtained from the sample data using the chosen estimator.

5

7. Sampling and sampling distributions

21. What is a sampling fraction, and to what extent does this affect the sampling variance?

Solution: The sampling fraction is the proportion of the population which is sampled. It is defined as n/N , where n is the sample size and N is the population size. When sampling without replacement the sampling variance of the sample mean is: Var(X̄) =

N − n σ2 × .

N −1 n

Provided the sampling fraction is small, the term: N −n ≈1 N −1 so can be ignored. Therefore, the precision depends effectively on the sample size only.

22. What is a standard error?

Solution: A standard error refers to the standard deviation of the sampling distribution. For example, consider the sample mean. Its standard error is: r q N − n σ2 × = σX̄ .

S.E.(X̄) = Var(X̄) = N −1 n As per Question 21, for small sampling fractions we can drop the (N − n)/(N − 1) term.

23. What is the sampling distribution of the sample mean, X̄, when: (a) the random sample is drawn from a population with a N (µ, σ 2 ) distribution? (b) the random sample is not drawn from a population with a normal distribution?

Solution: (a) When sampling from N (µ, σ 2 ), the sampling distribution of the sample mean is such that:

σ2 X̄ ∼ N µ, . n (b) When taking a random sample of size n from a non-normal population with finite mean µ and finite variance σ 2 , then:

σ2 X̄ ∼ N µ, n approximately, as n → ∞.

24. How large does the sample size need to be for the central limit theorem to provide a good approximation to the sampling distribution of X̄? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends. The larger the sample size n, the better the normal approximation provided by the CLT is. In practice, we have various rules of thumb for what is ‘large enough’ for the approximation to be ‘accurate enough’. This also depends on the population distribution of X. For example:

6

7.1. Exercises

• for symmetric distributions, even small n is enough • for very skewed distributions, a larger n is required.

For many distributions, n ≥ 30 is sufficient for the approximation to be reasonably accurate.

25. What is the purpose of the finite population correction factor?

Solution: When sampling n observations without replacement from a population of size N , then: Var(X̄) =

n σ2 1− n N

and: σ S.E.(X̄) = √ × n

r n 1− .

N

The factor 1 − n/N is known as the finite population correction factor (FPCF). If n < 0.1N , then the FPCF can be ignored as it is negligible. The purpose of the FPCF is to correct for the increased precision when of the sample mean as we sample a greater proportion of the population.

26. ‘The sample proportion is just a sample mean’. Do you agree? Briefly justify your view.

Solution: The sample proportion is a special case of the sample mean, where the data are binary (0s and 1s).

The sum of the data is the total number of ‘successes’ (the 1s) since the 0s do not affect the total.

Division by n then returns the sample proportion.

The sample proportion estimates the population proportion, π, while the sample mean estimates the population mean, µ.

27. What is the (approximate?) sampling distribution of the sample proportion?

Solution: The sample proportion, P , is the estimator of the population proportion, π, such that:

π(1 − π) P ∼ N π, n approximately, as n → ∞, by the central limit theorem.

Note that P is a special case of X̄ when Xi = Bernoulli(π), i.e. when all observations are binary.

28. ‘All sampling distributions are normal distributions.’ Do you agree? Briefly justify your view.

Solution: Disagree. A counterexample is sufficient here. For example, by taking a simple function of S 2 , that is: (n − 1)S 2 ∼ χ2n−1 . σ2 whose sampling distribution is a chi-squared distribution with n − 1 degrees of freedom, i.e. χ2n−1 .

7

7. Sampling and sampling distributions

29. A random variable, X, can take the values 1, 2 and 3, each with equal probability. List all possible samples of size 2 which may be chosen when order matters, without replacement, from this population, and hence construct the sampling distribution of the sample mean, X̄.

Solution: Each possible sample has an equal probability of occurrence of 1/6. The sampling distribution of X̄ is: Possible samples

(1, 2) (2, 1)

(1, 3) (3, 1)

(2, 3) (3, 2)

Sample mean, X̄ = x̄

1.5

2

2.5

Relative frequency, P (X̄ = x̄)

1/3

1/3

1/3

30. A perfectly-machined regular tetrahedral (pyramid-shaped) die has four faces labelled 1 to 4. It is thrown twice onto a level surface and after each throw the number on the face which is downward is recorded. If the recorded values are x1 and x2 , when order matters, then the observed sample mean is x̄ = (x1 + x2 )/2. Write out the sampling distribution of the sample mean as a random quantity over repeated double throws.

Solution: Each possible sample has an equal probability of occurrence of 1/16. The sampling distribution of X̄ is: Possible samples

Sample mean, X̄ = x̄ Relative frequency, P (X̄ = x̄)

(1, 1)

(1, 2) (2, 1)

(1, 3) (3, 1) (2, 2)

(1, 4) (4, 1) (2, 3) (3, 2)

(2, 4) (4, 2) (3, 3)

(3, 4) (4, 3)

(4, 4)

1

1.5

2

2.5

3

3.5

4

1/16

1/8

3/16

1/4

3/16

1/8

1/16

31. Consider the population below with N = 4 elements: A 3

B 6

C 9

D 12

(a) Calculate the population mean and variance. (b) Write down the sampling distribution of the sample mean for samples of size n = 2 drawn without replacement, where order does not matter. (c) Using the result in (b), calculate the mean of the sampling distribution. (d) Using the result in (b), calculate the variance of the sampling distribution.

8

7.1. Exercises

Solution: (a) The population mean is: N P

µ=

xi

i=1

=

N

3 + 6 + 9 + 12 = 7.5.

4

The population variance is: N P

x2i

σ 2 = i=1 N

− µ2 =

9 + 36 + 81 + 144 − (7.5)2 = 11.25.

4

(b) The sampling distribution of the sample mean for samples of size n = 2 is: Sample AB AC AD BC BD CD

Values 3 6 3 9 3 12 6 9 6 12 9 12

X̄ = x̄ 4.5 6 7.5 7.5 9 10.5

P (X̄ = x̄) 1/6 1/6 1/6 1/6 1/6 1/6

(c) The mean of the sampling distribution is: E(X̄) =

4.5 + 6 + 7.5 + 7.5 + 9 + 10.5 45 = = 7.5.

6 6

(d) The variance of the sampling distribution is: Var(X̄) =

(4.5)2 + (6)2 + (7.5)2 + (7.5)2 + (9)2 + (10.5)2 − (7.5)2 = 3.75.

6

32. The weights of a large group of manufactured components have mean 8.2 kg and standard deviation 2.2 kg. What is the probability that a random selection of 80 components will have a mean weight between 8.3 kg and 8.4 kg? State any assumptions you make.

Solution: We are not told that the population is normal, but n is ‘large’ so we can apply the central limit theorem. The sampling distribution of X̄ is, approximately:

σ2 (2.2)2 X̄ ∼ N µ, = N 8.2, . n 80 Hence:

P (8.3 ≤ X̄ ≤ 8.4) = P

8.3 − 8.2 8.4 − 8.2 √ ≤Z≤ √ 2.2/ 80 2.2/ 80

= P (0.41 ≤ Z ≤ 0.81) = 0.1319 using =NORM.S.DIST(0.81,1)-NORM.S.DIST(0.41,1).

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7. Sampling and sampling distributions

33. A random sample of 25 audits is to be taken from a company’s total audits, and the average value of these audits is to be calculated. (a) Explain what is meant by the sampling distribution of this average and discuss its relationship to the population mean. (b) Is it reasonable to assume that this sampling distribution is normally distributed? (c) If the population of all audits has a mean of £54 and a standard deviation of £10, find the probability that: i. the sample mean will be greater than £60 ii. the sample mean will be within 5% of the population mean.

Solution: (a) The sample average (mean) is formed from 25 observations which are subject to sampling variability, hence the average is also subject to this variability. Its sampling distribution describes its probability properties. If a large number of such averages were independently sampled, then their histogram would be the (approximate) sampling distribution. (b) It is reasonable to assume that this sampling distribution is normal due to the central limit theorem (CLT) – although the sample size is rather small. (c) If n = 25, µ = 54 and σ = 10, then the CLT says that, approximately:

100 σ2 = N 54, .

X̄ ∼ N µ, n 25 i. We have: P (X̄ > 60) = P

60 − 54 Z>p 100/25

! = P (Z > 3) = 0.00135

using =1-NORM.S.DIST(3,1). ii. We are asked for:

P (0.95 × 54 < X̄ < 1.05 × 54) = P

−0.05 × 54 0.05 × 54 <Z< 2 2

= P (−1.35 < Z < 1.35) = 0.8230 using =NORM.S.DIST(1.35,1)-NORM.S.DIST(-1.35,1).

34. A country is about to hold a referendum about leaving the European Union. A survey of a random sample of adult citizens of the country is conducted. In the sample, n respondents say that they plan to vote in the referendum. These n respondents are then asked whether they plan to vote ‘Yes’ or ‘No’. Define X = 1 if such a person plans to vote ‘Yes’, and X = 0 if such a person plans to vote ‘No’.

Suppose that in the whole population 45% of those people who plan to vote are currently planning to vote Yes, and hence the referendum result would show a majority opposing leaving the European Union.

10

7.1. Exercises

(a) Let X̄ =

n P

Xi /n denote the proportion of the n voters in the sample who plan to vote Yes.

i=1

What is the central limit theorem approximation of the sampling distribution of X̄ here? (b) If there are n = 50 likely voters in the sample, what is the probability that X̄ > 0.5? (Such an opinion poll would suggest a majority supporting leaving the European Union in the referendum.) (c) How large should n be so that there is less than a 1% chance that X̄ > 0.5 in the random sample? (This means less than a 1% chance of the opinion poll incorrectly predicting a majority supporting leaving the European Union in the referendum.)

Solution: (a) Here the individual responses, the Xi s, follow a Bernoulli distribution with probability parameter π = 0.45. The mean of this distribution is 0.45, and the variance is 0.45 × 0.55.

Therefore, the central limit theorem (CLT) approximation of the sampling distribution of X̄ is:

2 !

0.4975 0.45 × 0.55 √ = N 0.45, .

X̄ ∼ N 0.45, n n (b) When n = 50, the CLT approximation from (a) is X̄ ∼ N (0.45, (0.0704)2 ). With this, we get:

X̄ − 0.45 0.5 − 0.45 P (X̄ > 0.5) = P > = P (Z > 0.71) = 0.2389 0.0704 0.0704 using =1-NORM.S.DIST(0.71,1). (c) Here we need the smallest integer n such that:

√ X̄ − 0.45 0.5 − 0.45 √ > √ P (X̄ > 0.5) = P = P (Z > 0.1005 × n) < 0.01.

0.4975/ n 0.4975/ n Using =NORM.S.INV(0.99), the smallest z such that P (Z > z) < 0.01 is z = 2.33. Therefore, we need:

2 √ 2.33 0.1005 × n ≥ 2.33 ⇔ n ≥ = 537.5 0.1005 which means that we need at least n = 538 likely voters in the sample.

35. Suppose that we plan to take a random sample of size n from a normal distribution with mean µ and standard deviation σ = 2. (a) Suppose µ = 4 and n = 20. i. What is the probability that X̄ is greater than 5? ii. What is the probability that X̄ is less than 3? iii. What is P (|X̄ − µ| ≤ 1) in this case? (b) How large should n be in order that P (|X̄ − µ| ≤ 0.5) ≥ 0.95 for every possible value of µ? (c) It is claimed that the true value of µ is 5 in a population. A random sample of size n = 100 is collected from this population, and the mean for this sample is x̄ = 5.8. Based on the result in (b), what would you conclude from this value of X̄?

11

7. Sampling and sampling distributions

Solution: (a) Denote the random sample {X1 , X2 , . . . , Xn }. We know that the sampling distribution of X̄ is N (µ, σ 2 /n), which here is N (4, 22 /20) = N (4, 0.2). i. The probability we need is:

P (X̄ > 5) = P

X̄ − 4 5−4 √ > √ 0.2 0.2

= P (Z > 2.24) = 0.0126

using =1-NORM.S.DIST(2.24,1). ii. P (X̄ < 3) is obtained similarly. Note that this leads to P (Z < −2.24) = 0.0126, which is equal to P (X̄ > 5) = P (Z > 2.24). This is because 5 is 1 unit above the mean µ = 4, and 3 is 1 unit below the mean, and because the normal distribution is symmetric around its mean.

One way of expressing this is: P (X̄ − µ > 1) = P (X̄ − µ < −1) = 0.0126 for µ = 4. iii. From above, we have: P (X̄ − µ > 1) + P (X̄ − µ < −1) = P (|X̄ − µ| > 1) = 2 × 0.0126 = 0.0252 and, therefore: P (|X̄ − µ| ≤ 1) = 1 − 2 × 0.0126 = 0.9748.

In other words, the probability is 0.9748 that the sample mean is within 1 unit of the true population mean µ = 4. (b) We can use the same ideas as in (a). Since X̄ ∼ N (µ, 4/n) then: P (|X̄ − µ| ≤ 0.5) = 1 − 2 × P (X̄ − µ > 0.5) X̄ − µ 0.5 =1−2×P p >p 4/n 4/n √ = 1 − 2 × P (Z > 0.25 × n)

!

≥ 0.95 which holds if:

√

0.05 = 0.025.

2 √ Using =NORM.S.INV(0.975), we see that this is true when 0.25 × n ≥ 1.96, i.e. when n ≥ (1.96/0.25)2 = 61.5. Rounding up to the nearest integer, we get n ≥ 62. The sample size should be at least 62 for us to be 95% sure that the sample mean will be within 0.5 units of the true mean µ.

P (Z > 0.25 ×

n) ≤

(c) Here n > 62, yet X̄ is further than 0.5 units from the claimed mean of µ = 5. Based on the result in (b), this would be quite unlikely if µ is really 5. One explanation of this apparent contradiction is that µ is not really equal to 5. This kind of reasoning will be the basis of statistical hypothesis testing, which will be discussed in Chapter 10.

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Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 8: Opinion research

ii

Chapter 8 Opinion research 8.1

Exercises

8.1.1

Test your understanding

1. What is the difference between ‘measurement’ and ‘scaling’ ?

Solution: Measurement is the assignment of numbers to characteristics of individuals (or objects) which adhere to prespecified rules that are standardised and applied uniformly over time. Scaling involves creating a continuum.

2. In a survey, is it better to ask a respondent for their age or age group? Briefing justify your view.

Solution: Well, James, it depends. Knowing a respondent’s age is more specific than knowing their age group.

However, some respondents may be too sensitive to reveal their age so the risk of non-response is greater. Revealing your age group is less specific. Saying, for example, that you are in the 18–29 age group means you could be as ‘young’ as 18, or as ‘old’ as 29.

Note if you know someone’s age you can always assign them to an age group. However, if you only know their age group you cannot determine their exact age.

3. When choosing the wording for anchors of a scale, what should you keep in mind?

Solution: Care should be taken over the descriptors used for anchors of a scale. Wording that is too extreme, such as ‘completely agree’ or ‘absolutely agree’ may be off-putting for some respondents, i.e. there can be an aversion to extremist labels, reducing the likely response alternatives chosen by respondents, and hence limiting the variation in the data obtained (a bad thing). Other factors to consider include clarity and cultural sensitivity.

4. When developing an itemised rating scale, outline the arguments for and against: (a) an odd number of points versus an even number of points (b) few points versus many points.

Solution: (a) Odd: allows a neutral option (for), does not force a non-neutral response (against). Even: can force a non-neutral response (for), does not allow a neutral option (against).

1

8. Opinion research

(b) Few: keeps a scale simple (for), restricts the amount of variation in collected data (against).

Many: increases the amount of variation in collected data (for), makes the scale more complicated (against).

5. What are the analytical consequences of working with (a) nominal variables, (b) ordinal variables, and (c) measurable variables?

Solution: The levels of measurement of variables will impact the kinds of statistical methods which can be applied to them. With nominal scale variables, we are extremely limited with the kinds of statistical analysis we can perform, such as frequency counts, percentages and modal categories. For ordinal scale variables, all of the permissible statistical techniques applicable to nominal data extend to ordinal data, but we can also extend it to looking at percentiles, quartiles and also the median, but not the mean. For measurable variables, valid statistical techniques now extend to the mean and standard deviation.

6. Provide business examples of variables of each different measurement level.

Solution: Summary of measurement levels, their properties and business examples: Measurement level Nominal Ordinal Interval Ratio

Main properties Identification and classification Rank-ordered categories Differences between values meaningful, no fixed zero point Fixed zero point, ratios have meaning

Business examples Name, gender product type Rank-order of brands or television channels Attitude scores, indices Market share, sales, income

7. Distinguish between ‘comparative scales’ and ‘non-comparative scales’.

Solution: In a comparative scale there is a direct comparison of stimuli, such as brands. We seek respondents to report their preferences for one brand over another. So it will be a comparison between these stimuli to say which one is strictly preferred to another (the respondent could be indifferent though). There will be a rank order of these stimuli objects giving rise to data with an ordinal level of measurement.

Whereas for a non-comparative scale the brands would be considered by the respondents such that they are rated independently of one another along some scale, giving rise to an interval level of measurement (rather than ratio, as typically there won’t be a fixed zero point). So, from an analytical perspective, since we tend to favour interval or ratio level variables over nominal or ordinal ones, then on that basis alone, we would have a preference for non-comparative scales.

2

8.1. Exercises

8. When comparing n brands using a paired comparison scale, how many distinct paired comparisons are required?

Solution: With n brands there are:

n2 − n n(n − 1) = 2 2

paired comparisons.

9. What is the objective of constant sum scaling?

Solution: Constant sum scaling allows analysts to understand the relative importance or allocation of different attributes or items in the eyes of respondents. Respondents are given a fixed sum of points, and they are asked to distribute these points among various attributes or options based on their perceived importance or preference. This scaling technique allows analysts to capture the trade-offs individuals make when allocating their points among different alternatives, providing insights into their preferences and priorities.

10. What is a Likert scale and when would it be useful? Provide an example of such a scale.

Solution: A Likert scale is commonly used in surveys to measure attitudes, opinions or preferences of respondents towards a particular statement or item. It consists of multiple statements or items, and respondents are asked to indicate their level of agreement or disagreement on a fixed scale, typically ranging from ‘strongly agree’ to ‘strongly disagree’. It provides a structured and quantifiable way to measure subjective constructs, making it easier to analyse and compare responses across respondents.

For example, a company is conducting a customer satisfaction survey to assess the quality of its customer service. They could use a Likert scale to ask customers to rate statements like: • ‘The customer service representatives were helpful.’

• ‘I am satisfied with the resolution of my issue.’

• ‘I would recommend this company to others.’

The Likert scale responses would range from ‘strongly agree’ to ‘strongly disagree’, helping the company gauge the overall satisfaction and identify areas for improvement.

11. What is a semantic differential scale and when would it be useful? Provide an example of such a scale.

Solution: A semantic differential scale measures the perceived meaning of an object, product, or concept based on polar opposite adjectives. Respondents are asked to rate the object or concept on a set of bipolar adjective pairs which represent different dimensions. The scale captures the respondents’ attitudes or feelings toward the subject being rated.

For example, a company wants to evaluate its new product’s brand image among consumers. They could use a semantic differential scale with pairs of adjectives like:

3

8. Opinion research

• ‘modern – outdated’ • ‘reliable – unreliable’ • ‘affordable – expensive’ • ‘innovative – conventional’ Respondents would indicate where they perceive the brand on each dimension, providing insights into how the brand is perceived by consumers and identifying potential areas for brand positioning and improvement.

12. What is a Stapel scale and when would it be useful? Provide an example of such a scale.

Solution: A Stapel scale is used in surveys to measure the direction and intensity of attitudes or opinions towards a particular subject, concept, or product. Unlike other rating scales, the Stapel scale uses a single vertical line with numerical values on one side (positive) and the other side (negative), but there are no specific anchor points. This would be useful when analysts want to quickly capture respondents’ attitudes or opinions on a single dimension without the need for elaborate rating scales. It is often used when survey space is limited and to reduce respondent burden.

For example, a company is conducting a product satisfaction survey for a newly launched mobile app. They could use a Stapel scale to ask users about their overall satisfaction with the app, with values ranging from −5 (strongly dissatisfied) to +5 (strongly satisfied) on a single line. This scale would provide a quick and straightforward way to gauge the app’s general satisfaction level among users.

8.1.2

Case study corner

Suppose you are a marketing executive at the company of your choice. You have been tasked to create a market research survey to investigate consumer satisfaction levels. Create a short survey to fulfil this task, and prepare an executive summary explaining your choice of survey questions.

Suggested approach: Objective: The market research survey aims to investigate consumer satisfaction levels with our company’s products and services and identify areas for improvement.

1. On a scale of 1 to 10, how satisfied are you with our products/services? (Likert scale: 1 = Not at all satisfied, 10 = Extremely satisfied.)

2. What factors influenced your decision to purchase our products/services? (Select all that apply.) (a) Price (b) Quality (c) Customer service (d) Brand reputation (e) Product features (f) Convenience (g) Others (please specify).

4

8.1. Exercises

3. How often do you use our products/services? (Select one.) (a) Daily (b) Weekly (c) Monthly (d) Occasionally (e) Rarely.

4. How likely are you to recommend our products/services to others? (Likert scale: 1 = Not likely at all, 10 = Extremely likely.)

5. Which aspect of our products/services do you like the most? (Free text response.)

6. Which aspect of our products/services needs improvement? (Free text response.)

7. How satisfied are you with our customer service? (Likert scale: 1 = Not at all satisfied, 10 = Extremely satisfied.)

8. How do you usually contact our customer service? (Select one.) (a) Phone (b) Email (c) Live chat (d) In-person visit (e) Others (please specify).

9. Have you faced any issues with our products/services in the past six months? If yes, please describe.

(Free text response.)

10. How would you rate the value for money you receive from our products/services? (Likert scale: 1 = Poor value for money, 10 = Excellent value for money.)

11. How well do our products/services meet your needs and expectations? (Likert scale: 1 = Not at all, 10 = Completely.)

12. Overall, how satisfied are you with your experience with our company? (Likert scale: 1 = Not at all satisfied, 10 = Extremely satisfied.)

Executive Summary The survey includes a combination of rating scale questions and open-ended questions to gain insights into consumer satisfaction levels comprehensively. Questions 1, 4, and 12 use a 10-point Likert scale to gauge overall satisfaction, likelihood to recommend, and overall experience. These questions provide a quantifiable measure of satisfaction.

Questions 2, 5, and 6 allow respondents to share their opinions and provide specific feedback about what influenced their purchase decision, what they like most, and what needs improvement. This qualitative data will offer deeper insights into consumer preferences and pain points.

Questions 3, 7, 8, 9, 10, and 11 assess different aspects of the customer journey, such as frequency of use, satisfaction with customer service, preferred communication channels, product issues, value for money,

5

8. Opinion research

and meeting customer needs and expectations. These questions help identify specific areas of strength and weakness in our products/services and customer support.

By combining quantitative and qualitative data, this survey will provide a comprehensive understanding of consumer satisfaction levels and actionable insights for improving our products/services and enhancing overall customer experience.

6

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 9: Estimation

ii

Chapter 9 Estimation 9.1

Exercises

9.1.1

Test your understanding

1. What is the difference between a point estimate and a point estimator?

Solution: A point estimator is a statistic, that is, a known function of data, used to calculate an estimate of a population parameter. A point estimate is the specific numerical value obtained from the sample data using the chosen estimator.

2. How is the bias of an estimator defined?

Solution: We define the bias of an estimator θb as: b = E(θ) b −θ Bias(θ) that is, as the difference between an estimator’s expected value and the parameter it is estimating.

It can be thought of as the average estimation error. b and θ such that θb is: 3. If θb is an estimator of θ, provide an expression involving E(θ) (a) a positively biased estimator of θ (b) an unbiased estimator of θ (c) a negatively biased estimator of θ.

Solution: (a) A positively biased estimator of θ satisfies: b −θ >0 E(θ)

⇔

b > θ.

E(θ)

⇔

b = θ.

E(θ)

⇔

b < θ.

E(θ)

(b) An unbiased estimator of θ satisfies: b −θ =0 E(θ) (c) A negatively biased estimator of θ satisfies: b −θ <0 E(θ)

1

9. Estimation

4. Would you prefer the variance of an estimator to be large or small? Briefly justify your view.

Solution: Ceteris paribus, the variance of an estimator should be as small as possible, indicating greater precision.

5. What is meant by the bias–variance trade-off when evaluating the performance of estimators?

Solution: Ceteris paribus, we prefer zero bias; ceteris paribus, we prefer low variance. However, in practice, other things are usually never equal, so we typically face a bias–variance trade-off. So, we seek a way to combine bias and variance properties of estimators which will allow us to choose between competing estimators. One such approach is the mean squared error.

6. Provide a formula for the mean squared error of an estimator, and explain the purpose of the mean squared error.

Solution: The mean squared error (MSE) of an estimator is the average squared error. Formally, this is defined as: b = E((θb − θ)2 ) = Var(θ) b + (Bias(θ)) b 2.

MSE(θ) When faced with several competing estimators, we prefer the estimator with the smallest MSE. So, although an unbiased estimator is intuitively appealing, it is perfectly possible that a biased estimator might be preferred if the ‘cost’ of the bias is offset by a substantial reduction in the variance. Hence the MSE provides us with a formal criterion to assess the trade-off between the bias and variance of different estimators.

7. Is unbiasedness a variant or an invariant property? Briefly explain why.

Solution: Unbiasedness is not an invariant property (hence it is a variant property), by which we mean that if θb is an unbiased estimator of θ there are important functions of θb which are not unbiased. For b = θ, it follows that: example, if E(θ) b = E(θb2 ) − (E(θ)) b 2>0 Var(θ)

=⇒

b + θ2 > θ2 .

E(θb2 ) = Var(θ)

Unbiased estimators of functions of parameters, when the parameter is drawn from the population distribution, can, however, often be found by an appropriate adjustment known as a bias correction.

8. What is meant by the ‘coverage probability’ of an interval estimator?

Solution: The coverage probability of an interval estimator refers to the probability that the true population parameter lies within the interval estimated from repeated random samples. For example, if a 95% confidence interval is constructed, the coverage probability is the percentage of such intervals that would contain the true parameter in repeated sampling.

2

9.1. Exercises

9. When sampling from a normal distribution, how are the endpoints of a 95% confidence interval for µ calculated: (a) when σ 2 is known? (b) when σ 2 is unknown?

Solution: √ (a) When σ 2 is known, a 95% confidence interval for µ has endpoints X̄ ± 1.96 × σ/ n. Hence the reported confidence interval would be:

σ σ x̄ − 1.96 × √ , x̄ + 1.96 × √ . n n √ (b) When σ 2 is unknown, a 95% confidence interval for µ has endpoints X̄ ± t0.025, n−1 × S/ n, leading to a reported confidence interval of the form:

S S x̄ − t0.025, n−1 × √ , x̄ + t0.025, n−1 × √ n n where t0.025, n−1 is the t-value which cuts off 2.5% probability in the upper tail of the t distribution with n − 1 degrees of freedom.

10. What is the best confidence level to use in interval estimation? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends! The best confidence level to use in interval estimation depends on the specific context and the level of uncertainty that is acceptable in the decision-making process.

Commonly used confidence levels are 90%, 95%, and 99%.

Higher confidence levels provide wider intervals and more certainty that the true parameter is captured. However, wider intervals may be less precise and may require larger sample sizes. Lower confidence levels provide narrower intervals but carry higher risk of not capturing the true parameter. The choice of confidence level should consider the trade-off between precision and risk, the desired level of confidence in the results, and the available resources for data collection.

Additionally, factors like the variability in the population, prior knowledge about the parameter, and the potential consequences of estimation errors also influence the selection of the appropriate confidence level. Therefore, it’s essential to carefully consider the specific requirements and constraints of each scenario when deciding on the confidence level for interval estimation.

11. Discuss any similarities and any differences between the standard normal distribution and the Student’s t distribution.

Solution: Both the standard normal distribution and Student’s t distribution are bell-shaped and symmetric.

However, the main difference is that the t distribution has heavier tails, making it more robust for small sample sizes. Additionally, the standard normal distribution has a mean of 0 and variance of 1, while for the t distribution the existence of the mean, and value of the variance, depend on the degrees of freedom.

3

9. Estimation

12. How are the endpoints of a 100(1 − α)% confidence interval for a single population proportion calculated?

Solution: An approximate 100(1 − α)% confidence interval for a single proportion has endpoints: r p ± zα/2 ×

p(1 − p) n

r ⇒

p − zα/2 ×

p(1 − p) , p + zα/2 × n

r

p(1 − p) n

!

where zα/2 is the z-value which cuts off 100α/2% probability in the upper tail of the standard normal distribution, obtained using NORM.S.INV.

