Instructor Solution Manual For Statistics A Tool for Social Research and Data Analysis, 5th Edition

TABLE OF CONTENTS Preface.............................................................................................................................. iv Chapter 1: Introduction .................................................................................................. 1-1 Chapter 2: Basic Descriptive Statistics: Percentages, Ratios and Rates, Tables, Charts, and Graphs ......................................................................................................... 2-1 Chapter 3: Measures of Central Tendency and Dispersion ........................................... 3-1 Chapter 4: The Normal Curve........................................................................................ 4-1 Chapter 5: Introduction to Inferential Statistics: Sampling and the Sampling Distribution .................................................................................................... 5-1 Chapter 6: Estimation Procedures for Sample Means and Proportions ......................... 6-1 Chapter 7: Hypothesis Testing with Nominal and Ordinal Variables: Chi Square ....... 7-1 Chapter 8: Measures of Association for Variables Measured at the Nominal Level..... 8-1 Chapter 9: Measures of Association for Variables Measured at the Ordinal Level ...... 9-1 Chapter 10: Hypothesis Testing with Means and Proportions: The One-Sample Case.................................................................................................. 10-1 Chapter 11: Hypothesis Testing with Means and Proportions: The Two-Sample Case ................................................................................................. 11-1 Chapter 12: Hypothesis Testing with More Than Two Means: One-Way Analysis of Variance ................................................................................................................... 12-1 Chapter 13: Hypothesis Testing and Measures of Association for Variables Measured at the Interval-Ratio Level ........................................................................................... 13-1 Chapter 14: Partial Correlation and Multiple Regression and Correlation .................. 14-1

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CHAPTER 1 INTRODUCTION SUMMARY Statistics: A Tool for Social Research, Fifth Canadian Edition, begins by explaining the role of statistics in the research process. The discussion is guided by ―The Wheel of Science,‖ as conceptualized by Walter Wallace (Figure 1.1). The text consistently presents statistics in the context of the research enterprise; that is, statistics are always presented not as ends in themselves, but as useful devices or tools for answering questions and testing theories.

The chapter also distinguishes between descriptive and inferential statistics, and between univariate, bivariate, and multivariate statistics. The distinction between discrete and continuous variables and the concept of level of measurement are presented as well. The latter is stressed throughout the text as an organizational device and as a major criterion for selecting statistics appropriately. Exercises are provided at the end of the chapter for reviewing the characteristics of the three levels of measurement used in this text.

NEW TO THE FIFTH CANADIAN EDITION The ―You are the Researcher‖ section has been updated with new survey data, the 2018 GSS and CCHS.

WHAT CAN I DO IN CLASS? 1. Provide students with a table consisting of a variable (Column 1), the variable’s response categories (Column 2), whether it is independent or dependent (Column 3), and the level of measurement (Column 4). Ask students to identify the variables as independent or dependent, as well as the level of measurement for each given variable. Go over the answers in class. For example, gender is an independent nominal-level variable and health is a dependent variable measured at either the interval-ratio (body weight in kilograms), ordinal (poor, good, or excellent self-rated health), or nominal (have a family doctor, yes or no) level.

2. Show a list of examples of populations and samples, and have students identify each as such. For example, we want to know if more male students or more female students visit the local pub on campus in a given week. a. Sample: The number of male and female visitors you observe visiting during a one-hour period b. Population: Everyone who visits the local pub on campus in a given week

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WHAT CAN I DO ONLINE? 1. Have students look up examples online of variables that can be identified as having more than one level of measurement, depending on how they are categorized. For example, household income can be reported as a continuous variable or grouped into finite categories (see, for example, https://www150.statcan.gc.ca/n1/en/catalogue/98-400-X2016097).

2. Have students search a recent news story or report online and identify and list variables that could be used to conduct a quantitative analysis (see, for example, Angus Reid Institute’s report on term care facilities in Canada during the COVID19 pandemic, https://angusreid.org/long-term-care-covid/).

CHAPTER 2 BASIC DESCRIPTIVE STATISTICS Percentages, Ratios and Rates, Tables, Charts, and Graphs SUMMARY This chapter covers relatively simple descriptive devices: percentages and proportions, ratios, rates, frequency distributions, pie and bar charts, histograms, and frequency polygons. The emphasis is on frequency distributions and the construction and interpretation of these tables for variables measured at each of the three different levels.

Instructors may want to supplement this material with additional examples of each technique and/or graphs and charts, especially those created by software such as Microsoft Excel.

The underlying theme of this chapter is the need to present results clearly, to communicate results accurately and concisely but without losing too much detail. The problem used in Section 2.6: Charts and Graphs, for example, is intended to contrast the anarchy of raw, unorganized data with the clarity and simplicity of frequency distributions and various graphs.

NEW TO THE FIFTH CANADIAN EDITION Most tables, graphs, and application boxes have been updated to better reflect the underlying relevance and importance of using basic descriptive statistics to help understand society. The SPSS exercises have been updated with the new Canadian survey data used in the textbook.

ANSWERS TO TEXT PROBLEMS 2.1 Copyright © 2023 Cengage Learning Canada, Inc.

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a. Complex A: (5/20) × 100 = 25.00% Complex B: (10/20) × 100 = 50.00% b. Complex A: 4:5 = 0.80 Complex B: 6:10 = 0.60 c. Complex A: (0/20) = 0.00 Complex B: (1/20) = 0.05 d. (6/(4 + 6)) = (6/10) = 60.00% e. Complex A: 8:5 = 1.60 Complex B: 2:10 = 0.20

2.2 a. 42.36% (see below for detailed solution) b. 0.47 (see below for detailed solution c. 1.37 (see below for detailed solution) d. 51.83% e. 0.03 f. 0.08 g. 25.47% h. 0.68 i. 4.34 j. 0.03

2.3 Bank robbery rate = (47/211,732) × 100,000 = 22.20 Homicide rate = (13/211,732) × 100,000 = 6.14 Auto theft rate = (23/211,732) × 100,000 = 10.86

2.4 State New Jersey Iowa Alabama Texas California

1997 4.20 1.82 10.29 6.83 7.99

2019 3.23 2.54 11.97 5.92 4.54

Province Nova Scotia Quebec Ontario Manitoba

1997 2.56 1.80 1.56 2.72

2019 0.62 0.91 1.69 5.26

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British Columbia

2.90

1.77

2.5 Answers may vary.

2.6 Answers may vary.

2.7 Answers may vary.

2.8 Answers may vary.

2.9 Answers may vary.

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2.10 Response Time, 2000 21 minutes or more 16–20 minutes 11–15 minutes 6–10 minutes Less than 6 minutes

Frequency (f) 35 75 180 375 275 940

Percentage 3.72 7.98 19.15 39.89 29.26 100.00

Response Time, 2020 21 minutes or more 16–20 minutes 11–15 minutes 6–10 minutes Less than 6 minutes

Frequency (f) 45 95 155 350 250 895

Percentage 5.03 10.61 17.32 39.11 27.93 100.00

2.11 40 and Under Value 1 2 3 4 5 6

Frequency 2 2 2 1 4 4 15

Percentage 13.33 13.33 13.33 6.67 26.67 26.67 100.00

Cumulative Percentage 13.33 26.66 39.99 46.66 73.33 100.00

Over 40 Value 1 2 3 4 5 6 7 8

Frequency 1 4 2 2 1 0 3 3 16

Percentage 6.25 25