13. What three factors affect the minimum required sample size when estimating: (a) a single population mean? (b) a single population proportion?

Solution: (a) The three factors are: • the confidence level, which affects the zα/2 value (under the control of the analyst) • the tolerance on the estimation error, e (under the control of the analyst) • the variance in the population, σ 2 (not under the control of the analyst). (b) The three factors are: • the confidence level, which affects the zα/2 value (under the control of the analyst) • the tolerance on the estimation error, e (under the control of the analyst) • the variance in the population, which influences the value of p (not under the control of the analyst).

14. How are the endpoints of a 100(1 − α)% confidence interval for the difference between two population proportions calculated?

Solution: With point estimates for π1 and π2 of p1 = r1 /n1 and p2 = r2 /n2 , respectively, an approximate 100(1 − α)% confidence interval for the difference between two population proportions has endpoints: s p1 (1 − p1 ) p2 (1 − p2 ) p1 − p2 ± zα/2 × + n1 n2 where zα/2 is the z-value which cuts off 100α/2% probability in the upper tail of the standard normal distribution, obtained using NORM.S.INV.

4

9.1. Exercises

15. What is the purpose of the pooled variance estimator? How is it calculated?

Solution: The pooled variance estimator is used when we assume a common population variance for two normally-distributed populations. The pooled variance estimator, where S12 and S22 are sample variances from samples of size n1 and n2 , respectively, is: Sp2 =

(n1 − 1)S12 + (n2 − 1)S22 n1 + n2 − 2

on n1 + n2 − 2 degrees of freedom, where the subscript ‘p’ denotes ‘pooled’.

16. How are the endpoints of a 100(1 − α)% confidence interval for the difference between two population means calculated: (a) when σ12 and σ22 are known? (b) when σ12 and σ22 unknown? (c) when working with paired (dependent) samples?

Solution: (a) If the population variances σ12 and σ22 are known, a 100(1 − α)% confidence interval for µ1 − µ2 has endpoints: s x̄1 − x̄2 ± zα/2 ×

σ12 σ22 + n1 n2

where zα/2 is the z-value which cuts off 100α/2% probability in the upper tail of the standard normal distribution, obtained using NORM.S.INV. (b) If the population variances σ12 and σ22 are unknown, provided sample sizes n1 and n2 are large (greater than 30), an approximate 100(1 − α)% confidence interval for µ1 − µ2 has endpoints: s s2 s21 + 2 x̄1 − x̄2 ± zα/2 × n1 n2 where zα/2 is the z-value which cuts off 100α/2% probability in the upper tail of the standard normal distribution, obtained using NORM.S.INV. If the population variances σ12 and σ22 are unknown but assumed equal, a 100(1 − α)% confidence interval for µ1 − µ2 has endpoints: s

1 1 2 x̄1 − x̄2 ± tα/2, n1 +n2 −2 × sp + n1 n2 where s2p is the estimate from the pooled variance estimator (9.10), and where tα/2, n1 +n2 −2 is the t-value which cuts off 100α/2% probability in the upper tail of the Student’s t distribution with n1 + n2 − 2 degrees of freedom, obtained using T.INV. (c) When working with paired (dependent) samples, we use the t distribution when constructing a 100(1 − α)% confidence interval with endpoints:

sd sd s x̄d ± tα/2, n−1 × √ ⇒ x̄d − tα/2, n−1 × √ , x̄ + tα/2, n−1 × √ n n n

5

9. Estimation

where tα/2, n−1 is the t-value which cuts off 100α/2% probability in the upper tail of the t distribution with n − 1 degrees of freedom, obtained from T.INV. Note that x̄d and sd are the sample mean and sample standard deviation, respectively, of the differenced data.

17. A business requires an inexpensive check on the value of stock in its warehouse. In order to do this, a random sample of 50 items is taken and valued. The average value of these is computed to be £320.41, with a (sample) standard deviation of £40.60. It is known that there are 9875 items in the total stock. (a) Estimate the total value of the stock to the nearest £10,000. (b) Calculate a 95% confidence interval for the mean value of all items and hence determine a 95% confidence interval for the total value of the stock. (c) You are told this confidence interval is too wide for decision purposes. Assess how many more items would need to be sampled to obtain an interval with the same degree of confidence, but with half the width.

Solution: (a) The total value of the stock is 9875µ, where µ is the mean value of an item of stock. We know that X̄ is the obvious estimator of µ, so 9875X̄ is the obvious estimator of 9875µ. Therefore, an estimate of the total value of the stock is 9875 × 320.41 = £3,160,000 (to the nearest £10,000). (b) In this question n = 50 is large and σ 2 is unknown, so a 95% confidence interval for µ is: 40.6 s x̄ ± 1.96 × √ = 320.41 ± 1.96 × √ = 320.41 ± 11.25 n 50

⇒

(£309.16, £331.66).

Note that because n is large we have used the standard normal approximation. It is more accurate to use a t distribution on 49 degrees of freedom. This gives a 95% confidence interval of (£308.87, £331.95) – not much of a difference.

To obtain a 95% confidence interval for the total value of the stock, 9875µ, multiply the interval endpoints by 9875. This gives (to the nearest £10,000): (£3,050,000, £3,280,000). (c) Increasing√the sample size by a factor of k reduces the width of the confidence interval by a factor of k. Hence increasing the sample √ size by a factor of 4 will reduce the width of the confidence interval by a factor of 2 (= 4). Therefore, we need to increase the sample size from 50 to 4 × 50 = 200, i.e. collect another 150 observations.

18. Two advertising companies each give quotations for nine different campaigns. Their quotations (in thousands of pounds) are shown in the following table. Calculate a 95% confidence interval for the true difference between average quotations. Can you deduce from this interval if one company is more expensive on average than the other?

Company A B

6

1 39 46

2 24 26

3 36 32

4 42 39

5 45 51

6 30 34

7 38 37

8 32 41

9 39 44

9.1. Exercises

Solution: The data are paired observations for n = 9 campaigns. First look at the differences: 7,

2,

−4,

−3,

6,

4,

−1,

9 and 5.

The sample mean of these differences is 2.78 and the sample standard deviation is 4.58. (Make sure you know how to do these calculations!) Therefore, a 95% confidence interval for the true difference, µd , is: 4.58 2.78 ± 2.306 × √ ⇒ (−0.74, 6.30) 9 using =T.INV(0.975,8). Zero is in this confidence interval so we cannot conclude that one company is more expensive than the other on average. It is possible that the differences in quotations are due to random variation.

19. In order to assess the impact of an advertising campaign, a restaurateur monitors his daily revenue before and after the campaign. The table below shows some sample statistics of daily sales calculated over a period of 60 days prior to the campaign, and 45 days after the campaign.

Determine a 95% confidence interval for the increase in average daily sales due to the campaign. Is there strong evidence that the campaign has increased sales?

Number of days Mean daily sales Standard deviation

Before campaign 60 £503 £21

After campaign 45 £559 £29

Solution: As both sample sizes are quite large there is no need to use a pooled estimator of the variance. A 95% confidence interval for the difference in means is: r (21)2 (29)2 559 − 503 ± 1.96 × + = 56 ± 10.0 ⇒ (46.0, 66.0).

60 45 Zero is well below the interval lower endpoint, so there is strong evidence that the campaign has increased sales.

20. In a market research study to compare two chocolate bar wrappings, 30 out of 50 children preferred a gold wrapping to a silver wrapping, whereas 25 out of 40 adults preferred the silver wrapping.

Compute a 95% confidence interval for the difference in the true proportions in favour of gold wrapping between the two groups.

It is decided to take further samples of children and adults so that we finally sample the same number of adults and children such that the final estimator of the difference between proportions is within 0.06 of the true difference (with 95% confidence). How many more adults and children need to be sampled?

Solution: For ‘Children − Adults’, the estimate of the difference in population proportions is 30/50 − 15/40 = 0.225. The estimated standard error of this estimate is: r 0.6 × 0.4 0.375 × 0.625 + = 0.1032.

50 40

7

9. Estimation

A 95% confidence interval is 0.225 ± 1.96 × 0.1032, giving (0.023, 0.427).

Let n be the required common sample size. We require: r 0.6 × 0.4 + 0.375 × 0.625 .

0.06 = 1.96 × n Solving, gives n = 507. Hence we must sample 457 more children and 467 more adults.

21. A factory has 1000 workers. A simple random sample of 100 of these had weekly salaries with a (sample) mean of £315 and a (sample) standard deviation of £20. (a) Calculate a 90% confidence interval for the mean weekly salary of all workers in the factory. (b) How many more workers should be sampled if it is required that the estimate is to be within £3 of the true average (again, with 90% confidence)?

Note that this means a tolerance of £3 – equivalent to a confidence interval width of £6.

Solution: (a) We have n = 100, x̄ = 315 and s = 20. Also, we are sampling 10% of the population and so it is necessary to use the finite population correction factor (see Chapter 7). The estimated standard error is: r r n 20 100 s √ × 1− =√ = 1.897. × 1− N 1,000 n 100 Hence a 90% confidence interval for the true mean has endpoints 315 ± 1.645 × 1.897, giving (311.9, 318.1). (b) For the tolerance to be 3, we require: 20 3 ≥ 1.645 × √ × n

r 1−

n .

1,000

Solving this gives n ≥ 107.36, so we round up to get n = 108. Hence we need to take a further sample of 8 workers.

22. A company recently redesigned its website to increase its appeal to younger consumers (those aged under 30). Random samples of younger consumers were surveyed before and after the redesign to see if they had a positive opinion of the website. The results were as follows: Sample size Before redesign After redesign

125 150

Respondents with a positive opinion 75 111

(a) Compute a 98% confidence interval for the difference between the proportions of younger consumers with a positive view before and after the website redesign. Briefly interpret your confidence interval. (b) Briefly discuss any concerns had the sample sizes been 25 and 30, instead of 125 and 150, respectively.

8

9.1. Exercises

Solution: (a) Since p2 = 0.74 and p1 = 0.60, we use: s p2 − p1 ± z0.01 ×

p1 (1 − p1 ) p2 (1 − p2 ) + n1 n2

giving us: r 0.74 − 0.60 ± 2.326 ×

0.60 × 0.40 0.74 × 0.26 + 125 150

⇒

(0.0084, 0.2716).

Interpretation: We are 98% confident that the increase in positive opinion of the website redesign among consumers under 30 is between 0.8% and 27.16%. (b) A discussion expected about the applicability of the central limit theorem approximation to normality with ‘small’ sample sizes of 25 and 30 relative the the ‘large’ sample sizes of 125 and 150.

23. A pharmaceutical company has developed a new vaccine to immunise against Covid-19 and needs to determine the vaccine’s efficacy. A randomised experiment was conducted to address this question.

The study randomly allocated 200 people to either a group where the vaccine was administered, or a group where a placebo was given. These people were monitored and the numbers of those who did not go on to contract Covid-19 (and those who did) were recorded. The results are summarised below:

Vaccine group Placebo group

Developed Covid-19 21 48

Did not develop Covid-19 79 52

(a) Provide a 90% confidence interval for the difference in the proportions of those who did not develop Covid-19 between those who took the vaccine and the placebo group. (b) Explain, with reasons, whether 200 people in this study are a sufficient number.

Solution: (a) Since p1 = 79/(21 + 79) = 0.79 and p2 = 52/(48 + 52) = 0.52, we use: s p1 (1 − p1 ) p2 (1 − p2 ) p1 − p2 ± z0.05 × + n1 n2 giving us: r 0.79 − 0.52 ± 1.645 ×

0.79 × 0.21 0.52 × 0.48 + 100 100

⇒

(0.1640, 0.3760).

(b) A discussion expected about the relationship between sample size accuracy of estimates.

Clinical trials are likely to be based on larger sample sizes in practice as these are often matters of life and death!

9

9. Estimation

24. A random sample is drawn from a normal distribution, N (µ, 36). You are told that a 90% confidence interval for the population mean is (6.78, 8.22). What was the size of the sample?

Solution: The sample mean, x̄, would be in the centre of the interval (6.78, 8.22). Adding the two endpoints and dividing by 2 gives: 6.78 + 8.22 = 7.5. x̄ = 2 To find the sample size, note that the standard error when estimating a single mean is: σ 6 √ =√ . n n Since this is a 100(1 − α)% = 90% confidence interval, then α = 0.10, so the confidence coefficient is zα/2 = z0.05 = 1.645 using =NORM.S.INV(0.95). Therefore, to determine n we need to solve: 6 1.645 × √ = 0.72. n The correct sample size is n ≈ 188 (depending on rounding).

25. Based on the central limit theorem, you are told that a 99% confidence interval for a population proportion is (0.5782, 0.7018). What was the size of the sample used?

Solution: The sample proportion, p, would be in the centre of the interval (0.5782; 0.7018). Adding the two endpoints and dividing by 2 gives: p=

0.5782 + 0.7018 = 0.64.

2

To find the sample size, note that the (estimated) standard error when estimating a single proportion is: r r √ p(1 − p) 0.64 × 0.36 0.64 × 0.36 0.48 √ = = = √ . n n n n Since this is a 100(1 − α)% = 99% confidence interval, then α = 0.01, so the confidence coefficient is zα/2 = z0.005 = 2.576, using =NORM.S.INV(0.995). Therefore, to determine n we need to solve: 0.48 2.576 × √ = 0.0618. n The correct sample size is n ≈ 401 (depending on rounding).

26. Suppose a coin is to be tossed n times for the purpose of estimating π, where π = P (heads). (a) State the approximate sampling distribution of P , the sample proportion, and briefly state the condition on n which should be satisfied for the approximation to be reasonable. (b) How large must n be to guarantee that the width of the 99% confidence interval for π will be less than 0.02?

10

9.1. Exercises

Solution: (a) By the central limit theorem:

π(1 − π) σ2 = N π, P ∼ N µ, n n approximately, as n → ∞. (b) An approximate 99% confidence interval for π is: r p ± 2.576 ×

p(1 − p) n

in practice. Maximum value of p(1 − p) = p − p2 is: 1 − 2p = 0 Hence:

⇒ r

2.576 × Therefore: n≥

p = 0.5

⇒

p(1 − p) ≤ 2.576 × n

p(1 − p) ≤ 0.25. r

1 ≤ 0.01.

4n

(2.576)2 = 16,589.44, i.e. n ≥ 16,590.

4 × (0.01)2

27. A company wishes to investigate whether average brand awareness varies between older and younger people. Random samples of 15 adults over the age of 60 and 14 adults aged between 18 and 30 scored their brand awareness on a scale from 1 to 100. Summaries of these scores are presented below.

Over 60s 18–30s

Sample size 15 14

Sample mean 68.7 74.3

Sample variance 15.4 18.2

(a) Construct a 99% confidence interval for the difference between mean brand awareness scores and explain in one sentence how the confidence interval should be interpreted. (b) Is it possible that there is no difference between older and younger adults in terms of their average brand awareness? Briefly justify your view.

Solution: (a) For a 99% confidence interval, we use t0.005, 27 = 2.771. The confidence interval formula is then: s

1 1 x̄2 − x̄1 ± t0.005, 27 × s2p + n1 n2 giving us: s 74.3 − 68.7 ± 2.771 ×

16.75 ×

1 1 + 15 14

⇒

(1.386, 9.814).

We are 99% confident that µ2 − µ1 falls in this interval.

11

9. Estimation

(b) Although zero is not included in the confidence interval, the interval is very wide and it is based on small sample sizes, so it is not impossible that there is no difference. Resampling using larger sample sizes would be recommended.

28. Two market research companies each take random samples to assess the public’s attitude towards a particular clothing brand. If n represents the sample size and r the number of people with a positive view, the results of these independent surveys are as follows:

Company 1 Company 2

n 400 900

r 160 324

(a) If π is the population proportion with a positive view of the clothing brand, determine a 95% confidence interval to comment on the compatibility of the two companies’ results. (b) An observer decides to estimate π using both datasets and uses the estimator:

R2 R1 +b a n1 n2 where a and b are constants. Show that we must have b = 1 − a for this estimator to be unbiased and, subject to this, find a value of a which will make this estimator the most efficient (i.e. with the minimum standard error).

Solution: (a) Here we are estimating the difference between two proportions (if the two companies’ results are compatible, then there should be no difference between the proportions). As the sample sizes are large, the formula for a 95% confidence interval of the difference is: s p1 (1 − p1 ) p2 (1 − p2 ) + . p1 − p2 ± 1.96 × n1 n2 The estimates of the proportions are p1 = 160/400 = 0.4 and p2 = 324/900 = 0.36, so a 95% confidence interval is: r 0.4 × 0.6 0.36 × 0.64 0.4 − 0.36 ± 1.96 × + = 0.04 ± 0.057 ⇒ (−0.017, 0.097).

400 900 As zero is in this interval, it is plausible that the two proportions are the same, so the companies’ surveys are compatible. (b) Now:

R1 R2 E a +b = aπ + bπ = (a + b)π. n1 n2 If this is to be equal to π (for unbiasedness) we must have a + b = 1, i.e. we have b = 1 − a. The variance of this estimator is:

R1 R2 π(1 − π) π(1 − π) Var a +b = a2 + (1 − a)2 . n1 n2 400 900

12

9.1. Exercises

By differentiating with respect to a and equating to zero we obtain (after some manipulation): a=

4 = 0.3077.

13

Note this question is an example of how a confidence interval (here, with 95% confidence) can be used to conduct an informal hypothesis test (two-tailed, with a 5% significance level). We would not reject a null hypothesis which stated that there was no difference. Hypothesis testing will be covered in Chapter 10.

29. Twelve independent observations are sampled from a population known to be normal, N (µX , 9σ 2 ).

For these, the sample mean is 12.6 and the sample standard deviation is 1.1.

Sixteen independent observations are sampled from a population known to be normal, N (µY , 8σ 2 ).

For these, the sample mean is 10.1 and the sample standard deviation is 1.8.

Determine a 95% confidence interval for µX − µY .

Solution: There are two important points to note about this question.

• Both the populations have (we are told) normal distributions, and hence the sampling distributions for the sample means are themselves normal (not just approximately normal).

• The population variances are (we are told) unequal, but they are both (different) multiples of some common variance, σ 2 , so we can estimate this value using a pooled variance technique, and then use this estimate to compute estimates of the variances for the population distributions, and hence for the sampling distributions.

The sampling distributions of X̄ and Ȳ are:

9σ 2 X̄ ∼ N µX , 12

8σ 2 and Ȳ ∼ N µY , 16

so the sampling distribution of X̄ − Ȳ is:

9σ 2 8σ 2 X̄ − Ȳ ∼ N µX − µY , + 12 16

3 1 2 + = N µX − µY , σ 4 2

5σ 2 = N µX − µY , .

4 2 Now SX is an unbiased estimator of Var(X) = 9σ 2 . Similarly, SY2 is an unbiased estimator of Var(Y ) = 8σ 2 .

2 Hence WX2 = SX /9 is an unbiased estimator of σ 2 , as is WY2 = SY2 /8.

Since the sample sizes are small (12 + 16 = 28 < 40) and the population variances are unknown, we use the standard pooling formula, but with WX2 and WY2 , to estimate σ 2 by: s2P =

2 (12 − 1)(1.1)2 /9 + (16 − 1)(1.8)2 /8 (nX − 1)wX + (nY − 1)wY2 = = 0.2905. nX + nY − 2 12 + 16 − 2

13

9. Estimation

So, noting that this is on 26 degrees of freedom, we find a 95% confidence interval for µX − µY is given by: r 5 ⇒ (1.261, 3.739).

12.6 − 10.1 ± 2.056 × 0.2905 × 4 using =T.INV(0.975,26).

30. Ten independent observations are sampled from a normal population, N (µX , 4σ 2 ). For these, the sample mean is 12.3 and the sample standard deviation is 1.2.

Fourteen independent observations are sampled from a normal population, N (µY , 8σ 2 ). For these, the sample mean is 10.1 and the sample standard deviation is 1.9.

Determine a 95% confidence interval for µX − µY .

Solution: The sampling distribution of X̄ − Ȳ is N (µX − µY , σ 2 (4/10 + 8/14)). Since we have two unbiased 2 estimators of σ 2 , SX /4 and SY2 /8, we can estimate σ 2 with a pooled estimate: (10 − 1)(1.2)2 /4 + (14 − 1)(1.9)2 /8 = 0.4139 (10 − 1) + (14 − 1) which has 22 degrees of freedom. A 95% confidence interval for µX − µY has endpoints: √ 12.3 − 10.1 ± 2.074 × 0.9714 × 0.4139 giving (0.885, 3.515), using =T.INV(0.975,22).

14

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 10: Hypothesis testing

ii

Chapter 10 Hypothesis testing 10.1

Exercises

10.1.1

Test your understanding

1. What is the difference between a null hypothesis and an alternative hypothesis?

Solution: A null hypothesis states that there is no significant difference or effect, while an alternative hypothesis suggests the presence of a significant effect. For example, the null hypothesis could be that a new marketing campaign has no impact on sales, and the alternative hypothesis could be that the campaign increases sales.

2. In hypothesis testing, we make a binary decision. What are the two possible decisions?

Solution: We either ‘reject H0 ’ or we do ‘not reject H0 ’. Note we never ‘accept’ a hypothesis!

3. If you served on a jury, how would you decide whether to find the defendant guilty or not guilty?

Solution: You should be objective and initially assume the null hypothesis (that the defendant is not guilty) is true. You should then listen to all the evidence presented in court and judge (as objectively as possible) how consistent the evidence is with respect to the defendant being not guilty. You should find the defendant ‘not guilty’ unless the evidence was sufficiently inconsistent with the null hypothesis ‘beyond a reasonable doubt’. Of course, jurors’ interpretations of what a ‘reasonable doubt’ is will likely differ!

4. What is a Type I error? What is a Type II error?

Solution: A Type I error, also known as a false positive, occurs when a null hypothesis which is true is incorrectly rejected. In other words, it is the incorrect conclusion that there is a significant effect or difference when there is none in the population.

A Type II error, also known as a false negative, occurs when a null hypothesis that is false is incorrectly not rejected. In other words, it is the incorrect conclusion that there is no significant effect or difference when there actually is one in the population.

1

10. Hypothesis testing

5. Which is worse, a Type I error or a Type II error? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends! Whether a Type I error or a Type II error is considered worse depends on the specific context and the associated costs or consequences of each error.

Consider a pharmaceutical company conducting clinical trials for a new drug. A Type I error occurs if the company incorrectly concludes that the drug is effective when it’s not. This could lead to significant costs and time invested in further developing and marketing a drug that ultimately fails to provide the expected benefits. Additionally, it may pose serious health risks to patients if an ineffective or unsafe drug is brought to the market.

Suppose a manufacturing company implements quality control checks on its products. A Type II error occurs when the quality control process fails to identify defective items. If these defective products are released to the market, it can lead to customer complaints, product recalls, financial losses, and damage to the company’s reputation.

6. In a hypothesis testing context, what is α and what is β?

Solution: α and β are both conditional probabilities. Specifically: P (H0 rejected | H0 is true) = α and P (H0 not rejected | H1 is true) = β.

7. What is ‘power’ in a hypothesis testing context and is it desirable?

Solution: Power is extremely desirable in hypothesis testing. It represents the probability of correctly rejecting a false null hypothesis. Specifically: Power = P (H0 rejected | H1 is true) = 1 − β.

Ceteris paribus, increasing the sample size will increase the power of the test. However, this would increase data collection costs.

8. Write out, in matrix form, the decision space in hypothesis testing.

Solution: We have:

True state of nature

2

H0 true H1 true

Decision made H0 not rejected H0 rejected Correct decision Type I error Type II error Correct decision

10.1. Exercises

9. What are common values for the significance level in hypothesis testing, and how do these affect the reported strength of evidence?

Solution: Common values for the significance level in hypothesis testing are α = 0.05 (a 5% significance level) and 0.01 (a 1% significance level). The lower the significance level, the stronger the evidence required to reject the null hypothesis, making it less likely to reject it. A higher significance level leads to a higher chance of rejecting the null hypothesis – whether correctly or incorrectly!

10. Define the p-value.

Solution: A p-value is the probability of the event that the ‘test statistic’ (a known function of our data) takes the observed value or more extreme (i.e. more unlikely) values under H0 . It is a measure of the discrepancy between the null hypothesis, H0 , and the data evidence.

• A ‘small’ p-value indicates that the data are inconsistent with H0 .

• A ‘large’ p-value indicates that the data are consistent with H0 .

11. Can we ever ‘accept’ a hypothesis? Briefly justify your view.

Solution: In hypothesis testing, we never technically ‘accept’ a hypothesis. Instead, we either reject the null hypothesis or fail to reject it based on the evidence from the sample data. Why? Well, hypothesis testing is based on probability and statistical inference. Even if we fail to reject the null hypothesis, it does not necessarily mean the null hypothesis is true; it only means we do not have sufficient evidence to reject it with the given sample data. The possibility of making a Type II error (failing to reject a false null hypothesis) also adds to the caution in using the term ‘accept’. Instead, we interpret the results as either supporting or not supporting the null hypothesis based on the available evidence.

12. What is the hypothesis testing decision rule using p-values?

Solution: When testing at the 100α% significance level, for α ∈ [0, 1], then: ( ≤ α then reject H0 if the p-value > α then do not reject H0 .

13. What is an effect size and how does it influence the p-value?

Solution: The effect size reflects the difference between what you would expect to observe if the null hypothesis is true and what is actually observed in a random experiment. Equality between our expectation and observation would equate to a zero effect size, which (while not proof that H0 is true) provides the most convincing evidence in favour of H0 – for example, if x̄ = µ0 . As the difference between our expectation and observation increases, the data evidence becomes increasingly inconsistent with H0 making us more likely to reject H0 . Hence as the effect size gets

3

10. Hypothesis testing

larger, the p-value gets smaller (and so is more likely to be below α, which would result in rejection of H0 ).

14. What is the sample size’s influence on the p-value?

Solution: For a non-zero effect size, the p-value decreases as the sample size increases. (A zero effect size would result in non-rejection of H0 , regardless of n.) Hence there is an inverse relationship between the sample size and the p-value.

15. Does statistical significance imply practical significance?

Solution: Statistically significant results are not necessarily significant in terms of being of practical importance. They might be significant only in the statistical sense. This is the case when your response to the results of the study is ‘who cares?’ or ‘so what?’.

There is always a possibility of statistical significance, but not practical significance with large sample sizes. On the contrary, with smaller samples, results may not be statistically significant even if they represent the truth about the population(s).

16. When sampling from N (µ, σ 2 ), what is the procedure for performing a hypothesis test for µ when: (a) σ 2 is known? (b) σ 2 is unknown?

Solution: (a) When testing H0 : µ = µ0 with σ 2 known, the test statistic is: Z=

X̄ − µ0 √ ∼ N (0, 1). σ/ n

Hence p-values are obtained from the standard normal distribution, i.e. using the NORM.S.DIST function in Excel. (b) When testing H0 : µ = µ0 with σ 2 unknown, the test statistic is: T =

X̄ − µ0 √ ∼ tn−1 .

S/ n

Hence p-values are obtained from the Student’s t distribution with n − 1 degrees of freedom, i.e. using the T.DIST function in Excel.

17. What is the procedure for performing a hypothesis test for a single proportion, π?

Solution: When testing H0 : π = π0 , the test statistic is: Z=p

P − π0 π0 (1 − π0 )/n

∼ N (0, 1) (approximately, for large n).

Validity requires nπ0 > 5 and n(1 − π0 ) > 5. Hence p-values are obtained from the standard normal distribution, i.e. using the NORM.S.DIST function in Excel.

4

10.1. Exercises

18. What is the procedure for performing a hypothesis test for the difference between: (a) two proportions? (b) two means? (Report all cases.)

Solution: (a) When testing H0 : π1 = π2 , the test statistic is: Z=p

P1 − P2 − (π1 − π2 ) P (1 − P )(1/n1 + 1/n2 )

∼ N (0, 1) (approximately, for large n1 and n2 )

where the pooled proportion estimator is: P =

R1 + R2 . n1 + n2

Hence p-values are obtained from the standard normal distribution, i.e. using the NORM.S.DIST function in Excel. (b) When testing H0 : µ1 = µ2 we have the following four cases. i. Unpaired samples – variances known. In this case, the test statistic is: Z=

X̄1 − X̄2 − (µ1 − µ2 ) p ∼ N (0, 1). σ12 /n1 + σ22 /n2

Hence p-values are obtained from the standard normal distribution, i.e. using NORM.S.DIST in Excel. ii. Unpaired samples – variances unknown and unequal. If the population variances σ12 and σ22 are unknown, provided sample sizes n1 and n2 are large (greater than 30): Z=

X̄1 − X̄2 − (µ1 − µ2 ) p ∼ N (0, 1) (approximately, for large n1 and n2 ).

S12 /n1 + S22 /n2

Hence p-values are obtained from the standard normal distribution, i.e. using NORM.S.DIST in Excel. iii. Unpaired samples – variances unknown and equal. If the population variances σ12 and σ22 are unknown but assumed equal: X̄1 − X̄2 − (µ1 − µ2 ) ∼ tn1 +n2 −2 T = q Sp2 (1/n1 + 1/n2 ) where Sp2 is the pooled variance estimator. Hence p-values are obtained from the Student’s t distribution with n1 + n2 − 2 degrees of freedom, i.e. using T.DIST in Excel. iv. Paired (dependent) samples. Using the sample mean and sample standard deviation of differenced data: X̄d − µd √ ∼ tn−1 .

T = Sd / n Hence p-values are obtained from the Student’s t distribution with n − 1 degrees of freedom, i.e. using T.DIST in Excel.

5

10. Hypothesis testing

19. Can a hypothesis test which is not significant at the 5% significance level be significant at the 10% significance level?

Solution: Yes, for any p-value in the interval (0.05, 0.10). For example, if the p-value is 0.07, then according to the p-value decision rule we would reject H0 at the 10% significance level, but not reject it at the 5% significance level.

20. When testing H0 : µ = 4 versus H1 : µ > 4, explain why an observed sample mean of x̄ = 3 would not result in the rejection of H0 . (Answer without conducting a full hypothesis test.)

Solution: It is an upper-tailed test with a negative test statistic value (since x̄ < µ0 because 3 < 4) hence the test will not be significant as the p-value will be greater than 0.50.

21. A random sample of size n = 8 is drawn from a normal distribution with an unknown mean µ and an unknown variance σ 2 < ∞. It is desired to test: H0 : µ = −3 versus H1 : µ 6= −3.

Suppose that the sample data are such that: x̄ = −4.74 and s2 = 2.69.

Conduct the hypothesis test by calculating the p-value.

Solution: Under H0 , the test statistic is (note the double negative!): T = The test statistic value is: t=

X̄ + 3 X̄ − (−3) √ √ ∼ t7 . = S/ 8 S/ 8

x̄ + 3 −4.74 + 3 √ = p = −3.001. s/ n 2.69/8

Using =2*T.DIST(-3.001,7,1), the p-value is 0.0199, hence we reject H0 at the 5%, but not 1%, significance level. Hence the test is moderately significant, i.e. there is moderate evidence that µ 6= −3.

22. Suppose H0 : µ = 240 is tested against H1 : µ < 240 at the 1% significance level using a random sample of n = 25 observations from N (µ, (50)2 ). What is the probability of a Type II error if µ = 220?

Solution: Using =NORM.S.INV(0.01), H0 should be rejected if: z=

6

x̄ − 240 √ ≤ −2.326 50/ 25

10.1. Exercises

or, equivalently, if: 50 x̄ ≤ 240 − 2.326 × √ = 216.74.

25 If µ = 220, the Type II error probability is:

216.74 − 220 √ = P (Z > −0.33) = 0.6293 P (X̄ > 216.74 | µ = 220) = P Z > 50/ 25 using =1-NORM.S.DIST(-0.33,1).

23. A paired-difference experiment under two conditions used a random sample of n = 61 adults and reported a sample mean difference of 2.326 and a standard deviation of 7.6 for the differences. The researchers reported a t statistic value of 2.390, when testing whether the means of the two conditions are the same. (a) Show how the researchers obtained the t statistic value of 2.390. (b) Calculate the p-value of the test and use the p-value to draw a conclusion about the significance of the test. Use a 5% significance level.

Solution: (a) The test statistic value is: 2.326 x̄d √ = √ = 2.390. sd / n 7.6/ 61 (b) Under H0 : µd = 0, the test statistic is: T =

X̄d − µd √ ∼ tn−1 Sd / n

⇒

T =

X̄d √ ∼ t60 .

Sd / 61

The p-value is, where T ∼ t60 , using =2*T.DIST(-2.39,60,1): 2 × P (T ≥ 2.390) = 2 × 0.01 = 0.02.

Since 0.02 < 0.05, the test is (moderately) significant at the 5% significance level.

24. Consider the test of H0 : µ = 60 versus H1 : µ 6= 60 using a 5% significance level. If the data consist of a random sample of size 16 from a normal distribution having σ = 4, determine the power of the test when: (a) µ = 57 (b) µ = 61.

Recall that the power of a test is the probability that the test will reject a false null hypothesis.

Solution: For n = 16, σ = 4 and α = 0.05, using =NORM.S.INV(0.025) H0 : µ = 60 should be rejected in favour of a two-sided H1 if: x̄ − 60 √ < −1.96 or 4/ 16

x̄ − 60 √ > 1.96.

4/ 16

7

10. Hypothesis testing

Rearranging, we reject H0 if: 4 4 x̄ < −1.96 × √ + 60 = 58.04 or x̄ > 1.96 × √ + 60 = 61.96.

16 16 Conditional on an arbitrary µ, this equates to a power of: P (X̄ ≤ 58.04 | µ) + P (X̄ ≥ 61.96 | µ).

(a) For µ = 57, we calculate:

61.96 − 57 58.04 − 57 √ √ +P Z ≥ = P (Z ≤ 1.04) + P (Z ≥ 4.96) = 0.8508 P Z≤ 4/ 16 4/ 16 using =NORM.S.DIST(1.04,1)+(1-NORM.S.DIST(4.96,1)). Hence the power is 0.8508. (b) For µ = 61, we calculate:

61.96 − 61 58.04 − 61 √ √ +P Z ≥ = P (Z ≤ −2.96) + P (Z ≥ 0.96) = 0.1701 P Z≤ 4/ 16 4/ 16 using =NORM.S.DIST(-2.96,1)+(1-NORM.S.DIST(0.96,1)). Hence the power is 0.1701.

25. In a wire-based nail manufacturing process the target length for cut wire is 22 cm. It is known that widths vary with a standard deviation equal to 0.08 cm. In order to monitor this process, a random sample of 50 separate wires is accurately measured and the process is regarded as operating satisfactorily (the null hypothesis) if the sample mean width lies between 21.97 cm and 22.03 cm, so that this is the decision procedure used (i.e. if the sample mean falls within this range then the null hypothesis is not rejected, otherwise the null hypothesis is rejected). (a) Determine the probability of a Type I error for this test. (b) Determine the probability of making a Type II error when the process is actually cutting to a length of 22.05 cm. (c) Find the probability of rejecting the null hypothesis when the true cutting length is 22.01 cm.

(This is the power of the test when the true mean is 22.01 cm.)

Solution: Each is a conditional probability. (a) Conditioning on µ = 22, we have: α = 1 − P (21.97 < X̄ < 22.03 | µ = 22)

22.03 − 22 21.97 − 22 √ √ <Z< =1−P 0.08/ 50 0.08/ 50 = 1 − P (−2.65 < Z < 2.65) = 0.008 using =1-(NORM.S.DIST(2.65,1)-NORM.S.DIST(-2.65,1)).

8

10.1. Exercises

(b) Conditioning on µ = 22.05, we have: β = P (21.97 < X̄ < 22.03 | µ = 22.05)

21.97 − 22.05 22.03 − 22.05 √ √ =P <Z< 0.08/ 50 0.08/ 50 = P (−7.07 < Z < −1.77) = 0.0384 using =NORM.S.DIST(-1.77,1)-NORM.S.DIST(-7.07,1). (c) Conditioning on µ = 22.01, we have: P (rejecting H0 | µ = 22.01) = 1 − P (21.97 < X̄ < 22.03 | µ = 22.01)

22.03 − 22.01 21.97 − 22.01 √ √ <Z< =1−P 0.08/ 50 0.08/ 50 = 1 − P (−3.53 < X < 1.77) = 0.0386 using =1-(NORM.S.DIST(1.77,1)-NORM.S.DIST(-3.53,1)) 26. An accounting firm wishes to test whether 6% of a large number of transactions contain errors against an alternative hypothesis that there are more than 6% errors. In order to test this claim, they examine a random sample of 200 transactions and find that exactly 18 of these are in error.

Calculate the p-value for the appropriate hypothesis test and comment on this value.

Solution: Here we are testing a proportion: H0 : π = 0.06 vs. H1 : π > 0.06 (i.e. that there are more than 6% errors). As this is a test for a single proportion and n is large, we compute the test statistic value: z=p

p−π

18/200 − 0.06 =p = 1.79. π(1 − π)/n 0.06 × 0.94/200

Using =1-NORM.S.DIST(1.79,1) returns a p-value of 0.0367, hence we can reject H0 at the 5%, but not 1%, significance level. There is moderate evidence that more than 6% of transactions contain errors.

27. In a particular city it is known, from past surveys, that 25% of households regularly use a washing powder named ‘Snolite’. After an advertising campaign, a survey of 300 randomly selected households showed that 100 had recently purchased Snolite. Is there evidence that the campaign has been successful? Use a 1% significance level.

Solution: Here we are testing a proportion: H0 : π = 0.25 vs. H1 : π > 0.25. Note that this is a one-sided test – we have reason to believe that the sales campaign has increased sales, and we believe this before

9

10. Hypothesis testing

collecting any data. As this is a test for a single proportion and n is large, we compute the test statistic value: 100/300 − 0.25 p−π = √ z=p = 3.33.

0.000625 π(1 − π)/n Using =1-NORM.S.DIST(3.33,1) returns a p-value of 0.0004, hence we can reject H0 at the 1% significance level. There is (very) strong evidence that more than 25% of households use Snolite. It appears that the campaign has been successful.

28. A museum conducts a survey of its visitors in order to assess the popularity of a device which is used to provide information on the museum exhibits. The device will be withdrawn if fewer than 30% of all of the museum’s visitors make use of it. Of a random sample of 80 visitors, 20 chose to use the device. (a) Carry out a test at the 5% significance level to see if the device should be withdrawn and state your conclusions. (b) What is the power of this test if the actual percentage of all visitors who would use this device is only 20%? (Remember that power is the complement of the Type II error probability.)

Solution: (a) Let π be the population proportion of visitors who would use the device. We test H0 : π = 0.30 vs. H1 : π < 0.30. p The sample proportion is p = 20/80 = 0.25. The standard error of the sample proportion is 0.30 × 0.70/80 = 0.0512. The test statistic value is: z=

0.25 − 0.30 = −0.976.

0.0512

For a one-sided (lower-tailed) test the p-value is P (Z ≤ −0.976) = 0.1645, using =NORM.S.DIST(-0.976,1). Hence the test is not significant – and not even at the 10% significance level. On the basis of the data, there is no reason to withdraw the device. (b) Using =NORM.S.INV(0.05), we would reject H0 at the 5% significance level if: p − 0.30 < −1.645 0.0512

⇒

p < 0.2157

that is, if the sample proportion, p, is less than 0.2157, we would obtain a test statistic value less than −1.645 which would have a p-value of less than 0.05. The power of the test is the probability of rejecting H0 (which happens when the sample proportion p < 0.2157) conditional on π = 0.20. This is given by: P (P < 0.2157 | π = 0.20). p When π = 0.20, the standard error of the sample proportion is 0.20 × 0.80/80 = 0.0447.

Hence the power when π = 0.20 is:

0.2157 − 0.20 P Z< = P (Z < 0.35) = 0.6368 0.0447 using =NORM.S.DIST(0.35,1).

29. A company recently redesigned its website to increase its appeal to younger consumers (those aged under 30). Random samples of younger consumers were surveyed before and after the redesign to see if they had a positive opinion of the website. The results were as follows:

10

10.1. Exercises

Sample size Before redesign After redesign

125 150

Respondents with a positive opinion 75 111

(a) Do the survey results indicate the redesign has been effective? Conduct a suitable hypothesis test, reporting the p-value and comment on your results. State any assumptions that you make. (b) Briefly discuss any concerns had the sample sizes been 25 and 30, instead of 125 and 150, respectively.

Solution: (a) Let period 1 indicate before the redesign, and period 2 after the redesign. We test: H0 : π1 = π2

vs. H1 : π1 < π2 .

We have p1 = 75/125 = 0.60, p2 = 111/150 = 0.74 and the pooled sample proportion is p = (75 + 111)/(125 + 150) = 0.6764. Under H0 we have: P1 − P2 p ∼ N (0, 1) P (1 − P )(1/n1 + 1/n2 ) approximately, for large n (which we assume here). The test statistic value is: 0.74 − 0.60 p

0.6764 × 0.3236 × (1/125 + 1/150)

= 2.47.

Using =1-NORM.S.DIST(2.47,1), the p-value is P (Z ≥ 2.47) = 0.0068, hence we reject H0 at the 1% significance level and conclude that there is strong evidence that the redesign was effective. (b) A discussion is expected here about the applicability of the central limit theorem approximation to normality with ‘small’ sample sizes of 25 and 30 relative the the ‘large’ sample sizes of 125 and 150.

30. A consumer report examined potential differences between two brands of tyres. The mean lifespan of the tyres is of primary concern. The available data, measured in thousands of miles, is provided below.

Brand X Brand Y

Sample size 14 15

Sample mean 23.6 24.8

Sample standard deviation 1.9 1.6

(a) Use an appropriate hypothesis test to determine whether there is evidence of a difference in the mean lifespans of the two brands. Comment on your findings. (b) State clearly any assumptions you made in part (a). (c) Repeat the procedure in part (a) to determine whether the mean lifespan of the tyres of brand Y is greater than that of the brand X tyres.

11

10. Hypothesis testing

(d) Based on your test conclusion in part (c), explain which type of error may have been committed, and what the practical consequences of the error would be.

Solution: (a) Using obvious notation, we test: H0 : µX = µY

vs. H1 : µX 6 µY .

Assuming equal variances, the pooled sample variance is: s2p =

13 × (1.9)2 + 14 × (1.6)2 = 3.07.

14 + 15 − 2

Under H0 we have: Ȳ − X̄ q ∼ tn1 +n2 −2 = t27 .

Sp2 (1/n1 + 1/n2 ) The test statistic value is:

24.8 − 23.6 p = 1.84.

3.07 × (1/14 + 1/15)

Using =2*T.DIST(-1.84,27,1), the p-value is 0.0768 hence we can only reject H0 at the 10% significance level. We conclude that there is weak evidence of a difference in the mean lifespans of the two brands. (b) The test in part (a) assumed: • normally-distributed populations • independent samples • equal population variances. (c) We now test: H0 : µX = µY

vs. H1 : µX < µY .

The test statistic value is still 1.84, but now the p-value is determined for this one-tailed test.

Using =1-T.DIST(1.84,27,1), the p-value is now 0.0384, hence we can reject H0 at the 5% significance level. We conclude that there is moderate evidence that the mean lifespan of the tyres of brand Y is greater than that of the brand X tyres. (d) Since we have rejected H0 , albeit at different significance levels, if we have committed an error in each case it would be a Type I error. This would mean we think there is a difference (in part (a)) when there is no difference, while in part (b) we would think that the mean lifespan of the tyres of brand Y is greater than that of the brand X tyres, when they are the same.

31. A pharmaceutical company has developed a new vaccine to immunise against Covid-19 and needs to determine the vaccine’s efficacy. A randomised experiment was conducted to address this question.

The study randomly allocated 200 people to either a group where the vaccine was administered, or a group where a placebo was given. These people were monitored and the numbers of those who did not go on to contract Covid-19 (and those who did) were recorded. The results are summarised below:

12

10.1. Exercises

Vaccine group Placebo group

Developed Covid-19 21 48

Did not develop Covid-19 79 52

(a) Carry out an appropriate hypothesis test at the 1% significance level to determine whether the proportion of people who did not develop Covid-19 is higher in the vaccine group than in the placebo group. State the test hypotheses, and specify your test statistic and its distribution under the null hypothesis. Comment on your findings and clearly state any assumptions you make. (b) Explain, with reasons, whether it is sufficient to have 200 people in this study.

Solution: (a) We test: H0 : π1 = π2

vs. H1 : π1 > π2 .

The pooled sample proportion is p = (79 + 52)/200 = 0.655, and the sample proportions are p1 = 79/(21 + 79) = 0.79 and p2 = 52/(48 + 52) = 0.52. Under H0 we have: P1 − P2 p ∼ N (0, 1) P (1 − P )(1/n1 + 1/n2 ) approximately, for large n (which we assume here). The test statistic value is: 0.79 − 0.52 p = 4.02.

0.655 × 0.345 × 1/50 Using =1-NORM.S.DIST(4.02,1), the p-value is 2.91 × 10−5 , hence we reject H0 at the 1% significance level and conclude that there is (very!) strong evidence that the vaccine is effective. (b) A discussion is expected here about the relationship between sample size and power. Clinical trials are likely to be based on much larger sample sizes, especially in terms of life and death matters such as vaccines.

32. A company wishes to investigate whether its average brand awareness varies between older and younger people. Random samples of 15 adults over the age of 60 and 14 adults aged between 18 and 30 scored their brand awareness on a scale from 1 to 100. Summaries of these scores are presented below.

Over 60s 18–30s

Sample size 15 14

Sample mean 68.7 74.3

Sample variance 15.4 18.2

(a) Use an appropriate hypothesis test to determine whether the mean brand awareness score is higher for the younger age group. State clearly the hypotheses, the test statistic and its distribution under the null hypothesis. Comment on your findings and clearly state any assumptions you make. (b) Is it possible that there is no difference between older and younger adults in terms of their average brand awareness? Briefly discuss.

13

10. Hypothesis testing

Solution: (a) Let group 1 = over 60s; group 2 = 18–30s. We test: H 0 : µ1 = µ2

vs. H1 : µ1 < µ2 .

Assuming equal variances, the pooled sample variance is: s2p =

14 × 15.4 + 13 × 18.2 = 16.75.

15 + 14 − 2

Under H0 we have: X̄2 − X̄1 q ∼ tn1 +n2 −2 = t27 .

Sp2 (1/n1 + 1/n2 ) The test statistic value is:

75.3 − 63.7 p = 3.68.

16.75 × (1/15 + 1/14)

Using =1-T.DIST(3.68,27,1), the p-value is 0.0005 hence we can reject H0 at the 1% significance level. We conclude that there is strong evidence that the mean brand awareness score is higher for the younger age group. (b) It is possible, although this would mean a Type I error occurred in part (a). The sample sizes are small, so for greater certainty it is worth resampling with larger sample sizes and test again, with a more powerful statistical test.

33. A survey is conducted to compare customer satisfaction in two branches of a bank. Various customers visiting the two branches were selected randomly, and asked if they were satisfied with the services provided by the branch. The results of this survey are shown in the following table.

Branch A Branch B

Sample size 153 188

Number satisfied 115 120

You are asked to consider an appropriate hypothesis test to determine whether there is a difference between the two bank branches regarding the proportion of satisfied customers. Perform a suitable test and comment on your findings. Specify the test statistic you use and its distribution under the null hypothesis.

Solution: Let πA denote the population proportion for branch A and πB denote the population proportion for branch B. The wording ‘whether there is a difference between. . .’ implies a two-sided test, hence the hypotheses can be written as: H0 : πA = πB

vs. H1 : πA 6= πB .

The test statistic formula is provided in the formula sheet: PA − PB ∼ N (0, 1) Z=p P (1 − P )(1/nA + 1/nB )

14

10.1. Exercises

approximately for large nA and nB , where the estimate of the common proportion is p = (115 + 120)/(153 + 188) = 0.6891, also pA = 0.7516 and pB = 0.6383. Hence the value of the test statistic is: 0.7516 − 0.6383 p = 2.2482 0.6891 × (1 − 0.6891) × (1/153 + 1/188) The sample sizes are large, hence the normal distribution can be used due to the central limit theorem.

Using =2*NORM.S.DIST(-2.2482,1), the p-value is 0.0246 hence we reject the null hypothesis at the 5%, but not 1%, significance level. We conclude that the test is moderately significant and that there is moderate evidence of a difference between the proportions of satisfied customers of the two branches.

34. A market research company has conducted a survey of adults in two large towns, either side of an international border, in order to judge attitudes towards a controversial internationally broadcast celebrity television programme. The following table shows some of the information obtained by the survey:

Sample size Sample number approving of the programme

Town A 50 26

Town B 50 22

(a) Conduct an appropriate hypothesis test, at the 5% significance level, of the claim that the population proportions approving the programme in the two towns are equal. (b) Would your conclusion be the same if, in both towns, the sample sizes had been 100 (with the same sample proportions of approvals)?

Solution: (a) We test H0 : πA = πB vs. H1 : πA 6= πB (note this is a two-sided test). We compute the sample proportions: pA = 26/50 = 0.52 and pB = 22/50 = 0.44. The estimate of the common pooled proportion is p = (26 + 22)/(50 + 50) = 0.48, hence the estimated standard error is: s

p(1 − p)

1 1 + nA nB

s

=

0.48 × 0.52 ×

1 1 + 50 50

= 0.09992.

The test statistic value is:

0.52 − 0.44 = 0.8006.

0.09992 Using =2*NORM.S.DIST(-0.8006,1), the p-value is 0.4234, hence the result is not significant and we do not reject H0 at the 5% significance level. There is no evidence to suggest there exists a difference in the population proportions approving of the programme.

(b) If nA = nB = 100, only the standard error is affected whose value would now be 0.0707 resulting in a test statistic value of 0.08/0.0707 = 1.1323, which again is not statistically significant (a p-value of 0.2575).

15

10. Hypothesis testing

35. The data in the following table show the numbers of daily parking offences in two areas of a city.

The day identifications are unknown and the recordings were not necessarily made on the same days. Is there evidence that the areas experience different numbers of offences?

Area A 38 38 29 45 42

Area B 32 38 22 30 34

Area A 33 27 32 32 34

Area B 28 32 34 24

Solution: It is clear that the data for the two areas cannot be regarded as paired, because there is no detail about the days to which the figures refer (and because there are different numbers of values for the two areas anyway). The sample sizes are 10 and 9 for A and B, respectively, so we need to decide whether we can use the combination of a pooled estimate of the variance and a t distribution test on the appropriate number of degrees of freedom.

So, start by calculating the sample means and standard deviations (or variances):

n Mean Standard deviation

A 10 35 5.68

B 9 30.44 5.08

Because the events to which the data refer are of the same type, it is reasonable to assume that the population variances of the numbers of offences are the same (especially since the sample standard deviations turn out to be quite similar. (It can be shown that, if the ratio of the sample variances is between roughly 1/3 and 3, then there is no reason to suppose that the population variances are different.)

We test H0 : µA − µB = 0 vs. H1 : µA − µB 6= 0. The pooled variance is: s2p =

(10 − 1) × (5.68)2 + (9 − 1) × (5.08)2 = 29.2243 (10 − 1) + (9 − 1)

Hence the estimated standard error is: s

√ 1 1 29.2243 × + = 5.406 × 0.21 = 2.484 10 9 on 17 degrees of freedom. The test statistic value is: t=

35 − 30.44 4.56 = = 1.836.

2.484 2.484

Using =2*(1-T.DIST(1.836,17,1)), the p-value is 0.0839 so we can only reject H0 at the 10% significance level, hence the evidence is weakly significant.

We cannot conclude with confidence that there is a difference in the number of parking offences between the two areas, but there is a suspicion that there may be such a difference. It would be preferable to examine more data in order to come to a firmer conclusion.

16

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 11: Bivariate analysis

ii

Chapter 11 Bivariate analysis 11.1

Exercises

11.1.1

Test your understanding

1. What type of exploratory analysis would you propose for: (a) two categorical variables? (b) one categorical variable and one measurable variable? (c) two measurable variables?

Solution: (a) For two categorical variables, contingency table or cross-tabulation is recommended. (b) For one categorical and one measurable variable, side-by-side boxplots are recommended. (c) For two measurable variables, a scatterplot is recommended.

2. In a hypothesis test of association, what are H0 and H1 ?

Solution: The null hypothesis, H0 , is that the two factors are not associated, while the alternative hypothesis, H1 , is that they are associated. Note H1 does not specify the nature of the association, just that there is an association (which we would need to investigate further if we end up rejecting H0 ).

3. How are expected frequencies calculated in cross-tabulations, and why are they calculated in this way?

Solution: In cross-tabulations, expected frequencies are calculated by assuming that the row and column variables are independent (that is, no association). They represent the frequencies that would be expected in each cell of the cross-tabulation if there were no association between the variables.

The expected frequency for a particular cell is calculated using the formula: row i total × column j total row i total × column j total = total number of observations n where i = 1, 2, . . . , r and j = 1, 2, . . . , c.

Eij =

The rationale behind calculating expected frequencies in this way is to provide a baseline for comparison with the observed frequencies (the actual counts in each cell). If the observed frequencies deviate significantly from the expected frequencies, it suggests that there is an association or relationship between the row and column variables.

1

11. Bivariate analysis

4. What condition is required on the expected frequencies so that the chi-squared distribution may be used when testing for association?

Solution: The required condition is that each expected frequency in the cross-tabulation should be at least 5.

When all expected frequencies are at least 5, the chi-squared test statistic follows an approximately chi-squared distribution, which is necessary for conducting a valid test of association between categorical variables. If this condition is not met, the chi-squared test may not provide accurate results.

5. State the test statistic for a test of association and its distribution under H0 .

Solution: Let the cross-tabulation have r rows and c columns, then formally the test statistic used for tests of association is: r X c X (Oij − Eij )2 ∼ χ2(r−1)(c−1) .

E ij i=1 j=1 6. Which Excel function can be used to calculate p-values for a test of association? Include the function’s arguments.

Solution: Test of association p-values are obtained from the χ2 distribution with (r − 1)(c − 1) degrees of freedom, i.e. in Excel using: =CHISQ.DIST.RT(test statistic value, degrees of freedom).

7. Name two ways of quantifying the strength of association between two categorical variables. When can each be used?

Solution: If we have a 2 × 2 cross-tabulation, i.e. when r = c = 2, then the we can use the phi coefficient, computed as: r χ2 φ= n where χ2 is the test statistic value and n is the total number of observations. φ ∈ [0, 1]. When φ = 0 (which is only when χ2 = 0), there is no association. As φ → 1, the stronger the association between the two factors.

The contingency coefficient is a more general version of the phi coefficient, applicable to cross-tabulations of any dimension. It is calculated as is: s χ2 C= χ2 + n where χ2 is the test statistic value and n is the total number of observations, as before. C ∈ [0, 1].

When C = 0 (which is only when χ2 = 0), there is no association. As C → 1, the stronger the association between the two factors. However, C does not always reach its maximum possible value

2

11.1. Exercises

of 1. The value of ‘max C’ depends on the dimensions of the cross-tabulation. As such, comparison of C values across multiple cross-tabulations is only advisable if the cross-tabulations are all of the same size.

8. Are correlation and causation the same? Briefly justify your view.

Solution: No, correlation and causation are not the same!

Correlation refers to a statistical relationship between two variables, where a change in one variable is associated with a change in another variable. Correlation measures the strength and direction of the (linear) relationship but does not imply a cause-and-effect relationship.

Causation refers to a cause-and-effect relationship between two variables, where a change in one variable directly influences the other variable. Causation implies that one variable is the direct cause of changes in the other variable.

While correlation can suggest the presence of a relationship between variables, it does not prove causation. Establishing causation requires more rigorous methods, such as experimental studies or well-designed causal inference techniques, to demonstrate that one variable actually causes changes in the other. Therefore, correlation is a weaker form of relationship than causation.

9. What are the properties of the correlation coefficient?

Solution: Both correlation coefficients (the parameter ρ, and its estimator r) have the following properties.

They: • are independent of the scale of measurement • are independent of the origin of measurement • are symmetric; that is, the correlation of X and Y is the same as the correlation of Y and X • can only take values between ±1, that is: −1 ≤ ρ ≤ 1 and

−1≤r ≤1

alternatively |ρ| ≤ 1 and |r| ≤ 1, i.e. correlation coefficients always have an absolute value less than or equal to 1.

10. What values of a correlation coefficient would you say constitute: (a) strong correlation? (b) moderate correlation? (c) weak correlation?

3

11. Bivariate analysis

Solution: This is subjective, but a suggested guide is:

11. What is the Spearman rank correlation? How is it calculated, and when is it used?

Solution: The Spearman rank correlation is a non-parametric measure of the strength and direction of the relationship between two ranked variables. It is used when the data is ordinal or the relationship between variables is not linear.

To compute rs we rank the xi and yi values in ascending order and use the ranks of the xi s and yi s, rather than their actual values. If there are no tied rankings of xi and yi , the Spearman rank correlation is: n P 6 d2i i=1 rs = 1 − n(n2 − 1) where the di s are the differences in the ranks between each xi and yi .

It is helpful to use the Spearman rank correlation if you are worried about the effect of extreme observations (outliers) in your sample. The possible values of rs are the same as for r, that is −1 ≤ rs ≤ 1, or equivalently |rs | ≤ 1.

12. When testing a correlation coefficient with H0 : ρ = 0, state the test statistic, its distribution under H0 and how the test statistic value is affected by the: (a) effect size (b) sample size.

Solution: Under H0 , the test statistic is:

√ r n−2 T = √ ∼ tn−2 .

1 − r2 Hence p-values are obtained from the Student’s t distribution with n − 2 degrees of freedom, i.e. using T.DIST in Excel.

Note that the (absolute) value of the test statistic is (i) increasing in r (the stronger the sample correlation coefficient, the larger the test statistic value, ceteris paribus), and (ii) increasing in n (the larger the sample size, the larger the test statistic value, ceteris paribus).

4

11.1. Exercises

Remark (i) reflects the effect size influence, and remark (ii) reflects the sample size influence. Note that n also affects the degrees of freedom of the tn−2 distribution, hence how fat/thin-tailed the test statistic distribution is, with implications for the magnitude of the p-value.

13. What is a ‘spurious correlation’ ? Provide an example.

Solution: A spurious correlation is a statistical relationship between two variables that appears to be significant but is actually caused by a third variable, unrelated to the variables being studied. It is a misleading association that arises due to coincidence rather than a causal relationship.

Suppose a company observes a strong positive correlation between the sales of ice cream and the number of drowning incidents during the summer months. This correlation might lead them to conclude that selling more ice cream somehow increases the risk of drowning. However, the spurious correlation occurs because both ice cream sales and drowning incidents are influenced by a common third variable, the temperature. In summer, high temperatures lead to increased ice cream sales and more people swimming, which in turn increases the risk of drowning. The correlation between ice cream sales and drowning incidents is spurious because there is no direct causal relationship between them; it is an indirect association due to the influence of temperature.

14. ‘Uncorrelated variables imply independent variables.’ Do you agree? Briefly justify your view.

Solution: Disagree. Variables being uncorrelated is a necessary, but not sufficient, condition for independent variables. Being uncorrelated only means that there is no linear relationship, but there could be a non-linear relationship which would mean the variables are not independent.

However, independent variables do imply uncorrelated variables, since independence is a stronger statistical property. When variables are independent, they have no relationship which, of course, means they have no linear relationship, hence the variables must be uncorrelated.

15. A 5 × 3 contingency table results in a test statistic value of 17.55. What is the corresponding p-value?

Solution: We have that r = 5 and c = 3, hence the p-value is determined with respect to a χ2 distribution with (r − 1)(c − 1) = (5 − 1)(3 − 1) = 8 degrees of freedom. In Excel, we use: =CHISQ.DIST.RT(17.55,8) which is 0.0249. Note this would be significant at the 5%, but not 1%, significance level.

16. The following table shows the numbers of car accidents in an urban area over a period of time.

These are classified by severity and by type of vehicle. Carry out a test for association on these data and draw conclusions.

Saloon Van Sports car

Severity of accident Minor Medium Major 29 39 16 15 24 12 7 20 12

5

11. Bivariate analysis

Solution: We test: H0 : There is no association between type of vehicle and severity of accident. vs.

H1 : There is an association between type of vehicle and severity of accident.

We construct the following contingency table:

Saloon

Van

Sports

O1j E1j (O1j − E1j )2 /E1j O2j E2j (O2j − E2j )2 /E2j O3j E3j (O3j − E3j )2 /E3j

Total The test statistic value is: 3 X 3 X (Oij − Eij )2 i=1 j=1

Eij

Severity of accident Minor Medium Major 29 39 16 24.6 40.1 19.3 0.78 0.03 0.57 15 24 12 15.0 24.3 11.7 0.00 0.00 0.00 7 20 12 11.4 18.6 9.0 1.72 0.10 1.03 51 83 40

Total 84 84 51 51 39 39 174

= 0.78 + 0.03 + · · · + 1.03 = 4.24.

Since r = c = 3, we have (r − 1)(c − 1) = (3 − 1)(3 − 1) = 4 degrees of freedom.

Using Excel, we have =CHISQ.DIST.RT(4.24,4) which returns a p-value of 0.3745. Hence the test is not significant so we have found no evidence whatsoever of an association between type of vehicle and severity of accident.

17. ‘Footfall’ is the daily number of customers per unit sales area of a shop, and ‘Ticket price’ is the average sale price of an item in the shop’s offer. Shops are classified as offering low-, medium- or high-price items and, during any sales period, as having low, medium or high footfall. An analyst studies a sample of shops and obtains the following frequency data: Low footfall Medium footfall High footfall

Low price 28 38 64

Medium price 42 86 75

High price 46 54 35

Perform an appropriate hypothesis test to check whether there is evidence of an association between footfall and ticket price. What do you conclude?

Solution: We test: H0 : No association between footfall and the ticket price vs.

H1 : Association between footfall and the ticket price

6

11.1. Exercises

The calculations can be summarised as: Footfall O1· E1· (O1· − E1· )2 /E1· O2· Medium E2· (O2· − E2· )2 /E2· O3· High E3· (O3· − E3· )2 /E3· Total Low

Ticket price Low Medium High 28 42 46 32.22 50.32 33.46 0.55 1.37 4.70 38 86 54 49.44 77.21 51.35 2.65 1.00 0.14 64 75 35 48.33 75.47 50.19 5.08 0.00 4.60 130 203 135

Total 116 116 178 178 174 174 468

The test statistic value is: 3 X 3 X (Oij − Eij )2 i=1 j=1

Eij

= 0.55 + 1.37 + · · · + 4.60 = 20.09.

Since r = c = 3, we have (r − 1)(c − 1) = (3 − 1)(3 − 1) = 4 degrees of freedom.

Using Excel, we have =CHISQ.DIST.RT(20.09,4) which returns a p-value of 0.0005. Hence the test is highly significant so we conclude that we have very strong evidence of an association between footfall and the ticket price.

18. A survey was conducted to investigate the relationship between hotels and the primary reason for not returning. The following table shows the results of this survey:

Hotel A Hotel B Hotel C Total

Primary reason for not returning Price of room Hotel location Room quality 23 37 40 35 16 49 46 42 12 104 95 101

Total 100 100 100 300

(a) Based on the data in the table, and without conducting a significance test, would you say there is an association between the hotel and the primary reason for not returning? (b) Perform an appropriate hypothesis test to check whether there is evidence of an association between the hotel and the primary reason for not returning. What do you conclude?

Solution: (a) There are some differences in the distributions within hotels. More specifically, price appears more of a concern for Hotel C guests than for Hotel A guests compared to room quality (46% vs. 23% and 12% vs. 40%, respectively). Hotel location seems to be of little concern to Hotel B guests (16%). Hence there seems to be an association between hotel and the primary reason for not returning, although this needs to be investigated further.

7

11. Bivariate analysis

(b) We test: H0 : No association between hotel and primary reason for not returning vs.

H1 : Association between hotel and primary reason for not returning.

The calculations can be summarised as:

Hotel A

Hotel B

Hotel C

O1· E1· (O1· − E1· )2 /E1· O2· E2· (O2· − E2· )2 /E2· O3· E3· (O3· − E3· )2 /E3·

Total

Primary reason for not returning Price of room Hotel location Room quality 23 37 40 34.67 31.67 33.67 3.93 0.90 1.19 35 16 49 34.67 31.67 33.67 0.00 7.75 6.98 46 42 12 34.67 31.67 33.67 3.71 3.37 13.94 104 95 101

Total 100 100 100 100 100 100 300

The test statistic value is: 3 X 3 X (Oij − Eij )2 i=1 j=1

Eij

= 3.93 + 0.90 + · · · + 13.94 = 41.77.

Since r = c = 3, we have (r − 1)(c − 1) = (3 − 1)(3 − 1) = 4 degrees of freedom.

Using Excel, we have =CHISQ.DIST.RT(41.77,4) which returns a p-value of 1.86 × 10−8 ! Hence the test is highly significant so we conclude that we have very strong evidence of an association between hotel and primary reason for not returning.

19. The following table shows the number of salespeople employed by a company and the corresponding values of sales (in thousands of pounds).

Salespeople (x) Sales (y)

210 206

209 200

219 204

225 215

232 222

221 216

220 210

233 218

200 201

215 212

205 204

227 212

Compute the sample correlation coefficient for these data and carry out a formal test for a relationship between the number of salespeople and sales.

Solution: Using the =CORREL(array1, array2) Excel function, we have that the sample correlation coefficient is r = 0.8716. Hence we test: H0 : ρ = 0 vs. H1 : ρ 6= 0.

The corresponding test statistic value is: √ √ r n−2 0.8716 × 10 t= √ =p = 5.62.

1 − r2 1 − (0.8716)2

8

11.1. Exercises

Using =2*T.DIST(-5.62,10,1), the p-value is 0.0002 hence we reject H0 at the 1% significance level, and we conclude that there is strong evidence that the number of salespeople and sales are correlated.

Note here we could legitimately have conducted an upper-tailed test, since plausibly we would expect a positive relationship (if any) between these variables before even observing the data.

20. For 15 paired observations, the sample correlation coefficient is calculated to be r = −0.6172. (a) Test H0 : ρ = 0 versus H1 : ρ < 0 at the 1% significance level. (b) Following the test in part (a), briefly explain whether it is worthwhile to test H0 : ρ = 0 versus H1 : ρ 6= 0 at the 1% significance level.

Solution: (a) We test H0 : ρ = 0 vs. H1 : ρ < 0. The test statistic value is: √ √ −0.6172 × 13 r n−2 = −2.828. =p t= √ 1 − r2 1 − (−0.6172)2 Using =T.DIST(-2.828,13,1), the p-value is 0.0071, hence we reject H0 at the 1% significance level and conclude that there is strong evidence that ρ < 0. (b) A two-tailed test would not affect the test statistic value, but would affect the p-value, which would be doubled, i.e. using =2*T.DIST(-2.828,13,1), the p-value would be 0.0142, which would not be significant at the 1% significance level, which is a different, but not contradictory, result (as not reject H0 6= accept H0 ).

However, it could be argued that if a negative correlation could reasonably be hypothesised a priori, then the lower-tailed test in part i. would be more powerful meaning the two-tailed test would not be desirable.

9

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 12: Analysis of variance

ii

Chapter 12 Analysis of variance 12.1

Exercises

12.1.1

Test your understanding

1. In an analytical sense, what do we mean by ‘understanding the causes of things’ ?

Solution: To understand the causes of things requires us to understand the causes of variation in a variable of interest. We seek to formally analyse and quantify the extent to which the variation we observe in a dependent variable (such as a customer’s intention to purchase) can be attributed to one or more independent variables (such as our marketing mix variables, the 4Ps). Our goal is literally to ‘understand the causes of things’, where our choice of dependent variables are the things we seek to understand.

2. Think of some examples of measurable dependent variables that a manager may be interested in understanding. For each, list possible categorical variables you believe may exert a causal influence, briefly justifying your view in each case.

Solution: Example 1: • Dependent variable: Sales revenue.

• Possible categorical variables: Marketing channel (online, offline), product category (electronics, apparel), region (North, South, East, West).

• Justification: Different marketing channels, product categories, and geographic regions can have varying consumer preferences and behaviours which influence sales revenue.

Example 2: • Dependent variable: Employee turnover rate.

• Possible categorical variables: Job satisfaction (satisfied, unsatisfied), remuneration (competitive, below market average), work–life balance (good, poor).

• Justification: Job satisfaction, remuneration, and work–life balance are known to impact employee retention and turnover decisions.

Example 3: • Dependent variable: Customer satisfaction score.

• Possible categorical variables: Customer service quality (high, low), product quality (excellent, average), price perception (affordable, expensive).

1

12. Analysis of variance

• Justification: Customer service, product quality, and perceived pricing are significant drivers of customer satisfaction.

Example 4: • Dependent variable: Website conversion rate.

• Possible categorical variables: Call-to-Action (CTA) button colour (red, blue), page layout (simple, complex), website loading time (fast, slow).

• Justification: The colour of CTA buttons, page layout, and website loading time can influence user engagement and hence conversion rates.

Example 5: • Dependent variable: Employee productivity.

• Possible categorical variables: Work environment (collaborative, individual), training (regular, infrequent), incentives (monetary, non-monetary).

• Justification: The work environment, training opportunities, and incentive programmes can impact employee motivation and productivity.

In each case, the dependent variable represents a measurable outcome or performance metric of interest to the manager. The possible categorical variables are factors which the manager believes may exert a causal influence on the dependent variable based on existing knowledge or prior research in the respective domains. Analysing the relationships between these variables can provide valuable insights for decision-making and process improvement in various organisational settings.

3. In a one-way ANOVA model, what assumptions are made about the observations of the dependent variable?

Solution: There are k independent random samples available from k normal distributions N (µj , σ 2 ), for j = 1, 2, . . . , k. We assume a constant variance, σ 2 , for each j = 1, 2, . . . , k. Denote by X1j , X2j , . . . , Xnj j the random sample with sample size nj from N (µj , σ 2 ), for j = 1, 2, . . . , k.

4. What are the null and alternative hypotheses in a one-way ANOVA? Solution: We test: H 0 : µ1 = µ2 = · · · = µk

vs. H1 : all µj s are not equal.

5. Explain the ANOVA decomposition (a) in words, and (b) using formulae.

Solution: (a) The ANOVA decomposition decomposes (hence the name!) the total variation observed in the dependent variable into the between-groups variation, i.e. the amount of the variation which can be attributed to the factor level, and the within-groups variation, i.e. the leftover or ‘residual’ variation which cannot be attributed to the factor level. Informally: total variation = explained variation + unexplained variation.

2

12.1. Exercises

(b) Using formuale, the ANOVA decomposition is: nj nj k X k k X X X X 2 2 (Xij − X̄) = nj (X̄·j − X̄) + (Xij − X̄·j )2 j=1 i=1

j=1

⇔

SSX = B + W.

j=1 i=1

6. How is explanatory power measured in a one-way ANOVA and how is the statistic used interpreted?

Solution: η 2 is the proportion of the variation in the dependent variable explained by an ANOVA model. For one-way ANOVA, it is: B explained variation η2 = = SSX total variation provided SSX > 0, i.e. to be defined we must have at least some variation in the dependent variable.

Since 0 ≤ B ≤ SSX (because B ≥ 0 and W = SSX − B ≥ 0), it must be that η 2 ∈ [0, 1], that is, its value lies between 0 and 1. As η 2 gets closer to 1, i.e. as η 2 → 1, the better the explanatory power of the factor.

7. ANOVA tests are performed using upper-tailed F tests. (a) Given a test statistic value, f , which Excel function could be used to calculate the p-value? (b) Using an appropriate sketch, illustrate the p-value graphically.

Solution: (a) In Excel, we could use =F.DIST.RT(test statistic value,k-1,n-k), where k is the number of groups (levels of the categorical variable) and n is the total number of observations of the dependent variable. (b) Suppose the test statistic value was 9, when there are k = 3 groups and n = 18. The p-value would then be P (F > 9) = 0.0027, where F2, 15 . A sketch is (right-hand side is zoomed in):

8. What information is provided in a one-way ANOVA table? What do the different entries represent?

Solution: It is common for results to be presented in a one-way ANOVA table as follows:

3

12. Analysis of variance

Source of Variation Between Groups Within Groups Total

SS B W SSX

df k−1 n−k n−1

MS B/(k − 1) W/(n − k)

F B/(k−1) W/(n−k)

P -value p

From this, the p-value is sufficient to establish whether there is evidence of a difference between the population means (i.e. if p < α). The rest of the table is just explaining how this p-value is determined. For reference the: • SS column is the ANOVA decomposition, reporting the values of B, W and SSX = B + W • df column reports the degrees of freedom of the test statistic F distribution: k − 1 = df1 and n − k = df2 , totalling n − 1 • M S column is the ‘mean sum of squares’, calculated as the respective row’s SS/df • F is the test statistic value, being the ratio of the between groups and within groups M S. Note you can see why k − 1 = df1 is the degrees of freedom in the numerator, and why n − k = df2 is the degrees of freedom in the denominator.

9. If you wanted to compute a 95% confidence interval for µj in a one-way ANOVA context, how would you proceed?

Solution: 95% confidence intervals for µj are given by: S X̄·j ± t0.025, n−k × √ nj

for j = 1, 2, . . . , k

where t0.025, n−k is the top 2.5th percentile of the Student’s tn−k distribution, which can be obtained in Excel using =T.INV.2T(0.05,degrees of freedom). This uses an estimator of σ (i.e. the square root of the common variance, σ 2 ) given by: r W σ b=S= n−k which is a pooled estimate, making use of all n =

k P

nj observations.

j=1

10. An alternative parameterisation of the one-way ANOVA model is: Xij = µ + βj + εij

for i = 1, 2, . . . , nj and j = 1, 2, . . . , k.

Explain each of the terms µ, βj and εij Solution: Xij = µ + βj + εij decomposes each observation of our dependent variable into three parts: • the average effect, µ • the factor effect (if it exists), βj • the ‘noise’, εij ∼ N (0, σ 2 ).

4

12.1. Exercises

11. How does a two-way ANOVA differ from a one-way ANOVA? Solution: A two-way ANOVA analyses the effects of two categorical factors on a continuous dependent variable, while a one-way ANOVA examines the effect of only one categorical factor.

12. What are the possible hypothesis tests which could be conducted in a two-way ANOVA? Solution: 1. To test for the significance of factor 1, the test statistic: F =

M SX1 SSX1 /(k1 − 1) = ∼ Fk1 −1, n−k1 k2 SSW /(n − k1 k2 ) M SW

under the null hypothesis of no factor 1 effect.

2. To test for the significance of factor 2, the test statistic: F =

M SX2 SSX2 /(k2 − 1) = ∼ Fk2 −1, n−k1 k2 SSW /(n − k1 k2 ) M SW

under the null hypothesis of no factor 2 effect.

3. To test for the significance of the interaction between factors 1 and 2, the test statistic: F =

SSX1 X2 /(k1 − 1)(k2 − 1) M SX1 = ∼ F(k1 −1)(k2 −1), n−k1 k2 SSW /(n − k1 k2 ) M SW

under the null hypothesis of no interaction effect.

13. In a two-way ANOVA how could you determine the relative importance of the two factors?

Solution: We can use ω 2 (omega-squared) to measure the proportion of the variation in the dependent variable which is related to a factor. The relative contribution of factor Xi is calculated as: 2 ωX = i

SSXi − ((ki − 1) × M SW ) .

SSY + M SW

2 The ωX s can be compared across factors to judge their relative importance. i

14. Explain the different types of interaction which could feature in a two-way ANOVA. You may find it helpful to sketch these, but also describe the nature of each interaction.

5

12. Analysis of variance

Solution: The possibilities are: • No interaction, which means that knowledge of the level of one factor has no effect on the other factor’s level effect on the dependent variable. A graphical example is:

• Ordinal interaction, which means that order is sustained such that the means of Y strictly increase (or decrease) as we pass through the factor levels in order. A graphical example is:

• Disordinal interaction without crossover, which means that order is not sustained such that the means of Y do not strictly increase (or decrease) as we pass through the factor levels in order of all factors. ‘Without crossover’ means that the lines do not cross. A graphical example is:

6

12.1. Exercises

• Disordinal interaction with crossover, which is as above except that with crossover the lines do cross. A graphical example is:

15. Three trainee salespeople were working on a trial basis. Salesperson A went in the field for 5 days and made a total of 440 sales. Salesperson B was tried for 7 days and made a total of 630 sales.

Salesperson C was tried for 10 days and made a total of 690 sales. Note these figures are total Pthat 2 sales, not daily averages. The sum of the squares of all 22 daily sales ( xi ) is 146,840. (a) Construct a one-way analysis of variance table. (b) Would you say there is a difference between the mean daily sales of the three salespeople?

Justify your answer. (c) Construct a 95% confidence interval for the mean difference between salesperson B and salesperson C. Would you say there is a difference?

Solution: (a) The means are 440/5 = 88, 630/7 = 90 and 690/10 = 69. We will perform a one-way ANOVA. First, we calculate the overall mean. This is: 440 + 630 + 690 = 80.

22 We can now calculate the sum of squares between salespeople. This is: 5 × (88 − 80)2 + 7 × (90 − 80)2 + 10 × (69 − 80)2 = 2230.

The total sum of squares is: 146,840 − 22 × (80)2 = 6040.

Here is the one-way ANOVA table: Source of Variation Between Groups Within Groups Total

SS 2230 3810 6040

df 2 19 21

MS F 1115 5.56 200.53

P -value 0.0126

(b) Using =F.DIST.RT(5.56,2,19) returns the p-value of 0.0126, hence we reject H0 : µ1 = µ2 = µ3 at the 5% significance level and conclude that there is evidence that the means are not equal.

7

12. Analysis of variance

(c) Under the assumption of equal variances, we must have that (see Section 9.9.3):

1 1 2 X̄B − X̄C ∼ N µB − µC , σ + . nB nC Since σ 2 is unknown, we will have to estimate it using σ b2 = W/(n − k) producing a point estimate of 200.53. Hence a 95% confidence interval for the mean difference between salesperson B and salesperson C would have the form (again, see Section 9.9.3): s

1 1 2 x̄B − x̄C ± t0.025, n−k × σ b + . nB nC Using the provided values, we have: s 90 − 69 ± 2.093 ×

200.53 ×

1 1 + 7 10

= 21 ± 14.61

⇒

(6.39, 35.61)

using =T.INV(0.975,19). Hence a 95% confidence interval is (6.39, 35.61). As zero is not included, there is evidence of a difference.

16. An executive of a prepared frozen meals company is interested in the amounts of money spent on such products by families in different income ranges. The table below lists the monthly expenditures (in dollars) on prepared frozen meals from 15 randomly selected families divided into three groups according to their incomes.

Under $15,000 45.2 60.1 52.8 31.7 33.6 39.4

$15,000 – $30,000 53.2 56.6 68.7 51.8 54.2

Over $30,000 52.7 73.6 63.3 51.8

(a) Based on these data, can we infer at the 5% significance level that the population mean expenditures on prepared frozen meals are the same for the three different income groups? (b) Report the one-way ANOVA table. (c) Construct 95% confidence intervals for the mean expenditures of the first (under $15,000) and the third (over $30,000) income groups.

Solution: (a) For this example, k = 3, n1 = 6, n2 = 5, n3 = 4 and n = n1 + n2 + n3 = 15.

We have x̄·1 = 43.8, x̄·2 = 56.9, x̄·3 = 60.35 and x̄ = 52.58. nj 3 P P Also, x2ij = 43,387.85. j=1 i=1

Hence the total variation is: SSX =

nj k X X j=1 i=1

8

2

(xij − x̄) =

nj 3 X X j=1 i=1

x2ij − nx̄2 = 43,387.85 − 41,469.85 = 1918.

12.1. Exercises

The within-groups variation is: w=

nj k X X

2

(xij − x̄·j ) =

j=1 i=1

nj 3 X X

x2ij −

j=1 i=1

3 X

nj x̄2·j = 43,387.85 − 42,267.18 = 1120.67.

j=1

Therefore, the between-groups variation is: b = SSX − w = 1918 − 1120.67 = 797.33.

To test H0 : µ1 = µ2 = µ3 , the test statistic value is: f=

797.33/2 b/(k − 1) = = 4.27. w/(n − k) 1,120.67/12

Using =F.DIST.RT(4.27,2,12) returns the p-value of 0.0398, hence we reject H0 at the 5% significance level, i.e. there exists evidence indicating that the population mean expenditures on frozen meals are not the same for the three different income groups. (b) The ANOVA table is as follows:

(c) A 95% confidence interval for µj is of the form: √ S 93.39 21.056 X̄·j ± t0.025, n−k × √ = X̄·j ± t0.025, 12 × √ = X̄·j ± √ . nj nj nj √ For j = 1, a 95% confidence interval is 43.8 ± 21.056/ 6 ⇒ (35.20, 52.40). √ For j = 3, a 95% confidence interval is 60.35 ± 21.056/ 4 ⇒ (49.82, 70.88).

17. Does the level of success of publicly traded companies affect the way their board members are paid?

The annual payments (in $000s) of randomly selected publicly-traded companies to their board members were recorded. The companies were divided into four quarters according to the returns in their stocks, and the payments from each quarter were grouped together. Some summary statistics are provided below.

9

12. Analysis of variance

Quarter Q1 Q2 Q3 Q4

Sample size 30 30 30 30

Sample mean 74.10 75.67 78.50 81.30

Sample standard deviation 15.81 13.57 15.28 15.59

(a) Can we infer that the amount of payment differs significantly across the four groups of companies? (b) Construct 95% confidence intervals for the mean payment of the first-quarter companies and the fourth-quarter companies.

Solution: (a) Here k = 4 and n1 = n2 = n3 = n4 = 30. We have x̄·1 = 74.10, x̄·2 = 75.67, x̄·3 = 78.50, x̄·4 = 81.30, also x̄ = (74.10 + 75.67 + 78.50 + 81.30)/4 = 77.3925. Hence the between-groups variation is: b=

k X

nj (x̄·j − x̄)2

j=1

= 30((74.10 − 77.3925)2 + (75.67 − 77.3925)2 + (78.50 − 77.3925)2 + (81.30 − 77.3925)2 ) = 909.08.

The within-groups variation is: nj k k X X X 2 (Xij − X̄·j ) = (nj − 1)s2j w= j=1 i=1

j=1

= 29((15.81)2 + (13.57)2 + (15.28)2 + (15.59)2 ) = 26,408.2 p p and the pooled estimate of σ is s = w/(n − k) = 26,408.2/(120 − 4) = 15.09.

Hence the test statistic value is: f=

b/(k − 1) 909.08/(4 − 1) = = 1.33. w/(n − k) (15.09)2

Under H0 : µ1 = µ2 = µ3 = µ4 , F ∼ Fk−1, n−k = F3, 116 . Using =F.DIST.RT(1.33,3,116) returns the p-value of 0.2680, hence we cannot reject H0 at any reasonable significance level. Hence there is no evidence to support the claim that payments among the four groups are significantly different. (b) A 95% confidence interval for µj is of the form: S 15.09 15.09 X̄·j ± t0.025, n−k × √ = X̄·j ± t0.025, 116 × √ = X̄·j ± 1.9806 × √ = X̄·j ± 5.46 nj 30 30 using =T.INV(0.975,116).

For j = 1, a 95% confidence interval is 74.10 ± 5.46 ⇒ (68.64, 79.56).

For j = 4, a 95% confidence interval is 81.30 ± 5.46 ⇒ (75.84, 86.76).

10

12.1. Exercises

18. Do the following data appear to violate the assumptions underlying one-way analysis of variance?

Explain why or why not.

A 1.78 8.26 3.57 4.69 2.13

Treatment B C D 8.41 0.57 9.45 5.61 3.04 8.47 3.90 2.67 7.69 3.77 1.66 8.53 1.08 2.09 10.04

Solution: We have s2A = 6.7992, s2B = 7.2737, s2C = 0.9247 and s2D = 0.8419. So we observe that although s2A ≈ s2B and s2C ≈ s2D , the sample variances for treatments are very different across all groups, suggesting that the assumption that σ 2 is the same for all treatment levels may not be true.

19. An indicator of the value of a stock relative to its earnings is its price-earnings ratio: the average of a given year’s high and low selling prices divided by its annual earnings. The following table provides the price-earnings ratios for a sample of 30 stocks, ten each from the financial, industrial and utility sectors of the New York Stock Exchange. Test at the 1% significance level whether the true mean price-earnings ratios for the three market sectors are the same. Use the ANOVA table format to summarise your results.

Financial 11.4 12.3 10.8 9.8 14.3 16.1 11.9 12.4 13.1 14.6

Industrial 9.4 18.4 15.9 21.6 17.1 20.2 18.6 22.9 18.6 19.9

Utility 15.4 16.3 10.9 19.3 15.1 12.7 16.8 14.3 13.8 15.0

Solution: For these n = 30 observations and k = 3 groups, we have x̄·1 = 12.67, x̄·2 = 18.26, x̄·3 = 14.96 and x̄ = 15.297. Also: 3 X 10 X x2ij = 7,382.47. j=1 i=1

Hence the total variation is: 3 X 10 X

x2ij − nx̄2 = 7,382.47 − 30 × (15.297)2 = 362.83.

j=1 i=1

11

12. Analysis of variance

The between-groups variation is: b=

3 X

nj x̄2·j − nx̄2 = 10 × ((12.67)2 + (18.26)2 + (14.96)2 ) − 30 × (15.297)2

j=1

= 157.94.

Therefore, w = 362.83 − 157.94 = 204.89. Hence the ANOVA table is:

To test the null hypothesis that the three types of stocks have equal price-earnings ratios, on average, we use =F.DIST.RT(10.40661,2,27) which returns a p-value of 0.000446, hence we reject H0 and conclude that there is strong evidence of a difference in the mean price-earnings ratios across the sectors.

20. The following table shows monthly percentage audience shares of three major networks’ evening news broadcasts in five major cities over three months. Perform a two-way ANOVA. Is either factor statistically significant at the 5% significance level? Is there an interaction? If so, what is its nature?

City A

B

C

D

E

12

BBC 21.3 21.8 21.5 20.6 20.2 20.9 24.1 25.2 24.6 23.6 23.8 23.5 21.8 21.5 21.4

ITV 17.8 18.4 17.7 17.5 17.9 17.4 16.1 15.4 16.2 18.3 18.5 18.6 17.0 17.3 17.5

Sky 20.2 21.0 20.9 20.1 19.7 19.8 19.4 19.1 19.8 20.8 21.2 21.3 28.7 28.3 29.2

12.1. Exercises

Solution: The two-way ANOVA table is:

Observing the p-values, it is clear that both factors as well as the interaction are highly significant.

Note that 3.43E-18 means 3.43 × 10−18 , hence a set of very(!) small p-values.

Plotting the means for each combination of factor levels produces:

It is clear that there is a disordinal interaction with crossover. Of particular note is that ITV has the lowest average audience share in every city, while the highest figure is the average audience share for Sky in city E, where it seems the sky’s the limit for Sky!

13

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 13: Linear regression

ii

Chapter 13 Linear regression 13.1

Exercises

13.1.1

Test your understanding

1. What is the difference between a dependent variable and an independent variable?

Solution: The dependent variable is the outcome or response variable that is being measured or observed in an experiment or study. It is affected by the independent variable. The independent variable is the variable that is manipulated or controlled by the analyst to observe its effect on the dependent variable.

Suppose a company wants to study the effect of different advertising strategies on sales revenue. In this case, the sales revenue is the dependent variable because it is what the company wants to measure and understand better. The independent variable is the advertising strategy, as the company can manipulate and control different advertising approaches (such as TV ads, social media ads, email campaigns) to observe their impact on sales revenue.

2. In a simple linear regression model, what do β0 and β1 represent?

Solution: β0 represents the y-intercept of the regression line, which is the predicted value of the dependent variable, y, when the independent variable, x, is equal to zero. It indicates the value of y when x has no effect. β1 represents the slope of the regression line, which shows the change in the dependent variable for a one-unit change in the independent variable. It indicates the rate of change in y for each unit increase in x.

3. Why do we include an error term in a regression model? Does it have any assumptions?

Solution: We include an error term to account for the variability or randomness in the relationship between y and the independent variable(s). The error term captures the differences between the observed values of the dependent variable and the values predicted by the regression model. It represents the part of the dependent variable that cannot be explained by the independent variable(s) included in the model.

We assume that εi ∼ N (0, σ 2 ) for all i = 1, 2, . . . , n.

1

13. Linear regression

4. Which optimisation technique is used to estimate β0 and β1 ?

Solution: We use an optimisation technique called least-squares estimation. Specifically, we seek to minimise the sum of the squared error terms: min

β0 ,β1

n X

ε2i = min β0 ,β1

i=1

n X

(yi − β0 − β1 xi )2 .

i=1

In words, we seek the values of β0 and β1 which minimise (the ‘min’) the above sum of squares function, hence the term ‘least squares’.

5. Does it make sense to suppress β0 to be equal to zero? Justify your view.

Solution: Suppressing β0 to be equal to zero in a regression model depends on the specific context and the underlying theory of the problem. In some cases, it may make sense to force the y-intercept to be zero, while in other situations, it might not be appropriate.

In many real-world scenarios, a y-intercept of zero might not have any meaningful interpretation. It is essential to let the model estimate the intercept based on the data to capture the actual starting point or baseline. Also, suppressing β0 to be zero can result in a poor-fitting model if the data does not support the assumption that the dependent variable is zero when the independent variables are zero.

6. What is a residual?

Solution: A residual, denoted ei = εbi , estimates of the unobserved error terms εi . It is computed as the difference between the observed value yi and its predicted value from the regression model, ybi , that is: ei = yi − ybi = yi − βb0 − βb1 x.

7. Explain the regression ANOVA decomposition.

Solution: We decompose the variation in y into its explained and unexplained components. It can be shown that the regression ANOVA decomposition is: n X i=1

(yi − ȳ)2 =

n X

βb12 (xi − x̄)2 +

i=1

n X

(yi − βb0 − βb1 xi )2

i=1

or, in words (where ‘SS’ stands for ‘sum of squares’): Total SS = Regression SS + Residual SS. Given our objective is to explain as much of the Total SS as possible, we see that we can decompose the Total SS into two components – the amount we are able to explain using the hypothesised model, called the Regression SS, and the remaining variation which we are unable to explain with the model, called the Residual SS.

2

13.1. Exercises

8. What useful information does an ANOVA regression table provide?

Solution: The standard format of an ANOVA regression table is: Source of variation

Degrees of freedom

Regression

1

Sum of squares n P βb2 (xi − x̄)2 1

Mean sum of squares

F statistic

p-value

(Regression SS)/1

f

P (F1, n−2 > f )

i=1

Residual

n−2

n P

(yi − βb0 − βb1 xi )2

i=1

Total

n−1

n P

(Residual SS)/(n − 2)

(yi − ȳ)2

i=1

where f = Regression MS/Residual MS, i.e. the ratio of the regression and residual mean sums of squares, respectively.

Some important points to note about this table are the following.

• There is only one degree of freedom for regression due to the presence of only one explanatory variable in the model (recall that k = 1).

• The ‘Mean sum of squares’ is the ‘Sum of squares’ divided by the ‘Degrees of freedom’.

• (Residual SS)/(n − 2) is the residual variance and is the estimate of σ 2 , i.e. the variance of the error term (recall that ε ∼ N (0, σ 2 )).

• f is a test statistic value for testing H0 : β12 = 0.

9. Explain how the coefficient of determination is calculated, and how it should be interpreted.

Solution: The coefficient of determination, denoted by R2 , measures the proportion of the total variability in the response variable which is explained by the model. It is defined as: n P βb12 (xi − x̄)2

R2 =

Regression SS = i=1P n Total SS

. (yi − ȳ)2

i=1

Clearly, 0 ≤ R2 ≤ 1 (since 0 ≤ Regression SS ≤ Total SS) and the closer R2 is to 1, the better the explanatory power of the model. In order to interpret R2 in words, we could say that the R2 value gives the ‘proportion of the total variation in the response variable which is explained by the regression model’. If R2 = 1, then we would have a perfect fit with all the data points lying exactly along a straight line. Obviously, this is highly unlikely to happen in practice!

10. Outline the F test for regression ANOVA. Solution: The test statistic for testing H0 : β1 = 0 vs. H1 : β1 6= 0 (which is equivalent to testing H0 : β12 = 0 vs. H1 : β12 > 0) is the ratio of the regression mean sum of squares to the residual mean sum of

3

13. Linear regression

squares (i.e. the residual variance), as given in the regression ANOVA table. Under H0 : β12 = 0, then: Regression MS ∼ F1, n−2 .

F = (Residual SS)/(n − 2) Hence p-values are obtained from the F distribution with 1 degree of freedom in the numerator and n − 2 degrees of freedom in the denominator, called ‘Significance F’ in the Excel regression ANOVA table. If H1 : β12 > 0 is true, it can be shown that E(F ) > 1, and for this reason this is always an upper-tailed test.

11. Outline the t tests for β0 and β1 , respectively.

Solution: For some constant, c, we test: H0 : β0 = c vs. H1 : β0 6= c.

Under H0 , the test statistic is: T =

βb0 − c standard error of βb0

∼ tn−2 .

Hence p-values are obtained from the Student’s t distribution with n − 2 degrees of freedom. A very special case is when c = 0, since if this is true then the intercept equals zero so that the (true) line passes through the origin. However, this would be rare in practice.

For some constant, c, we test: H0 : β1 = c vs. H1 : β1 6= c.

Under H0 , the test statistic is: T =

βb1 − c standard error of βb1

∼ tn−2 .

Hence p-values are obtained from the Student’s t distribution with n − 2 degrees of freedom. A very special case is when c = 0, since if this is true then there is no linear relationship between x and y.

12. Outline how to construct confidence intervals for β0 and β1 , respectively.

Solution: A 100(1 − α)% confidence interval for β0 has the general form: βb0 ± tα/2, n−2 × standard error of βb0 where tα/2, n−2 is the t-value which cuts off 100α/2% probability in the upper tail of the Student’s t distribution with n − 2 degrees of freedom.

A 100(1 − α)% confidence interval for β1 has the general form: βb1 ± tα/2, n−2 × standard error of βb1 where tα/2, n−2 is the t-value which cuts off 100α/2% probability in the upper tail of the Student’s t distribution with n − 2 degrees of freedom.

4

13.1. Exercises

13. Three equivalent tests were presented. When would you argue to use one over another?

Solution: The three equivalent tests are the following.

1. Test H0 : β1 = 0 vs. H1 : β1 6= 0 using the test statistic: βb1 standard error of βb1

∼ tn−2 .

2. Test H0 : β12 = 0 vs. H1 : β12 > 0 using the test statistic: Regression MS ∼ F1, n−2 .

(Residual SS)/(n − 2) 3. Test H0 : ρ = 0 vs. H1 : ρ 6= 0 using the test statistic: √ r n−2 √ ∼ tn−2 .

1 − r2 Test 1 (β1 6= 0) would be used when the primary interest is to determine whether there is a linear relationship between y and x. This two-tailed t test checks if the slope coefficient is significantly different from zero, indicating that the independent variable has a significant effect on the dependent variable and has the benefit of being able to be conducted as an upper-tailed (lower-tailed) test, when a positive (negative) effect of x on y is suspected a priori.

Test 2 (β12 > 0) is a generic test to assess the significance of the regression model (which for the simple linear regression model is identical to the two-tailed Test 1).

Test 3 (ρ 6= 0) would be used when the interest is in assessing whether there is a significant correlation between the two variables. This two-tailed t test with the correlation coefficient r as the test statistic checks if there is a significant linear association between the two variables.

14. For a given x, what is the difference between a confidence interval for E(y | x) and a prediction interval for y | x? Why is the prediction interval wider?

Solution: A confidence interval for E(y | x) = β0 + β1 x, is an interval estimator of an unknown parameter, i.e. for the expectation of y for a given value of x.

A prediction interval for y | x is an interval for a random variable, i.e. for y for a given value of x. \ Although E(y | x) = yb | x, i.e. the point estimate and the point prediction for a given x are the same, there is greater uncertainty in predicting the random variable y | x than there is in estimating the constant E(y | x). Intuitively, it is more challenging to predict a random variable as there is variability in the values it can take. This is reflected in the corresponding error bars in that the prediction interval is wider than the corresponding confidence interval.

15. Which is riskier, interpolation or extrapolation? Justify your view.

5

13. Linear regression

Solution: Interpolation involves estimating values within the range of observed data points. It assumes that the relationship between data points remains consistent within the observed range, making it relatively less risky as it relies on known data.

Extrapolation involves estimating values beyond the range of observed data points. It assumes that the relationship between data points extends beyond the observed range, which introduces greater uncertainty. Extrapolation relies on the assumption that the underlying pattern observed within the known data will hold in unobserved regions, which may not always be accurate. Hence extrapolation would be considered riskier than interpolation.

16. How can a simple linear regression model be used to estimate an elasticity?

Solution: Take logarithms of ‘quantity sold’ and ‘price’ and run the regression: log(quantity sold) = β0 + β1 log(price) + ε.

In this ‘log–log’ model, the interpretation of β1 is the percentage change in quantity for a 1 percentage point increase in price, i.e. this is the price elasticity of demand which can then be estimated.

17. In a simple linear regression model of the form y = β0 + β1 x + ε, where the dependent variable is income (in dollars) and the independent variable is age (in years), the value of β0 was estimated to be −203.56. (a) Interpret this estimate of β0 . (b) Explain why such an estimate could occur if you are told that age and income are highly correlated in the sample data used to run the regression.

Solution: (a) The estimate of β0 represents the estimated value of the dependent variable (income) when the independent variable (age) is zero. However, in this specific case, it’s important to note that the interpretation of β0 may not be practically meaningful since this suggests that an individual has a negative income when age is zero. (b) We are told that age and income are highly correlated (as we might expect). However, we would not expect any data for individuals of zero age, or even for any (young) child, i.e. no data in the region where x = 0. Hence the value of −203.56 is clearly based on extrapolation, so on that basis is not a cause for concern.

18. A car insurance company would like to examine the relationship between driving experience and insurance premium. For this reason, a random sample of ten drivers is taken and their years of driving experience (x) and monthly insurance premium (y, in pounds) is recorded. The data are shown in the table below.

Driver Driving experience (x) Insurance premium (y)

6

#1 6 66

#2 3 88

#3 11 51

#4 10 70

#5 15 44

#6 6 56

#7 25 42

#8 16 60

#9 #10 15 20 45 40

13.1. Exercises

(a) Determine the sample correlation coefficient and interpret its value. (b) Determine the least-squares line of y on x. (c) Based on the regression equation in part (b), what will be the predicted monthly insurance premium for a driver with 12 years of experience? Will you trust this value? Justify your answer.

Solution: (a) Using the CORREL Excel function, we have that r = −0.7872, hence there is a fairly strong, negative linear relationship between insurance premium and driving experience. The more experienced the driver, the lower their insurance premium. (b) Excel regression output is:

Hence the least-squares line is: \ = 78.432 − 1.751 × Experience.

Premium (c) For x = 12. We have: \ = 78.432 − 1.751 × 12 = 57.42 Premium

⇒

£57.42.

Since x = 12 is within the range of x data, this is interpolation and hence this predicted value can be trusted.

19. The data below show the yearly expenditure (in thousands of pounds) by a cosmetics company on advertising a particular brand of perfume.

Year (x) Expenditure (y)

1 170

2 170

3 275

4 340

5 435

6 510

7 740

8 832

(a) Fit a regression line to the data and report a 95% confidence interval for its slope. (b) Perform an F test using the ANOVA table and use this to compute the R2 statistic for the fit.

Comment on the goodness of fit of the linear regression model.

7

13. Linear regression

(c) Predict the expenditure for year 9 and calculate a 95% prediction interval for the actual expenditure.

Solution: (a) Excel regression output is:

The estimated regression coefficients are: βb0 = −9.786 and βb1 = 98.619.

The fitted line is: \ Expenditure = −9.786 + 98.619 × Year.

The standard error of βb1 is 9.521. Hence a 95% confidence interval for β1 is: 98.619 ± 2.447 × 9.521 ⇒ (75.321, 121.917) using =T.INV(0.975,6), and returned in the output above. (b) The ANOVA table is shown in the output above. We test: H0 : β12 = 0 vs. H1 : β12 > 0.

The test statistic value is: Regression MS 408,480.10 = = 107.28 (Residual SS)/(n − 2) 22,845.90/(8 − 2) with a p-value of 4.74 × 10−5 , hence we reject H0 and conclude that there is very strong evidence that β12 > 0.

For the coefficient of determination, i.e. R2 , we have: R2 =

Regression SS 408,480.10 = = 0.9470.

Total SS 431,326

Hence 94.70% of the variation in expenditure can be explained by the year. As R2 is very close to 1, the linear regression model provides a very good fit.

8

13.1. Exercises

(c) Substituting x = 9 we find the predicted year 9 expenditure (in thousands of pounds) is: −9.786 + 98.619 × 9 = 877.785 i.e. £877,785. The estimated standard error of this prediction is: v   u n P s u 2

(x − x) i u   204 u 2  i=1  = 3807.65 × 1 + = 78.227. b × 1 + P uσ n  t 8 × 42 2 n (xj − x̄) j=1

It follows that (using tn−2 = t6 ) a 95% prediction interval for the predicted profit is: 877.785 ± 2.447 × 78.227

⇒

(686.364, 1069.206)

⇒

(£686,364, £1069.206).

20. A company would like to predict how its trainees in sales will perform based on the results of an aptitude test that is given to them at the beginning of the training. The table below contains the test scores (x) and the values of the sales (y, in hundreds of dollars) for nine randomly selected trainees during the first month of working at the company.

Trainee x y

A 1.8 5.4

B 2.6 6.4

C 2.8 6.0

D 3.4 6.2

E 3.6 6.8

F 4.2 7.0

G 4.8 7.6

H 5.2 7.3

I 5.4 7.6

(a) Produce a scatter diagram of these data. (b) Determine the sample correlation coefficient. Interpret its value. (c) Report the least-squares line of y on x. (d) Based on the regression model above, what amount of sales during the first month would you expect from someone who scored 4 in this aptitude test? Would you trust this value? Justify your answer.

Solution: (a) The scatter diagram is:

9

13. Linear regression

(b) Using the CORREL Excel function, we have that r = 0.9491, hence there is a strong, positive linear relationship between sales and test scores, as seen via visual inspection of the scatter diagram. (c) Excel regression output is:

Hence the least-squares line is: [ = 4.520 + 0.580 × Scores.

Sales (d) For x = 4. We have: [ = 4.520 + 0.580 × 4 = 6.84 Sales

⇒

$684.

Since x = 4 is within the range of x data, this is interpolation and hence this predicted value can be trusted.

21. The director of a local tourism authority would like to know whether a family’s annual expenditure on recreation (y), measured in thousands of dollars, is related to their annual income (x), also measured in thousands of dollars. In order to explore this potential relationship, the variables x and y were recorded for 10 randomly selected families who visited the area last year. The results were as follows: Family x y

#1 41.2 2.4

#2 50.1 2.7

#3 52.0 2.8

#4 62.0 8.0

#5 44.5 3.1

#6 #7 #8 37.7 73.5 37.5 2.1 12.1 2.0

#9 #10 56.7 65.2 3.9 8.9

(a) Draw a scatter diagram of these data. (b) Determine the sample correlation coefficient. Interpret your findings. (c) Determine the least-squares line of y on x. (d) Do you find the analyses in parts (b) and (c) appropriate? Justify your answer and suggest any alternative ways to model the relationship between x and y.

10

13.1. Exercises

Solution: (a) The scatter diagram is:

(b) Using the CORREL Excel function, we have that r = 0.9222, hence there is a strong, positive linear relationship between annual recreation expenditure and annual income, as seen via visual inspection of the scatter diagram. (c) Excel regression output is:

Hence the least-squares line is: \ Expenditure = −9.107 + 0.267 × Income. (d) The scatter diagram suggests a non-linear curve rather than a straight line. Another, equivalent, way to note this is the presence of outliers. Hence a linear regression model does not seem to be a good model for the relationship between family annual income and family annual recreation expenditure. An alternative approach would be to consider transformations of the data, for example a logarithmic transformation.

11

13. Linear regression

22. The yearly profits (in thousands of pounds) made by a company over a period of eight consecutive years are shown below.

Year Profit

1 18

2 21

3 34

4 31

5 44

6 46

7 60

8 75

(a) Fit a straight line to these data and compute a 95% confidence interval for the ‘true’ yearly increase in profits. (b) The company accountant forecasts the profits for year 9 to be £90,000. Is this forecast reasonable if it is based on the above data?

Solution: The question implies that we want to explain changes in profitability by the passage of time. If we let x represent years and y represent profits (in thousands of pounds) then we need to perform a regression of y on x. (a) Excel regression output is:

The estimated regression coefficients are: βb0 = 6.679 and βb1 = 7.655.

The fitted line is: \ = 6.679 + 7.655 × Year.

Profit The standard error of βb1 is 0.816. Hence a 95% confidence interval for β1 is: 7.655 ± 2.447 × 0.816 ⇒ (5.658, 9.652) using =T.INV(0.975,6), and returned in the output above. (b) Substituting x = 9 we find the predicted year 9 profit (in thousands of pounds) is: 6.679 + 7.655 × 9 = 75.574

12

13.1. Exercises

i.e. £75,574. The estimated standard error of this prediction is: v   u n P u 2

(xi − x)  s u 204 u 2  i=1   b × 1 + P uσ n  = 27.978 × 1 + 8 × 42 = 6.706. t 2 n (xj − x̄) j=1

It follows that (using tn−2 = t6 ) a 95% prediction interval for the predicted profit is: 75.574 ± 2.447 × 6.706

⇒

(59.165, 91.983)

⇒

(£59,165, £91,983).

As £90,000 is in this prediction interval, we cannot reject the accountant’s forecast out of hand.

However, it is right at the top end of the prediction interval, and hence seems rather optimistic!

23. The manager of a store selling shoes is looking into the association between daily sales (in hundreds of dollars) in the store, y, and the number of customers who visited the store in that day, x. For this reason, in 10 days selected at random the variables x and y were recorded. They appear in the table below: Days # of customers (x) Sales (y)

#1 90 11.2

#2 92 11.1

#3 50 6.8

#4 74 9.2

#5 #6 78 88 9.4 10.1

#7 87 9.4

#8 51 7.7

#9 53 8.2

#10 42 6.1

(a) Draw a scatter diagram of these data. (b) Determine the sample correlation coefficient. Interpret your findings. (c) Determine the least-squares line of y on x. (d) Suppose that you observe more data and when you draw the corresponding scatter diagram a non-linear association is revealed. Discuss how this can be interpreted in the context of the problem.

Solution: (a) The scatter diagram is:

13

13. Linear regression

(b) Using the CORREL Excel function, we have that r = 0.9502, hence there is a strong, positive linear relationship between daily sales and number of customers, as seen via visual inspection of the scatter diagram. (c) Excel regression output is:

Hence the least-squares line is: [ = 3.031 + 0.084 × Customers.

Sales (d) Due to the non-linear shape of the association, alternative ways to model the relationship between y and x would be needed – for example, transforming the data with a logarithmic transformation.

It may be possible that if too many customers come in, not all of them will be willing to buy and some will just visit the shop for browsing.

24. A study into used cars examined whether there is a relationship between their prices (in thousands of dollars) and their mileages (in thousands of miles) when they were sold at auction. A random sample of n = 20 sales was collected and a linear regression was run.

An analyst summarised Excel output for the linear regression model as below.

Predictor Constant Mileage

Coefficient 17.000 −0.055

Residual MS = 0.3192

Standard error 0.3437 0.0093

t statistic 49.4606 −5.9792

p-value 1.10 × 10−20 1.18 × 10−5

R-sq = 66.51%.

(a) Calculate and interpret the sample correlation coefficient. (b) Write down the equation of the estimated regression line. (c) What conclusion can you draw about the connection between the two variables?

14

13.1. Exercises

(d) A used car which has been driven for 35,000 miles will soon be sold at auction. What is its expected selling price to the nearest dollar?

Solution: (a) The sample correlation coefficient, r, can be obtained from R2 , since R2 = r2 for the simple linear regression model. Hence: √ √ r = − R2 = − 0.6651 = −0.8155.

Note that r must be negative due to the negative estimated coefficient, βb1 = −0.055. This means there is a strong, negative, linear relationship between mileage and price. The higher the mileage, the lower the price. (b) The equation of the estimated regression line is: [ = 17.0 − 0.055 × Mileage.

Price (c) Some suggested comments are the following.

• R2 = 66.51% means that 66.51% of the variation in used car prices can be explained by mileage.

• An increase in mileage of 1,000 miles leads to a decrease in price of $55.

• The p-value of the coefficient of mileage is less than 0.05 indicating that mileage is a significant explanatory variable. (d) Given mileage data is in thousands of miles, 35,000 miles means x = 35. We have: [ = 17.0 − 0.055 × 35 = 15.075 Price

⇒

$15,075.

15

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 14: Multiple regression

ii

Chapter 14 Multiple regression 14.1

Exercises

14.1.1

Test your understanding

1. Provide a comparison of ANOVA and regression techniques in terms of their dependent and independent variables.

Solution: A comparison of ANOVA and regression techniques is:

Number of dependent variables and type Number of independent variable(s) and type

One-way ANOVA 1 quantitative 1 categorical

n-way ANOVA 1 quantitative ≥2 categorical

Simple linear regression 1 quantitative 1 quantitative

Multiple regression 1 quantitative ≥2 all quantitative or mix of quantitative and categorical

2. In the multiple linear regression model y = β0 + β1 x1 + β2 x2 + ε, provide brief interpretations of the parameters β0 , β1 and β2 .

Solution: • β0 is the intercept, i.e. the expected value of y when x1 = 0 and x2 = 0 (note that the intercept is often of no interest since it is rare for x1 = 0 and x2 = 0 in practice) • β1 is the change in y when x1 is changed by one unit, while controlling for x2 (that is, with x2 held constant) • β2 is the change in y when x2 is changed by one unit, while controlling for x1 (that is, with x1 held constant).

3. List the assumptions of the multiple linear regression model.

Solution: The assumptions for multiple linear regression can be summarised as follows.

1

14. Multiple regression

• y is accurately modelled as a linear combination of the xi s.

• The random error term, ε, is assumed to have a normal distribution with a mean of zero and a constant variance, σ 2 .

• The error terms, ε, are independent.

• The xi variables and error terms are uncorrelated.

4. Outline details of the joint test of significance in a multiple linear regression with k independent variables.

Solution: We test: H0 : β1 = β2 = · · · = βk = 0 vs. H1 : At least one of the βi s is not zero.

Under H0 , none of the independent variables affects y, whereas if we reject H0 then there is statistical evidence that at least one of the independent variables does affect y. Provided εi ∼ N (0, σ 2 ), then under H0 the test statistic is: F =

(Regression SS)/k ∼ Fk, n−k−1 .

(Residual SS)/(n − k − 1)

We reject H0 at the 100α% significance level if P (F > f ) < α, where F ∼ Fα, k, n−k−1 , i.e. if the p-value is less than α.

5. Is there any difference in (a) the regression ANOVA decomposition, and (b) R2 between the simple and multiple linear regression models?

Solution: (a) For both simple and multiple linear regression models, the ANOVA decomposition can be used to partition the total variation in the dependent variable into different sources of variability.

In a simple linear regression model (with one independent variable), the ANOVA table includes two sources of variability: • explained variation (Regression SS), i.e. the variability in y explained by the single independent variable.

• unexplained variation (Residual SS), i.e. the variability in y not explained by the independent variable and is captured by the error term.

In a multiple linear regression model (with more than one independent variable), the ANOVA table also includes two sources of variability: • explained variation (Regression SS), i.e. the variability in y explained by all the independent variables together.

• unexplained variation (Residual SS), i.e. the variability in y not jointly explained by the independent variables and is captured by the error term. (b) In both simple and multiple linear regression models, R2 is calculated as: R2 =

2

Regression SS Total SS

14.1. Exercises

but multiple linear regression accounts for the contribution of multiple independent variables to the overall explanation of variation in the dependent variable, rather than just one independent variable.

6. What is adjusted R2 used for and how is it calculated?

Solution: Adjusted R2 literally adjusts R2 to take into account the number of independent variables in the linear regression model. It is defined as: 2 =1− Radj

(n − i)(1 − R2 ) n−k

where: • i = 1 if there is an intercept (which is usually the case), and 0 otherwise (if the intercept is suppressed to be zero, i.e. if β0 = 0) • n = the number of observations used to fit the model • k = the number of βi parameters in the model.

7. What is Occam’s razor?

Solution: Occam’s razor is a principle in science and philosophy that suggests simpler explanations should be preferred over more complex ones when multiple hypotheses or theories can explain a phenomenon.

It states that unnecessary assumptions or entities should be avoided, and the simplest explanation that fits the evidence is often the most likely to be correct.

8. Explain the concept of omitted-variable bias.

Solution: Omitted-variable bias occurs in regression models when a relevant independent variable is not included, leading to biased and misleading coefficient estimates. This happens because the omitted variable might be correlated with both the dependent variable and the included independent variable, causing their effects to be confounded. The bias arises when the omitted variable introduces systematic variation in the dependent variable that is not accounted for, resulting in inaccurate parameter estimates. To minimise omitted-variable bias, analysts must identify and include all relevant variables that may influence the dependent variable, ensuring a more robust and accurate analysis.

9. What are the two types of variable misspecification in regression models, and what are their respective consequences?

Solution: 1. Suppose the true relationship is: y = β 0 + β 1 x1 + ε but the estimated model is: yb = βb0 + βb1 x1 + βb2 x2

3

14. Multiple regression

then while the coefficients are unbiased, the standard errors are inefficient (i.e. large) lowering precision.

2. Suppose the true relationship is: y = β 0 + β 1 x1 + β 2 x2 + ε but the estimated model is: yb = βb0 + βb1 x1 then this suffers from omitted-variable bias with invalid standard errors.

10. What are interactions and how are they constructed?

Solution: Interactions refer to the combined effect of two or more independent variables on the dependent variable, which is different from the sum of their individual effects. Interactions are constructed by multiplying the values of the interacting variables and including the interaction term in the regression model. They allow for studying how the relationship between the dependent and independent variables changes based on different combinations of the interacting variables.

Consider two quantitative variables, x1 and x2 . By taking the product we create an interaction variable, x1 x2 . We could then consider the multiple regression model: y = β0 + β1 x1 + β2 x2 + β3 x1 x2 + ε.

11. In the multiple regression model y = β0 + β1 x1 + β2 x2 + β3 x1 x2 + ε, what is the effect of x1 on y?

Solution: The coefficient β1 measures the change in y associated with a one-unit change in x1 , holding all other variables constant.

However, the effect of x1 on y is not fully captured by β1 alone due to the presence of the interaction term x1 x2 in the model. The interaction term implies that the effect of x1 on y also depends on the value of another independent variable x2 . In this case, the overall effect of x1 on y would be a combination of both the main effect (β1 ) and the interaction effect.

12. What is the purpose of a dummy variable and how is it defined? Does the choice of reference category matter?

Solution: Dummy variables represent categorical independent variables in regression analysis. They convert qualitative categories into binary variables (0 or 1). A dummy variable takes the value of 1 if the observation belongs to a particular category and 0 otherwise. The choice of reference category matters insofar as it becomes the baseline for comparison. The coefficients of other dummy variables represent the difference in relation to the reference category. However, while the choice of reference category is arbitrary, when the levels are ordinal (i.e. when they are in a ranked order), it seems reasonable to use the ‘lowest’ level as the reference category.

4

14.1. Exercises

13. If an independent variable is categorical with k = 3 levels, how would this be modelled in a multiple linear regression, and why?

Solution: Suppose we wanted to investigate whether customer satisfaction affected sales where we classify customer satisfaction into one of three levels: (i) satisfied, (ii) neutral, and (iii) dissatisfied. So, this categorical variable has k = 3 levels, which requires us to define k − 1 = 3 − 1 = 2 dummy variables.

Given the ordinal nature of customer satisfaction, we will choose ‘dissatisfied’ as the reference category. The two dummy variables would then be: ( ( 1 if a satisfied customer 1 if a neutral customer x1 = and x2 = 0 otherwise 0 otherwise such that knowing the values of x1 and x2 will allow us to unambiguously determine the satisfaction level: Customer satisfaction level Satisfied Neutral Dissatisfied

x1 1 0 0

x2 0 1 0

Suppose we considered the multiple linear regression model: y = β0 + β1 x1 + β2 x2 +

k X

β i xi + ε

i=3

where x3 , x4 , . . . , xk are other independent variables of interest, resulting in an estimated equation of: k X b b b yb = β0 + β1 x1 + β2 x2 + βbi xi . i=3

14. If an independent variable is categorical with k = 3 levels, explain how you would determine the estimated differences between all possible pairwise comparisons.

Solution: Using the example in Question 13, controlling for all other variables: • βb1 is the estimated difference in y of satisfied customers compared with dissatisfied customers (i.e. the comparison is with respect to the reference category) • βb2 is the estimated difference in y of neutral customers compared with dissatisfied customers.

If we wanted to compare satisfied and neutral customers, the estimated difference in y between these two types would be: βb1 − βb2 .

15. Explain what multicollinearity is, and why it is problematic.

Solution: Multicollinearity is a problem in multiple regression when two or more independent variables are highly correlated, making it challenging for the model to distinguish their individual effects. This

5

14. Multiple regression

creates instability and inflated standard errors of coefficient estimates, leading to difficulty in interpreting the significance of independent variables. Multicollinearity does not affect the overall predictive power of the model but hinders the ability to assess the unique contribution of each variable. It is problematic as it undermines the reliability of regression results and can lead to misleading conclusions.

16. Distinguish between the following model selection techniques: (a) forward selection (b) backward elimination (c) stepwise selection.

Solution: (a) Forward selection: In this model selection technique, the regression starts with an empty model and iteratively adds one independent variable at a time based on some criterion, such as the highest increase in model fit until the stopping criterion is met. It continues until all useful independent variables from a proposed set have been added, and the best model is chosen. (b) Backward elimination: This approach begins with the full model containing the full set of independent variables and then iteratively removes one independent variable at a time based on a criterion, such as the lowest decrease in model fit until the stopping criterion is met. It continues until no more non-useful independent variables can be removed, and the best model is chosen. (c) Stepwise selection: This combines both forward and backward techniques. It starts with an empty model and adds or removes one independent variable at a time based on the criterion. It proceeds in a stepwise manner, and at each step it decides whether to add or remove an independent variable based on the specified criterion, stopping when the criterion is met. The best model is then selected based on this process.

17. A credit card company wants to understand the variation in credit card charges accrued by its customers. While the company profits from these charges, it is concerned that customers with very high charges may be at greater risk of default – a bad outcome for both the customer and the company. It was decided to perform a multiple linear regression. A random sample of n = 1000 current credit card customers was obtained, with information on the following variables: • Annual charges, in pounds • Annual income, in thousands of pounds • Age • Household size • Gender (1 = female, 0 = male) • Exceeded credit limit in past 12 months? (1 = yes, 0 = no).

6

14.1. Exercises

A multiple linear regression was run and the results were:

(a) Write out the full regression model, including any assumptions, and the estimated model. (b) Interpret each estimated coefficient. (c) Perform a joint test of significance and test each independent variable separately. (d) Briefly discuss any changes you would recommend making to the model.

Solution: (a) The theoretical regression model is: Annual charge = β0 +β1 Annual income+β2 Age+β3 Household size+β4 Gender+β5 Credit limit+ε where we assume that ε ∼ N (0, σ 2 ).

The estimated regression model is: \ Annual charge = 1995.76 + 115.91 × Annual income − 21.06 × Age + 512.10 × Household size − 107.17 × Gender + 391.55 × Credit limit. (b) Addressing each estimated coefficient in turn: • βb0 = 1995.76: when all independent variables are equal to zero, the expected annual credit card charge is £1995.76.

• βb1 = 115.91: controlling for all other variables, an increase in annual income of £1,000 leads to a £115.91 increase in annual credit card charges.

• βb2 = −21.06: controlling for all other variables, an increase in a customer’s age by 1 year leads to a £21.06 decrease in annual credit card charges.

• βb3 = 512.10: controlling for all other variables, an increase in household size of 1 person leads to a £512.10 increase in annual credit card charges.

7

14. Multiple regression

• βb4 = −107.17: controlling for all other variables, a female’s annual credit card charge is £107.17 less than a male.

• βb5 = 391.55: controlling for all other variables, a credit card customer who has exceeded their credit limit in the past 12 months has an annual credit card charge £391.55 more than a customer who did not exceed their credit limit. (c) We test: H0 : β1 = β2 = · · · = β5 = 0 vs. H1 : At least one of the βi s is not zero.

The test statistic value is: f=

(Regression SS)/k = 94.89 (Residual SS)/(n − k − 1)

with a p-value of 9.09 × 10−82 . Hence the result is highly significant, with very(!) strong evidence that at least one of the βi s is not zero.

• Testing H0 : β1 = 0 vs. H1 : β1 > 0 (upper-tailed test if we assume higher income individuals spend more, so likely to incur higher credit card charges due to increased use), the p-value is 6.79015 × 10−79 /2 = 3.39508 × 10−79 hence highly significant – strong evidence of a positive income effect on annual credit card charges. Note that the p-values in Excel output are by default for two-tailed tests, hence division by 2 for this upper-tailed test.

• Testing H0 : β2 = 0 vs. H1 : β2 < 0 (lower-tailed test if we assume older people may spend less on credit cards than younger people (due to savings), so likely to incur lower credit card charges due to decreased use), the p-value is 0.15622/2 = 0.07811 hence weakly significant (i.e. reject H0 at the 10% significance level) – weak evidence of a negative age effect on annual credit card charges. Again, note that the p-values in Excel output are by default for two-tailed tests, hence division by 2 for this lower-tailed test.

• Testing H0 : β3 = 0 vs. H1 : β3 > 0 (upper-tailed test if we assume larger households spend more, so likely to incur higher credit card charges due to increased use), the p-value is 7.18653 × 10−12 /2 = 3.59327 × 10−12 hence highly significant – strong evidence of a positive household size effect on annual credit card charges.

• Testing H0 : β4 = 0 vs. H1 : β4 6= 0 (two-tailed test, unless you could argue why one gender would spend more than another), the p-value is 0.7347 hence this is not significant – no evidence of a gender effect on annual credit card charges. Note here we conduct a two-tailed test, so the reported Excel p-value is appropriate.

• Testing H0 : β5 = 0 vs. H1 : β5 > 0 (upper-tailed test if we assume those who exceed their credit limit would be penalised and hence on overage have higher credit card charges due to penalties), the p-value is 0.7068/2 = 0.3534 hence not significant – no evidence of a positive exceeding credit limit effect on annual credit card charges. (d) This is a deliberately open-ended question, so any reasonable model changes could be proposed – for example, suggestions for sensible additional independent variables, as well as proposing removing insignificant variables from the model, such as gender. To note, with R2 = 0.3231 the multiple linear regression model has quite low explanatory power (32.31% of the variation in annual credit card charges can be explained by the model) suggesting there are other important independent variables which have not been included in the model.

8

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 15: Time-series analysis and forecasting

ii

Chapter 15 Time-series analysis and forecasting 15.1

Exercises

15.1.1

Test your understanding

1. Why forecast?

Solution: Forecasting helps decision-makers reach better-informed decisions. Examples include inventory management, production and capacity planning. There are two types of problems: 1. Forecasting when an event occurs. Event timings, such as a recession or surge in consumer demand are important for managers to pre-empt.

2. Forecasting a time series. Business examples include sales revenues, stock price and product demand.

2. What is a time series?

Solution: A time series is a sequence of realised observations of a variable, X, recorded at equidistant points in time, denoted: {x1 , x2 , . . . , xt−1 , xt , xt+1 , . . .} i.e. the time series starts in period 1. By ‘equidistant’ we mean the observations must occur with the same frequency, such as daily, weekly, monthly, annually etc.

3. Provide examples of the possible frequencies of time-series data you would expect to encounter in business.

Solution: In business, time-series data can be collected at various frequencies depending on the nature of the business process, the data’s granularity, and the specific needs of analysis. Some examples of possible frequencies of time-series data encountered in business are the following.

• Hourly frequency: Call centre data, manufacturing process data, and some financial data may be collected at hourly intervals.

• Daily frequency: Sales data, website traffic, stock prices, weather data, social media interactions, and customer orders may be collected on a daily basis.

• Weekly frequency: Retail store footfall, employee attendance, and certain financial metrics may be collected on a weekly basis.

1

15. Time-series analysis and forecasting

• Monthly frequency: Monthly sales reports, utility bills, inventory levels, and financial statements are often reported on a monthly basis.

• Quarterly frequency: Earnings reports, financial statements for some businesses, and government economic indicators are often reported quarterly.

• Annual frequency: Annual financial reports, tax returns, and annual performance reviews may be collected on an annual basis.

4. What is the difference between forecasting and prediction?

Solution: Forecasting uses past data to effectively extrapolate into the future. Using time series analysis, regression and other statistical methods, quantitative estimates of the future can be calculated.

Forecasting techniques are scientific, based on data analysis and calculations.

Prediction can be done by anyone such that predictions are forward-looking statements about what may happen based on judgement, gut instinct or other arbitrary methods, with or without the use of data. Predictions are prone to bias.

5. What is a lead time? Explain what distinguishes whether a lead time is classified as: (a) short-run (b) medium-run (c) long-run.

Solution: Lead time refers to the amount of time between when the forecast is made and the future time period which is being forecast. (a) Short-term forecasting occurs when there is no expectation of a material change in the data-generating process (the process which results in the evolution of the time series) throughout the duration of the lead time. This means that the past may be extrapolated into the future. (b) If a material change is expected during the lead time, then we would be undertaking medium-run forecasting. (c) If we have minimal or no information about the data-generating process towards the end of the lead time then this would constitute long-run forecasting, which ventures into prediction territory as there will need to be a lot of expert opinion incorporated with whatever limited data are available.

6. Outline the features of: (a) extrapolative forecasting models (b) causal forecasting models.

2

15.1. Exercises

Solution: (a) Extrapolative forecasting models are based on the assumption that future patterns in the time series will continue the same as observed in the past. They do not consider the underlying causes or factors driving the data. In an extrapolative model we are basing the forecast only on the history of the time series itself. (b) Causal forecasting models explore the cause-and-effect relationships between the dependent variable (the forecasted variable) and independent variables (predictors) to make predictions.

They incorporate external factors and potential influences on the time series.

7. Which is better, an extrapolative or a causal forecasting model? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends! The choice between an extrapolative and a causal forecasting model depends on several factors, and there is no single right answer. Both approaches have their strengths and weaknesses, and hence there are inevitably trade-offs, with the suitability of each depending on the specific context and objectives of the forecasting task. As a steer, it would be wise to take into account some of the following.

• Data availability: If historic time-series data is readily available and represents stable and consistent patterns, an extrapolative model may be appropriate. However, if the data is limited, and there are external factors influencing the time series, a causal model might be more suitable.

• Understanding causality: If there is a clear understanding of the causal relationships and the factors which drive the time series, a causal model can provide valuable insights into how different variables interact and impact the forecasted variable. In contrast, an extrapolative model does not take causality into account.

• Lead time: For short-term forecasting, where historic patterns are likely to persist, extrapolative models can be effective. However, for long-term forecasting, where external factors and changes in causality may play a significant role, causal models may be more appropriate.

• Complexity of the system: If the system is complex, with multiple interacting variables and uncertain relationships, a causal model can better capture these complexities. However, for simple systems with stable historic patterns, extrapolative models would be more suitable.

• Accuracy and robustness: The performance of both types of models should be evaluated using appropriate metrics and validation methods. The better-performing model in terms of accuracy and robustness for the specific forecasting task should be chosen.

• Interpretability: Causal models offer more interpretability as they provide insights into the underlying relationships and drivers. This can be important for decision-making and understanding the dynamics of the system.

8. Summarise the trade-offs involved in developing forecasts.

Solution: Costs associated with forecasting are: • data collection costs

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15. Time-series analysis and forecasting

• software costs • development costs • forecast error costs.

Benefits associated with forecasting are: • better decisions from accurate forecasts • robust models (insensitive to changing conditions, which therefore give models long lifespans) • automated models, which are desirable.

Weighing up the costs and benefits can be subjective. Nonetheless, it is important to appreciate the trade-offs involved. Schematically, we can summarise the trade-offs between costs and amount of forecasting effort as shown below:

9. Describe each of the following four constituent parts of a time series. For each, provide a real-world example of a time series that you would expect would feature it, briefly justifying why. (a) Trend (b) Seasonal effects (c) Cycles (d) Random noise.

Solution: (a) Trend: this is the long-run direction of the time series. Trends can be linear or non-linear, such as exponential, and it is possible to use a regression-based model.

Example: Quarterly revenue of a successful e-commerce company. The revenue of a successful e-commerce company is likely to show a clear upward trend over time due to factors such as business growth, expanding customer base, and increasing sales. This increasing trend is expected to persist as the company continues to expand its operations and improve its market share. (b) Seasonal effects: for quarterly data these could relate to the traditional ‘seasons’ of spring, summer, autumn and winter (such as for electricity demand) or perhaps proxied by Q1, Q2, Q3 and Q4, but a ‘season’ would depend on the data frequency.

Example: Ice cream sales of a local ice cream parlour. Ice cream sales typically exhibit seasonal patterns, with higher demand during warmer months and lower demand during colder months.

4

15.1. Exercises

The seasonal effects are driven by weather conditions, reflecting people’s preference for ice cream in hot weather and reduced consumption during colder seasons. (c) Cycles: the long-run (business) cycles which are distinct from seasonal effects. Think economic booms and recessions.

Example: Housing market prices in a growing urban area. The housing market prices in a growing urban area may experience cyclical patterns due to factors like economic cycles, interest rates, and housing market demand. Periods of rising prices (boom) may alternate with periods of declining prices (bust), influenced by economic conditions and demand fluctuations. (d) Random noise: the random fluctuations which inevitably occur, as well as ‘blips’ and market crashes etc.

Example: Daily fluctuations in stock prices of a stable and well-established company. Stock prices can be influenced by various unpredictable factors, such as market sentiment, news events, and investor behaviour. While stock prices may exhibit long-term trends, short-term fluctuations are often driven by random noise, making it difficult to predict exact day-to-day movements.

10. What is the purpose of a moving average?

Solution: The purpose of a moving average is to smooth out fluctuations in time series data and highlight underlying trends or patterns. It calculates the average of a fixed number of consecutive data points, creating a new series that reduces noise and makes the overall trend more apparent, aiding in forecasting and analysis.

11. Explain how a five-point unweighted moving average is calculated. If instead you wanted to use a five-point weighted moving average, how would you choose the weights to use? Briefly justify your view.

Solution: A 5-point unweighted moving average is calculated as: 2

Xt−2 + Xt−1 + Xt + Xt+1 + Xt+2 1X Xt+i = .

Tt = 5 i=−2 5 Note that Tt is based on the average of the current value, Xt , and the two values immediately before, Xt−1 , Xt−2 , and after, Xt+1 , Xt+2 .

For a five-point weighted moving average, you assign weights to each data point. The choice of weights depends on the characteristics of the data and the importance given to different points. For instance, if recent data is considered more relevant, higher weights could be assigned to the most recent points. The choice of weights is subjective and should be based on the context and goals of the analysis.

12. What is a centred moving average? If you had quarterly data, explain how the centred moving average would be calculated.

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15. Time-series analysis and forecasting

Solution: If there are N periods, and N is even, then we require a centred moving average, since there is no midpoint. For quarterly data, N = 4, hence we have: Ttc =

Tt + Tt+1 (Xt−2 + Xt−1 + Xt + Xt+1 )/4 + (Xt−1 + Xt + Xt+1 + Xt+2 )/4 = 2 2 =

Xt−2 + 2(Xt−1 + Xt + Xt+1 ) + Xt+2 .

8

13. Why could it be argued that a multiplicative model is preferable to an additive model?

Solution: A multiplicative model may be preferred over an additive model when the relationship between the variables is not constant and changes proportionally with the level of the predictors. Multiplicative models are more suitable when the variance increases with the mean, making them a better fit for data with non-constant variability.

14. Write out the multiplicative model featuring trend and seasonality.

Solution: The time series observation Xi,j is modelled as: Xi,j = Ti,j Sj + εi,j for j = 1, 2, . . . , N , where S1 , S2 , . . . , SN are the seasonal effects such that: N X

Sj = N

j=1

and εi,j is the noise component. By assumption, E(εi,j ) = 0 for all i, j.

15. What is meant by deseasonalised data and how can a simple linear regression be run using it?

Solution: Deseasonalised data refers to time-series data from which the seasonal component has been removed, leaving only the underlying trend and irregular variations. It is obtained by dividing the original data by the corresponding seasonal factors. If we assume a linear trend, then we can run a simple linear regression of the deseasonalised data, Dt = Di,j , as the dependent variable on a time index, t, as the independent variable, where: Dt = Di,j =

Xi,j Sbj

i.e. we run the regression: Dt = β0 + β1 t + εt .

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15.1. Exercises

16. How can we assess forecast accuracy today when forecasts are of the future which hasn’t yet happened?

Solution: The accuracy of forecasts today can be judged based on forecast errors by comparing the actual bt , to see how well such ‘retrospective history of Xt values with the same-period forecasts, X forecasting’ performed.

17. Define a forecast error.

Solution: bt , the forecast error in For an actual historic value, Xt , and its respective same-period forecast, X period t is: bt Xt − X for t = 1, 2, . . . , T .

18. Explain how each of the following forecast accuracy measures is calculated and mention a distinctive feature of each. (a) Root mean squared error (b) Mean absolute error (c) Mean absolute scaled error (d) Mean absolute percentage error (e) Akaike information criterion.

Which is your favourite, and why?

Solution: (a) The root mean squared error (RMSE) is calculated by taking the (positive) square root of the average (mean) squared error (hence the name). The formula is: v u T u1 X bt )2 .

RMSE = t (Xt − X T t=1 Due to the squaring, large forecast errors are penalised heavily, i.e. when our model is wrong by a wide margin, we wish to ‘punish’ the model for such big mistakes. The smaller the RMSE the better. (b) The mean absolute error (MAE) is calculated by taking the average (mean) of the absolute errors, i.e. ignoring the forecast error sign (overestimates and underestimates are not distinguished). The formula is: T 1X bt |.

MAE = |Xt − X T t=1 Without any squaring, the MAE does not punish large forecast errors as heavily as RMSE. The smaller the MAE the better.

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15. Time-series analysis and forecasting

(c) The mean absolute scaled error (MASE) is a normalised metric which takes the form of a ratio comparing model forecast errors with the naive forecasts which are simply the previous value of bt = Xt−1 , i.e. naive forecasts assume no change between the current period the variable. Hence X and the next period – simple, but naive! So, a naive forecast model has a MASE of 1, with superior forecasting models having a MASE less than 1 (the smaller the better). The formula is: T

MASE =

1X bt | |Xt − X T t=1 T 1 X |Xt − Xt−1 | T − 1 t=2

.

The smaller the MASE the better. (d) The mean absolute percentage error (MAPE) considers the magnitude of the forecast error with respect to Xt in terms of a percentage. It is undefined if Xt = 0 in any period – a problem not encountered with MASE. The formula is: T bt 1 X Xt − X MAPE = 100 × .

T t=1 Xt

The smaller the MAPE the better. (e) The Akaike information criterion (AIC) works a bit like adjusted R2 in regression, in that we have to pay a ‘complexity penalty’ such that the more parameters the forecasting model has, we have to pay an estimation price. The formula is:   T P 2 b  t=1(Xt − Xt−1 )   + 2(k + 1) AIC = T × ln    T where k is the number of model parameters. The smaller the AIC the better.

19. Outline how exponential smoothing works.

Solution: In exponential smoothing, the estimate of the one-period-ahead trend, Tbt+1 , based on the observed series data up to the present period t, is given by: Tbt+1 = αXt + α2 Xt−1 + α3 Xt−2 + · · · where α is a smoothing constant/parameter, which takes values between 0 and 1. For any 0 < α < 1, we have that: α > α2 > α3 > · · · such that less weight is placed on more historic data. Effectively, as data becomes ‘older’ it is deemed less valuable and so it is essentially ‘discounted’ more heavily. It can be shown that: Tbt+1 = Tbt + α(Xt − Tbt ).

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15.1. Exercises

In practice, this means that the estimate of the level at time t + 1 (made in period t) equals the previous estimate Tbt (i.e. the estimate of the trend at time t made at time t − 1) plus α times the previous forecast error, Xt − Tbt (i.e. the difference between the realised observation in period t and its forecasted value made in the previous period).

We can represent this basic ‘updating equation’ as: new estimate = old estimate + smoothing parameter × (outcome − old estimate).

20. How would you choose the value of the smoothing parameter, α, in exponential smoothing? Briefly justify your view.

Solution: The smoothing parameter α can be chosen based on the solution to an optimisation problem, whereby we choose a suitable objective function. One option is to use the root mean squared error (RMSE) as our chosen quality metric. Since we would wish the RMSE to be as small as possible, this means we have a minimisation problem. Specifically: T

1X min (Tt − Tbt )2 0≤α≤1 T t=2 where: Tbt = αXt−1 + (1 − α)Tbt−1 .

21. What in-built forecast options does Tableau have?

Solution: Tableau’s built-in forecast options include: • Forecasting with trend models, such as exponential smoothing and linear trend, to project future values based on the historic trend of the data.

• Forecasting with seasonal models, which can capture seasonal patterns in the data, such as weekly, monthly, or yearly seasonality.

• Forecasting with confidence intervals, allowing you to visualise forecasted values along with upper and lower confidence intervals, providing a range of uncertainty around the forecast.

22. How does Tableau classify forecasts as being ‘Good’, ‘Ok’ or ‘Poor’ ?

Solution: Tableau provides forecast model summaries, including a classification of their quality. The categories are ‘Good’, ‘Ok’ and ‘Poor’ and are judged relative to the naı̈ve forecast based on the current observation only. Relative to the naı̈ve forecast: • ‘Good’ means that the forecast has less than half as much error • ‘Ok’ means that the forecast is likely to have less error • ‘Poor’ means the forecast has more error.

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15. Time-series analysis and forecasting

23. Suggest, with reasons, your recommended time-series model for forecasting: (a) gas demand (b) a stock market index.

Solution: (a) Gas demand often exhibits seasonality due to weather patterns and usage variations. A long-run trend may also be evident, hence a multiplicative model of trend and seasonality may be suitable. (b) Stock markets can be influenced by various external factors, so a causal forecasting model might seem appropriate.

24. Your manager has provided you with the following history of quarterly sales (in millions of pounds) over the past four years.

Year 1 Year 2 Year 3 Year 4

Q1 23.16 23.41 24.04 24.88

Q2 22.99 23.87 24.90 25.64

Q3 23.37 24.19 24.27 25.46

Q4 22.19 23.55 22.97 24.83

(a) Create sales forecasts for each quarter in years 5 and 6 using a multiplicative model featuring trend and seasonality. (b) Prepare a short memo for your manager explaining your approach and briefly discuss how reliable you expect the forecasts to be.

Solution: (a) Spreadsheet workings are:

The estimated trend line using deseasonalised data is: b i,j = 22.472 + 0.178t.

D

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15.1. Exercises

For years 5 and 6, the forecasts are (with seasonal adjustments applied): Year 5 Year 6

Q1 25.58 26.30

Q2 26.28 27.01

Q3 26.18 26.90

Q4 25.02 25.71

(b) Dear Manager, I have prepared sales forecasts for each quarter in years 5 and 6 using a forecasting model which takes into account the trend and seasonality in the historic sales data.

I applied the multiplicative model with both trend and seasonality components to the quarterly sales data from the past four years. The forecasts for each quarter in years 5 and 6 are provided in the table below: Year 5 Year 6

Q1 25.58 26.30

Q2 26.28 27.01

Q3 26.18 26.90

Q4 25.02 25.71

To compare these against historic data, and back-dated forecasts, we have:

Considering that the historic sales data has shown both a clear increasing trend and seasonality, it is appropriate to use these features in developing a forecasting model. However, please note that the accuracy and reliability of the forecasts may vary depending on the stability and continuity of the underlying patterns in sales. Changes in market conditions or other external factors might affect the reliability of the forecasts, especially for the longer-term forecasts.

If you have any further questions or require additional analysis, please feel free to reach out.

Sincerely, Forecasting Superstar

25. A wholesaler wants to use exponential smoothing to predict monthly sales. At the end of January, the wholesaler’s forecast for March sales was 11,000. In February, 15,000 units were sold, and during March 19,000 units were sold. (a) At the end of March, what is the wholesaler’s forecast for the total number of units which will be sold during the following May and June? Use a smoothing constant of 0.25.

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15. Time-series analysis and forecasting

(b) If the value of 0.25 for the smoothing constant could be changed, briefly explain how a different value could be determined.

Solution: (a) With α = 0.25 as the smoothing constant, we use: Tbt+1 = αXt + (1 − α)Tbt .

The forecast for the total number of units sold for May and June is 36,125 units. (b) The choice of the smoothing constant depends on the desired trade-off between responsiveness to recent changes and stability of the forecast. A higher value (closer to 1) will give more weight to recent observations, making the forecast more responsive to short-term fluctuations but potentially less stable. A lower value (closer to 0) will put more weight on historic observations, leading to a more stable forecast but potentially slower to adapt to recent changes.

To determine a different value for the smoothing constant, the wholesaler can use a technique like a grid search. By trying different values of the smoothing constant and evaluating the forecast accuracy (using measures like RMSE or MAE), they can identify the value that performs best for their specific forecasting needs. It’s essential to find a balance that provides the most accurate and reliable forecast for their sales prediction.

26. Source any time series of your choice online, and create forecasts for it!

Solution: This one is over to you!

12

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 16: Discriminant analysis

ii

Chapter 16 Discriminant analysis 16.1

Exercises

16.1.1

Test your understanding

1. In a binary classification context, what are the four possible types of classification?

Solution: The possible types of classification are shown in the matrix below:

Actual type ‘Negative’ ‘Positive’

Predicted type ‘Negative’ ‘Positive’ Correct prediction Incorrect prediction (true negative) (false positive) Incorrect prediction Correct prediction (false negative) (true positive)

2. In discriminant analysis which is worse, a false positive or a false negative? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends! Whether a false positive or a false negative is worse depends on the specific context and the consequences of making each type of error. Let’s consider both scenarios.

A false positive occurs when the discriminant analysis incorrectly classifies an observation into a group when it actually belongs to another group. In other words, it is a case of a positive prediction when the true condition is negative. This is worse when: • the cost of taking unnecessary actions or interventions is high.

• making false positive decisions may lead to significant financial losses or operational disruptions.

For example, in a credit card fraud detection system, a false positive occurs when a legitimate transaction is mistakenly flagged as fraudulent. This can result in an inconvenience for the customer and may lead to the customer’s account being temporarily blocked, affecting customer satisfaction and trust in the business.

A false negative occurs when the discriminant analysis fails to detect an observation that belongs to a particular group. In other words, it is a case of a negative prediction when the true condition is positive. This is worse when: • the cost of missing a critical event or condition is high.

• failing to identify important patterns or risks can lead to severe consequences.

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16. Discriminant analysis

For example, in medical diagnostics, a false negative may occur when a patient has a certain medical condition, but the diagnostic test fails to detect it. This can result in a delayed or incorrect diagnosis, leading to potentially severe health implications for the patient.

3. Can an object/case belong to more than one group in a discriminant analysis? Explain why or why not.

Solution: No, in discriminant analysis an object/case cannot belong to more than one group because the groups are mutually exclusive. At the classification stage, each object is assigned to the group (and only one group) with the highest posterior probability based on its feature values.

4. If you wanted to build a model to discriminate between defaulters and non-defaulters of a loan, which predictor variables would you recommend including in the model? Briefly justify your view.

Solution: Some recommended predictor variables are: • Credit score: A crucial factor in assessing creditworthiness, as lower credit scores are associated with higher default risk.

• Debt-to-income ratio: This measures the borrower’s ability to handle additional debt, and high ratios may indicate a higher likelihood of default.

• Employment status and income: Stable employment and higher income provide better repayment capacity and reduce default risk.

• Loan amount and term: Larger loans or longer repayment terms may pose higher risk if the borrower’s financial situation is uncertain.

• Past payment history: Delinquencies or late payments on prior loans indicate a higher risk of default.

• Purpose of loan: Different loan purposes (mortgage, personal, business) may have varying default rates, so considering the loan’s intended use is essential.

• Number of dependents: A higher number of dependents may impact the borrower’s ability to repay the loan.

• Age and education Level: These variables can reflect the borrower’s stability and earning potential.

Including the above predictor variables allows the model to capture various aspects of the borrower’s financial health and creditworthiness, enabling better discrimination between defaulters and non-defaulters. By using a combination of these variables, the model can provide a more comprehensive assessment of default risk and help lenders make informed decisions on loan approvals and terms.

5. For g groups and k predictor variables, how many discriminant functions will there be?

Solution: The number of discriminant functions will be min(g − 1, k), i.e. the smaller of the two values g − 1 and k.

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16.1. Exercises

6. What is a ‘good’ value of k, the number of predictor variables? Briefly justify your view.

Solution: Well, James, it depends! The larger the value of k, the greater the number of predictor variables and hence jointly these should be able to discriminate better. However, simpler models are appealing, which would favour a small value of k. In practice, k = 5 or 6 may be an appropriate compromise.

7. For the discriminant function Dj = β0 + β1 X1,j + β2 X2,j + · · · + βk Xk,j define all terms.

Solution: The discriminant function can be represented as: Dj = β0 + β1 X1,j + β2 X2,j + · · · + βk Xk,j where: • Dj is the discriminant score of the jth case (out of n cases), hence for j = 1, 2, . . . , n • X1 , X2 , . . . , Xk are the k predictor variables used to discriminate the groups • βi , for i = 0, 1, 2, . . . , k, are the discriminant function’s unstandardised coefficients, i.e. the weights attached to each predictor variable (plus the constant β0 ) to determine the discriminant score.

8. For g = 2 groups, how can a classification decision rule be developed based on the centroids?

Solution: The centroids are the averages (means) of the discriminant scores for all cases in each group. When g = 2, we simply use the average of the two centroids (D̄G1 and D̄G2 , one per group) as the cut-off point, c, where: D̄G1 + D̄G2 . c= 2 For any new case, say the (n + 1)st case, we compute the discriminant score: Dn+1 = B0 Xn+1 and classify this case to the nearest centroid by comparison to c.

9. What is the purpose of a confusion matrix?

Solution: The purpose of a confusion matrix is to evaluate the performance of a classification model by providing a summary of its predictions compared to the actual outcomes. It presents the counts of true positive, true negative, false positive, and false negative predictions, allowing us to calculate important metrics such as the overall predictive accuracy. The confusion matrix helps assess the model’s ability to correctly classify data into different classes and identify potential areas for improvement in the model’s predictions.

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16. Discriminant analysis

10. Consider the following confusion matrix for a two-group classification problem:

Actual group

Negative Positive Total

Predicted group Negative Positive n11 n12 n21 n22 n·1 n·2

Total n1· n2· n

Describe in words what each n value represents.

Solution: • n11 is the number of correct classifications of negative cases (true negatives).

• n12 is the number of incorrect classifications of negative cases (false positives).

• n21 is the number of incorrect classifications of positive cases (false negatives).

• n22 is the number of correct classifications of positive cases (true positives).

• n1· = n11 + n12 is the total number of negative cases.

• n2· = n21 + n22 is the total number of positive cases.

• n·1 = n11 + n21 is the total number of negative predictions.

• n·2 = n12 + n22 is the total number of positive predictions.

• n = n1· + n2· = n·1 + n·2 is the total number of cases (and hence also the number of predictions).

11. With respect to the confusion matrix in Question 10, explain how to calculate each of the following: (a) the overall predictive accuracy (b) the true positive rate (c) the true negative rate (d) the false positive rate (e) the false negative rate.

Solution: • The overall predictive accuracy:

n11 + n22 . n returns the proportion of correctly classified cases.

OPA =

• The true positive rate: TPR =

n22 n22 = n21 + n22 n2·

returns the proportion of correctly classified positive cases.

• The true negative rate: TNR =

n11 n11 = n11 + n12 n1·

returns the proportion of correctly classified negative cases.

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16.1. Exercises

• The false positive rate:

n12 n12 = n12 + n22 n·2 returns the proportion of incorrect positively classified cases.

FPR =

• The false negative rate:

n21 n21 = n11 + n21 n·1 returns the proportion of incorrect negatively classified cases.

FNR =

12. At a high level, how does leave-one-out cross-validation work? You should refer to training and test data in your response.

Solution: Leave-One-Out Cross-Validation (LOOCV) is a validation technique used to assess the performance of a model without biasing the results with a specific train–test split. In LOOCV, for each data point in the dataset, the model is trained on all other data points, leaving out the one under consideration (leave-one-out). The model’s performance is then evaluated by testing its prediction on the left-out data point. This process is repeated for all data points, and the average performance across all iterations provides an unbiased estimate of the model’s generalisation ability on unseen data. LOOCV minimises data wastage but can be computationally expensive for large datasets.

13. What is meant by ‘overfitting’ in a discriminant analysis context?

Solution: In discriminant analysis, overfitting refers to a situation where the model becomes excessively tailored to the training data, capturing noise and random variations rather than genuine underlying patterns. As a result, the model may perform well on the training data but poorly on new, unseen data. Overfitting can lead to misleading and unreliable predictions as the model lacks generalisation ability, compromising its effectiveness in classifying new observations accurately.

14. When trying to identify ‘good’ predictor variables for discriminant analysis only using descriptive statistics, what would we be looking out for?

Solution: When using descriptive statistics to identify ‘good’ predictor variables for discriminant analysis, we would be looking for variables which exhibit significant differences between different groups or classes. Key descriptive statistics to consider include means, standard deviations, and ranges for each group. ‘Good’ predictor variables should have distinct and meaningful variations between the groups, indicating their potential to discriminate between the classes effectively in the discriminant analysis model.

15. What is the difference between unstandardised coefficients and standardised coefficients? For what purpose is each type useful?

Solution: Unstandardised coefficients in discriminant analysis represent the raw regression coefficients obtained during model estimation. They reflect the change in the dependent variable associated with a one-unit change in the predictor variable, without considering the scale of the variables.

5

16. Discriminant analysis

Standardised coefficients are obtained by standardising the variables, making them comparable on the same scale. They represent the change in the dependent variable associated with one standard deviation change in the predictor variable.

Both unstandardised and standardised coefficients have their uses in discriminant analysis.

Unstandardised coefficients can be helpful for interpretability and understanding the practical implications of the model. Standardised coefficients are useful for comparing the relative importance of predictors and identifying which variables have a more significant effect on the outcome, regardless of their original scale. They also allow for easier comparison of coefficients between different models or datasets with varying scales of variables.

16. How would you check whether multicollinearity may be a concern in a discriminant analysis model?

Solution: To check for multicollinearity in a discriminant analysis model, you can calculate the correlation matrix for the predictor variables. High correlations between predictors indicate a potential problem of multicollinearity. Additionally, you can use exploratory data analysis and domain knowledge to identify potential redundant or highly correlated predictors.

17. What are the group centroids, and what purpose do they serve?

Solution: Group centroids are the average discriminant scores for each group. They are used to classify cases such that a new case is classified to the group whose centroid it is nearest. Centroids reflect a typical/average/representative member of the respective group, hence it is sensible to classify cases based on its similarity to such a typical/average/representative member.

18. Explain how discriminant scores would be calculated: (a) for a model discriminating between g = 2 groups (b) for a model discriminating between g = 3 groups.

Solution: (a) For g = 2 groups, the discriminant score of the jth case is: Dj = B0 Xj where B is the vector of unstandardised coefficients, and Xj is the multivariate observation for individual j. (b) For g = 3 groups, the pair of discriminant scores of the jth case is: D1,j = A0 Xj

and D2,j = B0 Xj .

where A and B are the vectors of unstandardised coefficients for the first and second discriminant functions, respectively, and Xj is the multivariate observation for individual j.

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16.1. Exercises

19. Would you expect the overall predictive accuracy to be higher or lower when based on cross-validated cases compared to original cases? Briefly justify your view.

Solution: In general, the overall predictive accuracy based on cross-validated cases is expected to be lower than that based on original cases. Cross-validation involves training the model on a subset of the data and testing on the remaining data, providing a more realistic estimate of the model’s generalisation ability. It helps to detect overfitting and provides a better indication of how well the model will perform on new, unseen data, resulting in a slightly lower accuracy.

20. If you had confusion matrices for several different discriminant analysis models (trying to classify the same distinct groups), how would you propose to compare the performances across the models?

Solution: It would depend on the practical consequences of different types of misclassification. For example, with g = 2, false positives and false negatives are the two possible misclassifications, but their practical severities are likely asymmetric, i.e. one is worse than the other. As such, the different performance measures should be carefully considered.

Ceteris paribus, we would want the highest overall predictive accuracy, but this might suggest a model over an alternative model with a lower false positive rate, say, which might be undesirable if we deem false positives to be especially egregious.

7

Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 17: Factor analysis

ii

Chapter 17 Factor analysis 17.1

Exercises

17.1.1

Test your understanding

1. What is a multivariate random variable? Provide an example.

Solution: A multivariate random variable is a random variable that consists of multiple components or variables, each of which can take on different values. In other words, it represents a collection of random variables that are jointly distributed.

Consider a retail business that operates multiple stores in different locations. The business can define a multivariate random variable to represent the monthly sales of products in each store. The components of the multivariate random variable would be the sales of individual products (clothing, electronics, accessories) in each store, and the joint distribution would capture the relationship and variability of sales across all stores and product categories. This information can help the business in making inventory decisions, identifying sales patterns, and optimising product assortments in each store.

2. For each of the following, what ‘factors’ influence your purchasing decision? Briefly explain why in each case. (a) Shopping for food. (b) Booking a holiday destination. (c) Choosing where to live.

Solution: (a) Factors influencing the purchasing decision for food would likely include: • Price: Affordability and budget considerations affect the choice of food products.

• Quality: Desired taste, freshness, and nutritional value impact the decision.

• Brand and reputation: Trust in well-known brands or trusted suppliers influences choices.

• Dietary preferences: Personal dietary needs, cultural, or health considerations play a role.

• Convenience: Proximity to stores and ease of accessibility affect shopping habits. (b) Factors influencing the choice of a holiday destination would likely include: • Budget: The available budget determines the range of destinations to consider.

• Activities and attractions: Desired activities, sights, and experiences influence the decision.

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17. Factor analysis

• Climate: Preferred weather conditions for the period of travel impact the choice.

• Safety and security: Perceptions of safety at the destination play a crucial role.

• Reviews and recommendations: Online reviews and word-of-mouth recommendations influence decisions. (c) Factors influencing the decision of where to live would likely include: • Location and proximity: Access to work, schools, amenities, and family/friends impacts the choice.

• Housing affordability: Affordability of homes and rent in the area affects the decision.

• Neighbourhood safety: Crime rates and safety perceptions are essential considerations.

• Quality of schools: Availability and quality of educational institutions matter for families with children.

• Lifestyle and community: The atmosphere, culture, and community vibe are significant factors.

3. Suppose you are a clothing retailer interested in people’s degree of fashion consciousness. How would you define fashion consciousness? Is it directly observable? If not, identify a plausible set of observed variables that you would expect to be highly correlated with fashion consciousness.

Solution: Fashion consciousness refers to an individual’s awareness, interest, and involvement in current fashion trends and styles. It is not directly observable but can be inferred from various observed variables. Some plausible observed variables highly correlated with fashion consciousness could include: • Frequency of clothing purchases: Fashion-conscious individuals are likely to buy clothes more frequently.

• Following fashion influencers: A strong interest in fashion may lead to following fashion influencers on social media.

• Clothing brands: Preference for trendy and fashionable clothing brands.

• Fashion magazines and websites: Regular reading of fashion magazines and visiting fashion websites.

• Social media engagement: Active engagement in fashion-related content and discussions on social media platforms.

4. In what sense does factor analysis facilitate ‘data reduction’ ?

Solution: Factor analysis facilitates ‘data reduction’ by identifying underlying latent factors that explain the shared variance among observed variables. It uncovers the common underlying dimensions in the data, reducing the original variables to a smaller set of factors. This allows for a more concise representation of the data while preserving the essential information captured by the identified factors. We reduce the ‘dimensionality’ of the dataset columnwise, such that the underlying factors are ‘recovered’ from the larger number of observed variables.

2

17.1. Exercises

5. Have you taken a Myers–Briggs personality test yet? If yes, do you agree with the result? If no, have a go! Whichever test you take, can you identify which personality trait is linked to each survey question?

Solution: Over to you!

6. To run a factor analysis, should the observed variables be correlated or uncorrelated? Briefly explain why.

Solution: To run a factor analysis successfully, the observed variables should be correlated, ideally with some pairs of variables highly correlated. Factor analysis is designed to identify underlying latent factors that explain the shared variance among the observed variables. If the observed variables are uncorrelated, there is no common variance to extract, and factor analysis will not be effective in identifying underlying factors!

7. How are observed variables ‘standardised’ ?

Solution: Each observed variable, Xj , for j = 1, 2, . . . , k, is standardised, denoted below as Xj0 , by subtracting its (sample) mean, X̄j , and dividing by its (sample) standard deviation, Sj . Hence: Xj0 =

Xj − X̄j .

Sj

8. What is a factor loading?

Solution: A factor loading in factor analysis represents the correlation between an observed variable and a latent factor. It indicates the strength and direction of the relationship between the observed variable and the underlying factor. Higher factor loadings (close to 1 or −1) indicate a stronger association between the variable and the factor, suggesting that the variable is more closely related to the factor being extracted. Factor loadings help interpret the influence of each variable on the identified factors.

9. What is a communality?

Solution: A communality refers to the proportion of variance in an observed variable that can be explained or accounted for by the underlying factors extracted from the data. It represents the common variance shared between the observed variable and all the factors. A high communality value (close to 1) indicates that the observed variable is well-represented by the factors, while a low value (close to 0) suggests that most of the variance in the variable is unique and not explained by the common factors.

The variance of each standardised variable, Xj0 , (which is 1, due to the standardisation) can be written as: 2 2 2 +Var(Vj ).

1 = Var(Xj0 ) = α1j + α2j + · · · + αmj {z } | communality

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17. Factor analysis

10. For what purpose could factor scores be used, and why?

Solution: Factor scores are used to represent the values of the underlying latent factors for each individual or observation in the dataset. These scores are obtained through the factor analysis process and can serve several purposes. Some important ones are the following.

• Dimension reduction: Factor scores provide a more compact representation of the data, reducing the number of variables to a smaller set of factors while retaining most of the original data’s information.

• Variable substitutes: Factor scores can be used as substitute variables in subsequent analyses (such as multiple regression and discriminant analysis), simplifying the modeling process and avoiding issues related to multicollinearity when dealing with highly correlated variables.

• Characteristic profiles: Factor scores allow for the comparison and profiling of individuals based on their positions along the latent dimensions, facilitating the identification of distinct groups or segments.

11. What is a ‘good’ value for the Kaiser–Meyer–Olkin statistic?

Solution: A ‘good’ value for the Kaiser–Meyer–Olkin (KMO) statistic is above 0.5. A KMO value greater than 0.5 indicates that the observed variables are suitable for factor analysis, suggesting an adequate level of common variance that can be extracted to identify underlying factors.

12. What does Bartlett’s test of sphericity actually test?

Solution: Bartlett’s test of sphericity tests (for k = 5):  ρ11 ρ12 ρ13 ρ14 ρ21 ρ22 ρ23 ρ24  H0 :  ρ31 ρ32 ρ33 ρ34 ρ41 ρ42 ρ43 ρ44 ρ51 ρ52 ρ53 ρ54

  ρ15 1   ρ25  0  ρ35   = 0 ρ45  0 ρ55 0

0 1 0 0 0

0 0 1 0 0

0 0 0 1 0

 0 0  0  0 1

i.e. that the true correlation matrix for the population is the ‘identity matrix’ with 0s off the leading diagonal (and 1s along the leading diagonal), which would correspond to uncorrelated observed variables in the population, against H1 that the true correlation matrix is not the identity matrix, meaning at least some of the observed variables are truly correlated (whether positively or negatively).

13. When reducing k observed variables to m underlying factors, with m < k, how can the value of m be determined?

Solution: The value of m, can be determined using various methods.

• Retain factors with eigenvalues greater than 1, as they explain more variance than a single observed (standardised) variable.

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17.1. Exercises

• Plot the eigenvalues in descending order using a scree plot and look for a point where the plot levels off (the ‘elbow’).

• Retain factors that cumulatively explain a significant proportion of the total variance (at least 60%).

• Consider the practical and theoretical meaning of the factors and select meaningful and interpretable solutions.

• Choose a smaller number of factors to achieve simpler and more parsimonious models.

The selection of the appropriate value of m should be based on a combination of these methods and good analyst judgement.

14. What does a scree plot show?

Solution: A scree plot shows the eigenvalues of the factors in descending order. It helps identify the point where the plot levels off, indicating the number of factors to retain in factor analysis.

15. At a high level, what is meant by ‘factor rotation’ ?

Solution: Factor rotation in factor analysis is a process of transforming the original factors to achieve a more interpretable and simple structure. It aims to maximise the variance of the loadings of each factor and reduce cross-loadings, making the factors easier to interpret and aligning them more closely with the underlying theoretical constructs they represent. The result is strong correlations (whether positive or negative) between the factors and some of the variables, with very weak correlations with the other variables, meaning there is little ambiguity about how to name the factors.

16. What is the connection, if any, between the rotated component matrix and the factor loading plot?

Solution: The rotated component matrix and the factor loading plot are closely connected. Both represent the relationships between observed variables and the rotated factors after the factor rotation process.

The rotated component matrix contains the factor loadings of each observed variable on the rotated factors. The loadings represent the strength and direction of the relationship between each variable and the underlying factors. After rotation, the loadings are repositioned to improve interpretability and simplify the factor structure.

The factor loading plot visually displays the relationships between observed variables and the rotated factors. Each variable is represented as a point in the plot, with its position indicating its loading on the factors. Variables that have high loadings on a particular factor are closer to that factor’s axis in the plot.

The factor loading plot helps the analyst to visualise and interpret the factor structure, showing the pattern of relationships between variables and factors after rotation. It complements the information provided by the rotated component matrix, making it easier to understand the underlying factor structure and identify groupings of variables that are strongly associated with specific factors.

5

17. Factor analysis

17. How are factors scores calculated?

Solution: Factor scores are calculated using: Fi = β1i X10 + β2i X20 + · · · + βki Xk0 for i = 1, 2, . . . , m, where m is the number of factors, and k is the number of observed variables. β1i , β2i , . . ., βki are from the factor score coefficient matrix, and X10 , X20 , . . ., Xk0 are the standardised observed variable values for the specific case for which the factor scores are being calculated.

18. When might you recommend use of a surrogate variable instead of an extracted factor?

Solution: A surrogate variable is recommended when an extracted factor is not interpretable or lacks clear theoretical meaning. Surrogate variables are used as proxies for underlying factors when the extracted factors are difficult to interpret or do not represent meaningful constructs in the context of the analysis.

19. In the context of factor analysis, what is a ‘residual’ and how is it calculated?

Solution: Residuals represent the differences between the original correlations (the rij s), and the reproduced correlations (the r̃ij s) such that (due to symmetry of the correlation matrix): residualij = rij − r̃ij = rji − r̃ji = residualji .

20. On what basis might a residual be deemed ‘good’ or ‘bad’ ? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends! If the factor analysis model is good, the reproduced correlations should be very close to the original correlations. To judge whether to classify a residual as ‘good’ versus ‘bad’, let’s first consider the range of possible residual values. Since correlation coefficients always take values in [−1, 1], the minimum possible value of a residual is when: residualij = rij − r̃ij = −1 − (+1) = −2 and the maximum possible value of a residual is when: residualij = rij − r̃ij = 1 − (−1) = 2.

A ‘perfect’ residual would be when: residualij = rij − r̃ij = 0

⇔

rij = r̃ij .

SPSS adjudicates that a ‘good’ residual is one with an absolute value ≤ 0.05, and a ‘bad’ residual is one with an absolute value > 0.05, which we can visualise as shown below.

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17.1. Exercises

This seems a very narrow range of tolerance for ‘good’ residuals, so we may wish to be a bit more flexible and accommodating in practice.

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Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 18: Cluster analysis

ii

Chapter 18 Cluster analysis 18.1

Exercises

18.1.1

Test your understanding

1. Identify possible forms of exploratory analysis which could be undertaken prior to more complex techniques.

Solution: Good starting points are to: • examine the distribution of each variable to detect any outliers (extreme observations, whose presence may exert distorting effects if not excluded), using histograms or boxplots (univariate) and scatterplots (bivariate) • compare statistical attributes of distributions, using descriptive statistics such as the mean and standard deviation of variables • standardise variables if comparisons are sought and the variables have very different scales of measurement • explore possible relationships between variables, using cross-tabulations, side-by-side boxplots, scatterplots – if causality is expected, then analysis of variance and linear regression techniques can be used.

2. In what sense is cluster analysis ‘data reduction’ ?

Solution: Cluster analysis is considered ‘data reduction’ in the sense that it groups similar observations together based on their characteristics, reducing the complex dataset into a smaller set of clusters.

Each cluster represents a condensed representation of similar data points, simplifying the analysis and aiding in understanding patterns and relationships within the data.

While factor analysis focuses on variable reduction, cluster analysis looks at observation reduction by searching for natural groups of observations. Whereas factor analysis looked at data reduction columnwise, cluster analysis is concerned with data reduction rowwise. With n (individual) observations, we hope to combine these into homogeneous groups, known as ‘clusters’.

3. How similar or dissimilar should clusters be: (a) within each cluster? (b) between each cluster?

1

18. Cluster analysis

Solution: (a) Clusters should be homogeneous within each cluster, so that a cluster consists of similar cases. (b) Clusters should be heterogeneous between clusters, so that each cluster is sufficiently distinct.

Hence members of one cluster should be dissimilar from members in all other clusters.

4. How do hierarchical and non-hierarchical clustering procedures differ? Outline the algorithm for each approach.

Solution: Hierarchical clustering is a step-by-step approach that merges data points into nested clusters, creating a tree-like structure. Non-hierarchical clustering directly assigns data points to clusters without forming a hierarchical structure, typically using algorithms like k-means clustering.

The algorithm by which hierarchical clustering works is as follows.

1. Begin with c = n clusters, each of size 1, and an n × n distance matrix or similarity matrix.

2. Use the matrix to find the nearest pair of cases (using the distance matrix) or the most similar pair of cases (using the similarity matrix) and combine these into the same cluster.

3. Update the distance/similarity matrix. Rows and columns of the two merged cases are replaced with a single row and column for the newly formed cluster.

4. Repeat stages 2 and 3 a total of (n − 1) times, culminating in c = 1 cluster containing all n cases.

The algorithm by which non-hierarchical clustering works is as follows.

1. Cases are initially partitioned into k starting clusters, or k centroids (starting ‘seeds’) are initially specified.

2. Allocate each case to the cluster with the nearest centroid.

3. Update the centroids.

4. Repeat stages 2 and 3 until there are no more re-allocations of cases.

5. What is Euclidean distance and how is it calculated?

Solution: Suppose each case is measured on k different observed variables, X1 , X2 , . . . , Xk . The Euclidean distance between any two cases, i and j, denoted δij , whose measurements are (xi1 , xi2 , . . . , xik ) and (xj1 , xj2 , . . . , xjk ), respectively, is: v u k uX δij = t (xip − xjp )2 . p=1

6. Are there any alternative distance measures to Euclidean distance? If so, what are they and how are they calculated?

Solution: Yes! Two possible alternatives (these are not exhaustive) to Euclidean distance are:

2

18.1. Exercises

• squared Euclidean distance: k X δij = (xip − xjp )2 p=1

• Chebyshev distance: δij = max |xip − xjk |. ∀p

7. How would you construct a similarity matrix for binary data?

Solution: Consider two customers, Customer 1 and Customer 2, and we determine the frequency of each contingency pair:

Customer 1 (= 0) Customer 1 (= 1)

Customer 2 (= 0) 1 (f0,0 ) 2 (f0,1 )

Customer 2 (= 1) 1 (f1,0 ) 1 (f1,1 )

Similarity between any two cases is identified by how many common attributes they share, i.e. when both are type ‘0’ or ‘1’. A natural choice of similarity measure would be: f0,0 + f1,1 . f0,0 + f1,0 + f0,1 + f1,1 Using this approach, the measure of similarity between Customer 1 and Customer 2 is 2/5 = 0.40.

Similarity measures for all pairs of customers can then be presented in a similarity matrix. For example:

Customer 1 Customer 2 Customer 3 Customer 4

Customer 1 1 0.40 0.60 0.20

Customer 2 0.40 1 0.40 0.40

Customer 3 0.60 0.40 1 0.60

Customer 4 0.20 0.40 0.60 1

8. What is the difference between a similarity matrix and a dissimilarity matrix? Provide an example of each.

Solution: A similarity matrix and a dissimilarity matrix are both used to quantify relationships between objects in a dataset, but they represent different aspects: A similarity matrix quantifies the degree of similarity between pairs of objects. Higher values in the matrix indicate greater similarity, and lower values indicate less similarity. A similarity matrix is typically symmetric, with identical values along the diagonal. For example, in an e-commerce business, a similarity matrix can be created based on customer behaviour, capturing the similarity between pairs of customers in terms of their purchase history, product preferences, and browsing patterns. High similarity values between customers may suggest potential cross-selling opportunities or targeting similar products to a group of customers.

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18. Cluster analysis

A dissimilarity matrix quantifies the degree of dissimilarity or distance between pairs of objects.

Lower values in the matrix indicate greater similarity, and higher values indicate greater dissimilarity. A dissimilarity matrix is also symmetric, with zero values along the diagonal. For example, in inventory management, a dissimilarity matrix can be constructed based on the characteristics of products stocked by a retailer. The matrix can measure the dissimilarity between products in terms of size, colour, features, or demand patterns. This information can be used for product clustering, categorisation, and optimal placement in stores.

9. Distinguish between the following clustering techniques: (a) single linkage (b) complete linkage (c) average linkage (d) Ward’s method (e) centroid clustering.

Solution: (a) Single linkage uses a ‘nearest neighbours’ approach. The distance between two clusters is measured by taking the shortest distance between members of two clusters, as illustrated below.

(b) Complete linkage uses a ‘furthest neighbours’ approach. The distance between two clusters is measured by taking the furthest distance between members of two clusters, as illustrated below.

(c) Average linkage uses an average approach, hence the name! The distance between two clusters is measured by taking the average distance (the arithmetic mean) between members of two clusters, as illustrated below.

4

18.1. Exercises

(d) Ward’s method works by minimising intra-cluster variation, i.e. cases are combined to ensure the smallest possible increase in the new cluster’s variance. Recall we seek homogeneity (limited variation) within each cluster, but at each stage of hierarchical clustering we inevitably merge more distant cases each time, thereby increasing this within, or ‘intra’, cluster variation – Ward’s method works to minimise this increase. (e) Centroid clustering uses the ‘nearest centroid’ approach. The distance between two clusters is measured by taking the distance between the cluster centroids, i.e. the averages of the variables.

10. What use is an agglomeration schedule?

Solution: An agglomeration schedule is used in hierarchical clustering to record the sequence of merging clusters as the algorithm progresses. It provides a visual representation of how clusters are formed and their similarity levels, aiding in determining the appropriate number of clusters and understanding the hierarchical relationships among data points.

11. What use is an icicle plot?

Solution: The icicle plot shows the cluster compositions for all possible cluster solutions. The vertical axis details the number of clusters and the individual cases are at the top of the plot . The icicles indicate the split between cluster solutions in terms of which observations are included in each cluster.

12. What use is a dendrogram?

Solution: A dendrogram is used to visualise the hierarchical relationships among objects in a dataset in hierarchical clustering. It represents the sequence of merging clusters as they form a tree-like structure, with each branch indicating the similarity or dissimilarity between clusters. Dendrograms help identify the optimal number of clusters and understand the grouping patterns within the data, making them valuable for analysis.

13. How is the ‘right’ number of clusters determined? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends! Determining the ‘right’ number of clusters is not always straightforward.

However, the following are some suggested approaches.

• Review the agglomeration schedule’s ‘Coefficient’ column. When there is a ‘large’ increase, at that stage we are combining somewhat distant clusters and so we may wish to halt the agglomeration just before this happens – otherwise we combine distant cases which we might be unwilling to consider as being sufficiently ‘similar’ or ‘homogeneous’.

• Review the cluster membership of each considered solution. We could use the relative cluster sizes as a criterion for deciding on the appropriate number of clusters if we had a specific requirement for equal/similar/minimum cluster sizes or proportions.

• The icicle plot is useful for visualising hierarchical relationships and identifying patterns and groupings.

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18. Cluster analysis

• Examine the dendrogram which clearly shows the proximity when cases are combined.

Judgement is required, but we can use the rescaled distance axis to judge which vertical lines (points of merging) are ‘acceptable’.

• Clusters must be of practical use, so check whether each cluster has a clear characteristic profile, which can be obtained by examining the group centroids.

• Consider the specific business problem you are trying to solve and the practical interpretability of the clusters.

14. How would you go about building a characteristic profile for candidate clusters?

Solution: In SPSS, use Analyze > Compare Means > Means..., then move the variables used for clustering into the ‘Dependent List:’, and move the different cluster solutions to the ‘Independent List:’.

Under ‘Options...’ keep the default ‘Cell Statistics’ of ‘Mean’ (to give the centroid values), ‘Number of Cases’ (to give the cluster sizes) and ‘Standard Deviation’ (to judge the degree of homogeneity/heterogeneity of the clusters). Click ‘Continue’, thereafter click ‘OK’.

15. Should you only perform cluster analysis once, with only one distance measure and one clustering technique? Briefly justify your view.

Solution: Experimenting with different clustering procedures and distance measures is recommended to see if there is stability in the solutions. If the same cluster solutions emerge each time, the analyst can have greater confidence that these are genuine clusters.

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Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 19: Constrained optimisation models

ii

Chapter 19 Constrained optimisation models 19.1

Exercises

19.1.1

Test your understanding

1. What are the three types of optimisation problems? Provide a real-world example of each.

Solution: Optimisation problems can be one of three types: • maximising a value – for example, a manufacturer choosing the optimal product mix to maximise profits • minimising a value – for example, a logistics company choosing delivery routes to minimise transportation costs • achieving a fixed value – for example, allocating a fixed research and development (R&D) budget across competing projects.

2. Explain the role each of the following has in an optimisation model: (a) decision variables (b) objective function (c) constraints.

Solution: (a) Decision variables – these are the variables whose values are under the control of the decision-maker, i.e. they are free to choose these. Decision variables are critical in optimisation models, since they determine everything else. For example, in manufacturing, the quantities of different goods to produce would be decision variables since management can decide the quantity amounts of each good. (b) Objective function – the function which returns the key variable to be optimised, whether maximised, minimised, or imposed to be a fixed value. For example, a manufacturer targets maximising its total profit (which in part will be affected by its production levels) which can be calculated using a profit function (where profit equals total revenue minus total cost). (c) Constraints – perhaps physical (such as factory space), logical (such as non-negative quantities) or economic restrictions (such as budget to be spent on raw materials), which vary depending on the nature of the problem. Constraints must be satisfied, either binding (with no slack) or non-binding (with some slack).

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19. Constrained optimisation models

3. Following on from Question 2, what terminology does Excel use when working with optimisation models?

Solution: Excel has its own terminology when working with optimisation models as follows.

• Changing cells – spreadsheet cells which contain values of the decision variables.

• Objective cell – spreadsheet cell which contains the formula of the objective function.

• Constraints – restrictions imposed on the values in the changing cells.

• Non-negativity – a special type of constraint forcing specified cells to contain non-negative numbers (useful for real-world variables which cannot be negative, like quantities).

4. What is the difference between a feasible solution and a feasible set?

Solution: A feasible solution is a solution which satisfies all of the constraints. The feasible set is the set of all feasible solutions, such that any feasible solution is viable in practice.

5. In what sense is the ‘optimal solution’ to a constrained optimisation problem optimal?

Solution: The ‘optimal solution’ to a constrained optimisation problem is optimal in the sense that it satisfies all the constraints of the problem while maximising or minimising the objective function. It is the best possible solution among all feasible solutions that adhere to the given constraints. The optimal solution may not be the absolute best solution in an unrestricted sense, but it is the best attainable solution within the specified constraints.

6. For a product mix model involving two products, say X and Y : (a) what are the decision variables? (b) what is the objective function? (c) what sorts of constraints could feature?

Note you may assume any parameter values you wish.

Solution: (a) Define the decision variables as follows.

• Let x be the quantity of product X to produce.

• Let y be the quantity of product Y to produce. (b) The objective function represents the goal of the optimisation problem. Let’s assume here that the objective is to maximise profit. If pX and pY represent the selling prices of products X and Y , respectively, and cX and cY represent their respective production costs, then the objective function can be written as: max profit = pX x + pY y −(cX x + cY y ). | {z } | {z } x,y total revenue

2

total cost

19.1. Exercises

(c) Constraints represent the limitations or restrictions on the decision variables. Some possible constraints are the following.

• Production capacity constraints: x and y must not exceed certain production capacities.

For example: x ≤ 1000 and y ≤ 800.

• Demand constraints: The quantities of x and y produced should not exceed the demand for these products. For example: x + y ≤ 1500.

• Non-negativity constraints: x and y cannot be negative, as they represent quantities of products produced. Hence: x ≥ 0 and y ≥ 0.

7. What is meant by a ‘binding solution’ ?

Solution: A ‘binding solution’ refers to a specific solution to an optimisation problem where one or more of the constraints are binding, meaning they are satisfied with equality (i.e. they are met at their upper or lower limits). In such a solution, the constraints play a key role in determining the optimal values of the decision variables. If any of the constraints were relaxed (made non-binding), the optimal solution would change. Binding solutions are significant because they highlight the constraints that are critical in shaping the optimal outcome.

8. Explain the purpose of Solver.

Solution: Solver in Excel is an add-in tool used for optimisation and solving complex quantitative problems.

Its purpose is to find the optimal values for a set of decision variables while adhering to specified constraints and maximising or minimising an objective function. Solver can handle linear and nonlinear programming problems, making it useful for various tasks. By automating the iterative process of adjusting variables, Solver allows analysts to efficiently find the best solution to real-world problems, making Excel a powerful tool for decision-making and problem-solving tasks.

In short, Solver solves problems!

9. In a sensitivity analysis report, what do the following terms mean? (a) Allowable increase. (b) Allowable decrease. (c) Reduced cost. (d) Shadow price.

Solution: (a) The allowable increase represents the maximum amount by which the coefficient of an objective function (such as profit or cost) can increase while maintaining the current optimal solution. It indicates the amount by which the objective function coefficient can increase before a new variable enters the optimal solution, or an existing variable leaves the solution.

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19. Constrained optimisation models

(b) The allowable decrease represents the maximum amount by which the coefficient of an objective function can decrease while maintaining the current optimal solution. It indicates the amount by which the objective function coefficient can decrease before a variable currently in the optimal solution becomes zero. (c) The reduced cost is applicable for any decision variable with an optimal solution value of zero.

In such instances, the reduced cost tells us by how much the respective objective coefficient must change until the associated decision variable has a non-zero solution. (d) The shadow price indicates the change in the optimal value of the objective function when the right-hand side of a constraint changes by one unit.

10. What is the purpose of conjoint analysis?

Solution: Conjoint analysis allows the analyst to understand and quantify how consumers make decisions and evaluate products or services based on various attributes. It is used in market research to assess the relative importance of different product features and determine the optimal product configuration which maximises a customer’s preference. By presenting respondents with hypothetical product profiles, conjoint analysis captures their trade-offs and preferences, allowing businesses to identify the most appealing attributes and design products which align with customer preferences. It helps businesses make data-driven decisions, improve product development, and optimise marketing strategies to meet customer needs effectively.

11. If a product has n = 10 characteristics, each with four levels, what is the total number of combinations of different levels?

Solution: We have n = 10 characteristics, such that the ith characteristic, for i = 1, 2, . . . , 10, has ci = 4 different levels. The total number of combinations of different levels is: n Y i=1

ci =

10 Y

4 = 410 = 1,048,576

i=1

which is a lot!

12. The conjoint analysis model is: preference rating of X =

ci n X X

αij xij

i=1 j=1

Define all terms in this model.

Solution: In the conjoint analysis model: • X = a specific bundle of characteristics • αij = the part-worth contribution towards the preference rating due to the jth level (for j = 1, 2, . . . , ci ) of the ith characteristic (for i = 1, 2, . . . , n)

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19.1. Exercises

• xij = 1 if the jth level of the ith characteristic is present, and 0 otherwise • ci = the number of levels of the ith characteristic • n = the number of product characteristics.

13. How is the importance of a characteristic defined, and how can importance weightings for each characteristic be derived from the importance values?

Solution: The importance of the ith characteristic, denoted Ii , is defined in terms of the range of the part-worth contributions, αij , across the levels of the ith characteristic, such that: Ii = max(αij ) − min(αij ) for i = 1, 2, . . . , n. The relative importance of the ith characteristic is obtained by normalising Ii by calculating its importance weighting, Wi , such that: Ii Wi = P n Ii

where

n X

Wi = 1.

i=1

i=1

14. For a conjoint analysis with n = 2 characteristics each with four levels, determine a full set of dummy variable codes which could be employed to run a dummy variable regression.

Solution: We need to define two triplets of dummy variables. ‘Two’ because there are two characteristics, and ‘triplets’ since each characteristic has four levels, with one (the first, say) used as the reference category, leaving each of the other three requiring a dummy variable. We define: • X1 , X2 , X3 = the dummy variables representing the first characteristic • X4 , X5 , X6 = the dummy variables representing the second characteristic.

Using the first level as the reference category in each case, the dummy variable codes are therefore:

Level 1 (reference category) Level 2 Level 3 Level 4

First set of dummies X1 X2 X3 0 0 0 1 0 0 0 1 0 0 0 1

Second set of dummies X4 X 5 X6 0 0 0 1 0 0 0 1 0 0 0 1

15. For a conjoint analysis with n = 5 characteristics, each with three levels, what is the total number of dummy variables required to run the dummy variable regression?

Solution: The total number of required dummy variables when n = 5 and ci = 3, for i = 1, 2, . . . , n, is given by: 5 n X X k= (ci − 1) = (3 − 1) = 2 + 2 + 2 + 2 + 2 = 10. i=1

i=1

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19. Constrained optimisation models

16. For a conjoint analysis with n = 3 characteristics, each with three levels, write down the full set of equations which would need to be solved in order to estimate all part-worths.

Solution: For n = 3 characteristics, each with three levels, we model preference ratings as: preference rating of X =

ci 3 X X

αij xij

i=1 j=1

= α11 x11 + α12 x12 + α13 x13 + α21 x21 + α22 x22 + α23 x23 + α31 x31 + α32 x32 + α33 x33 .

Hence the full set of equations to be solved will be as follows, where βb1 , βb2 , . . ., βb6 will be coefficient estimates from a dummy variable regression.

• For the first characteristic we have: α b12 − α b11 = βb1 ,

α b13 − α b11 = βb2

and α b11 + α b12 + α b13 = 0.

These three equations can be solved simultaneously to obtain the part-worth estimates α b11 , α b12 and α b13 .

• For the second characteristic we have: α b22 − α b21 = βb3 ,

α b23 − α b21 = βb4

and α b21 + α b22 + α b23 = 0.

These three equations can be solved simultaneously to obtain the part-worth estimates α b21 , α b22 and α b23 .

• For the third characteristic we have: α b32 − α b31 = βb5

α b33 − α b31 = βb6

and α b31 + α b32 + α b33 = 0.

These three equations can be solved simultaneously to obtain the part-worth estimates α b31 , α b32 and α b33 .

17. A profit-maximising manufacturer is deciding on its optimal product mix between two types of table: light tables (x) and heavy tables (y). Each table is made exclusively from (sustainably sourced) wood, with no additional factor inputs. A light table is made using 30 kg of wood, while a heavy table is made from 40 kg of wood. The manufacturer has 8000 kg of wood available for production.

Light tables generate a profit of £60 each, while heavy tables generate a profit of £100 each. Given estimates about product demand from consumers, the number of light tables should be at least double the number of heavy tables. (a) Formulate the manufacturer’s decision as a linear programming problem, explaining the purpose of each equation. (b) Solve this linear programming problem graphically and by solving a pair of simultaneous equations, making sure you clearly indicate the set of feasible production bundles on your graph.

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19.1. Exercises

(c) Briefly discuss any recommendation(s) that you would make to this table manufacturer so that it could increase profits beyond the amount you calculated in part (b).

Solution: (a) The linear programming problem is to: max 60x + 100y x,y

subject to the wood constraint: 30x + 40y ≤ 8000

⇒

y ≤ 200 − 0.75x

the production constraint: x ≥ 2y

⇒

y ≤ 0.5x

and the non-negativity constraint: x ≥ 0 and y ≥ 0. (b) The optimal product mix must be a corner solution with constraints binding: y = 0.5x and y = 200 − 0.75x

⇒

x = 160 and y = 80

producing an optimal profit of: £60 × 160 + £100 × 80 = £17,600.

Graphically:

(c) This is a deliberately open-ended question. Some suggested recommendations could be: • increasing profit margins per chair based on marketing efforts • increase supply of wood, subject to the price of wood • relax production constraint.

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19. Constrained optimisation models

18. Consider Example 19.7. For all 27 mobile phone combinations, record your personal preference ratings for each (on a 0–100 scale), and then run a full conjoint analysis to establish the relative importance of each characteristic for you. Are the results consistent with your own opinion about these characteristics?

Solution: Over to you!

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Business Analytics: Applied Modelling and Prediction

Solutions to end-of-chapter exercises

Chapter 20: Monte Carlo simulation

ii

Chapter 20 Monte Carlo simulation 20.1

Exercises

20.1.1

Test your understanding

1. What is the purpose of a simulation model?

Solution: The purpose of a simulation model is to replicate real-world processes or systems in a virtual environment to understand their behaviour, explore different scenarios, and make informed decisions. Simulation models use statistical techniques to model complex systems, allowing analysts to study their dynamic interactions, identify bottlenecks, and test various strategies without the risk and cost associated with real-world experimentation. By simulating multiple scenarios, companies can optimise processes, improve resource allocation, assess risks, and predict outcomes.

2. Produce a flow diagram illustrating a Monte Carlo simulation.

Solution: A flow diagram illustrating a Monte Carlo simulation is:

Probability distributions for uncertain inputs

⇓ Simulation model

⇓ Probability distributions for important outputs

3. How are the real world and abstract world related? What happens in the abstract world?

Solution: The real world and abstract world are related through the process of abstraction, where real-world phenomena are simplified and represented using models, concepts, or symbols. The abstract world is a conceptual representation of the real world, where complex systems or processes are broken down into manageable components. In the abstract world, simulations, mathematical models, and theories are used to study and predict the behaviour of real-world systems under different conditions.

Through abstraction, analysts can gain insights, test hypotheses, and make predictions without the complexities and constraints of the real world, enabling informed decision-making and

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20. Monte Carlo simulation

problem-solving. For example, suppose we have to design a schedule of arrivals and departures for a new airport. We would link the real world and the abstract world as follows:

4. What is the difference between a static simulation model and a dynamic simulation model?

Solution: A static simulation model represents a system or process at a specific point in time, without considering changes over time. It captures a snapshot of the system, and the inputs and outputs are not influenced by time-dependent variables or feedback loops. For example, a static simulation model can be used to analyse market share for different products at a specific moment. It considers factors like customer preferences, pricing, and competitors’ offerings, but the model does not incorporate changes in these variables over time.

A dynamic simulation model incorporates the element of time, modeling the system’s behaviour and outcomes as they evolve over time. It factors in feedback loops, delays, and time-dependent variables to simulate real-world processes with temporal changes. For example, a dynamic simulation model can be used to optimise supply chain processes, considering factors like production rates, inventory levels, transportation times, and customer demand. As time progresses, the model adapts to changes in demand, inventory levels, and production schedules, allowing businesses to understand how their supply chain behaves under varying conditions.

5. List the six steps for conducting a Monte Carlo simulation in practice.

Solution: 1. Sample from a random number generator for each input variable.

2. Represent each possible input variable value to a range of (random) numbers according to the above probability distributions.

3. For each input variable generate a random number, then from the random number map it to the respective revenue/cost.

4. Calculate the outcome variable (which here will be profit = revenue − cost) and record it.

5. Repeat ‘3.’ and ‘4.’ until the desired number of simulations is reached.

6. Draw a histogram of the outcome variable values and compute the mean and standard deviation, which represent the expected outcome and risk of the outcome, respectively.

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20.1. Exercises

6. On what basis would you choose a probability distribution for an input variable in a simulation model?

Solution: It is necessary to choose an appropriate probability distribution for each specific problem based on the real-world attributes, or stylised facts, of each variable in practice. General characteristics of probability distributions include: • discrete vs. continuous (or even a mixture of each) • symmetric vs. skewed • restricted sample space vs. unrestricted (such as x ≥ 0, i.e. positive-valued only, vs. −∞ < x < ∞).

Usually we do not try to invent new distributions from scratch. Instead, we use one of many existing standard distributions. There is a large number of such distributions, such that for most purposes we can find a suitable standard distribution. Note, however, that there is no single ‘right answer’ as to which distribution(s) to use. You must decide the most appropriate probability distribution which, in your opinion, most closely captures the variable’s actual real-world attributes (its stylised facts).

7. What does the Excel function =RAND() do?

Solution: The Excel function =RAND() generates a random decimal number between 0 and 1. Each time the worksheet is recalculated or a change is made on the sheet, the function will produce a new random value. By default, the generated numbers are uniformly distributed, meaning each value within the range (0, 1) has an equal chance of being selected.

8. Why is the Excel function =RAND() useful in simulation models?

Solution: This function is useful for various purposes, such as generating random numbers for simulations, creating random samples, or adding randomness to spreadsheet models.

9. What are pseudo-random numbers?

Solution: Pseudo-random numbers are numbers generated by algorithms that appear to be random but are actually determined by a starting value called a seed. When using the same seed, the sequence of numbers will be identical, making them predictable and repeatable. Although not truly random, pseudo-random numbers are widely used in computer simulations and other applications where reproducibility is essential.

10. What is a quantile function and how can it be used to simulate random samples from a specified probability distribution?

Solution: For any continuous random variable X, its distribution function, F (x), is strictly increasing over its sample space, S. The inverse function, F −1 (x), known as the quantile function, is strictly increasing on [0, 1], such that F −1 (x) maps [0, 1] to S.

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20. Monte Carlo simulation

Let U ∼ Uniform[0, 1], i.e. a continuous uniform random variable over [0, 1]. Its distribution function is:   0 for u < 0 FU (u) = u for 0 ≤ u ≤ 1   1 for u > 1.

Let X = FX−1 (U ). Its distribution function is: FX (x) = FU ((FX−1 )−1 (x)) for x ∈ S which can be used to simulate random samples from a required probability distribution by simulating values of u from Uniform[0, 1], and then view FX−1 (u) as a random drawing from FX (x).

11. State the in-built Excel quantile functions for the normal and binomial distributions. Include their arguments.

Solution: =NORM.INV(probability, mean, standard dev) returns the inverse of the normal cumulative distribution for the specified mean and standard deviation, where: • probability is a probability corresponding to the normal distribution, a number between 0 and 1 inclusive.

• mean is the arithmetic mean of the distribution.

• standard dev is the standard deviation of the distribution, a positive number. =NORM.S.INV(probability) returns the inverse of the standard normal cumulative distribution (has a mean of zero and a standard deviation of one), where: • probability is a probability corresponding to the normal distribution, a number between 0 and 1 inclusive. =BINOM.INV(trials,probability s,alpha) returns the smallest value for which the cumulative binomial distribution is less than or equal to a criterion value, where: • trials is the number of independent trials.

• probability s is the probability of success on each trial.

• alpha is the criterion value.

12. When might you wish to employ adaptive expectations in a simulation model? Provide a practical example.

Solution: Adaptive expectations are employed in simulation models when individuals adjust their expectations based on past experiences and observed outcomes. In business, this approach is useful for modelling situations where stakeholders, such as consumers or investors, update their beliefs about future outcomes based on historical data. For example, in financial forecasting, adaptive expectations can be used to simulate how investors update their expectations of stock prices based on past price movements and market performance.

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20.1. Exercises

13. What is the ‘right’ number of simulations to perform? ‘Well, James, it depends’ may feature in your answer, but explain what it depends on!

Solution: Well, James, it depends! The ‘right’ number of simulations to perform depends on several factors.

• The more precise the results needed, the more simulations are required.

• Complex models may require more simulations to capture all possible scenarios accurately.

• Limited time and computational resources may restrict the number of simulations.

• Highly uncertain systems may require more simulations to account for the variability.

• Higher confidence levels may require more simulations to reduce uncertainty.

• A balance between accuracy and computational cost must be considered.

Performing a sensitivity analysis on the simulation can help determine an appropriate number of simulations to achieve reliable results within resource constraints.

14. Consider the simulation case study in Section 20.6. Discuss any changes you would propose making to this simulation model, briefly justifying your choice(s).

Solution: This is a deliberately open-ended question such that there is no single ‘right’ answer. Possible changes could include (some of) the following.

• Experiment with different probability distributions for the input variables by way of a sensitivity analysis.

• Explicitly model the amount of competition that the company faces.

• Include pricing information and model the expected price elasticities of demand.

• Model marketing expenditure and its expected impact on demand.

15. A business wants to conduct a Monte Carlo simulation to model its profits over the next 5 years.

The model will include the following two input variables: • consumer demand for its most popular product each year • whether a new competitor begins trading each year.

For each of these two variables suggest an appropriate probability distribution that you would recommend is used to model it, briefly justifying your choice of distribution in each case.

Solution: For the consumer demand variable, a recommended probability distribution is the normal distribution, assuming demand is influenced by various factors and follows a bell-shaped curve around the mean value. The normal distribution is commonly used for continuous variables and is suitable when the demand fluctuates randomly around an average value with equal probabilities of being above or below the mean.

For the new competitor variable, a recommended probability distribution is the binomial distribution because the variable is discrete (either a competitor starts trading or not) and involves a fixed number of trials (each year). The binomial distribution is appropriate for modelling the number of successes (competitors starting) out of a fixed number of trials (years), assuming the events are independent and have the same probability of occurrence in each trial.

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20. Monte Carlo simulation

16. You are employed as a financial analyst and your team has run many simulation models to estimate the profitability of various projects. If you had to choose just two measures of the distribution of any important output variable, such as net profit, to report, which two would you choose and why?

Are there any drawbacks to only using your two chosen measures?

Solution: Each output variable in a simulation model has an entire distribution of values. If we can only use two measures you would probably want a measure of central tendency and a measure of dispersion.

For the former, perhaps the median (half the time, net profit is above the median; half the time it is below) and the standard deviation to quantify risk.

Unsurprisingly, two measures cannot fully characterise the entire distribution. These two ignore skewness and kurtosis, which may be important in some cases. Also, these measures may not be sufficient to describe complex and asymmetric distributions, and further exploration through visualisations or other summary statistics might be necessary for a more comprehensive understanding of the output variable’s distribution.

17. For each probability distribution in Appendix B, identify one possible real-world input variable of a simulation model which could arguably be assumed to follow the distribution.

Solution: (a) Discrete uniform distribution The number of days it takes for a delivery to arrive at a warehouse. Assuming a fixed and known minimum and maximum time frame for deliveries, such as 3 to 7 days, the distribution could be used to model the uncertainty of arrival times within this range. (b) Bernoulli distribution The success or failure of a marketing campaign. The variable could take the value 1 if the campaign succeeds (for example, reaching a sales target) and 0 if it fails (for example, not meeting the target). (c) Binomial distribution The number of defective products in a production batch. The variable could model the number of successes (defective products) out of a fixed number of trials (total products produced) given a constant probability of defects. (d) Poisson distribution The number of customer arrivals at a retail store during a specific time interval (such as per hour). The variable could model the random arrival process of customers, assuming a constant rate of arrivals per unit of time. (e) Geometric distribution The number of attempts required to sell a product to a potential customer. The variable could model the probability of achieving the first success (making a sale) after a sequence of trials (customer interactions) with constant success probability. (f) Continuous uniform distribution The demand for a product within a certain price range. Assuming that demand is evenly distributed between the minimum and maximum price points, the distribution could be used to model demand uncertainty over this price range.

6

20.1. Exercises

(g) Exponential distribution The time between equipment failures in a manufacturing plant. The variable could model the random occurrence of failures, assuming a constant failure rate over time. (h) Normal distribution The daily sales revenue of a retail store. The variable could capture the continuous and symmetrical distribution of sales, which is often influenced by numerous factors and fluctuations.

18. The triangular distribution is a continuous probability distribution which has three parameters: • the minimum, a • the maximum, b • the mode, c such that a ≤ c ≤ b. Explain why this could be a useful probability distribution for an input variable in a simulation model.

Solution: The triangular distribution is a useful probability distribution for an input variable in a simulation model because it allows for a simple and intuitive representation of uncertainty when limited information about the variable is available. The three parameters, minimum (a), maximum (b), and mode (c), provide a convenient way to define the range of possible values and the most likely value for the variable.

In many real-world situations, we may have some knowledge or constraints about the possible values of an input variable without precise data on its distribution. The triangular distribution allows us to account for this uncertainty in a realistic manner. By specifying the minimum, maximum, and mode, we can model a variable that is bounded within a certain range and where the most likely value is at a specific point within that range.

Furthermore, the triangular distribution is relatively easy to work with in simulations, making it computationally efficient. It provides a continuous representation of uncertainty. Its simplicity and flexibility make it a practical choice for representing uncertain input variables in simulation models when detailed data on the distribution is not available.

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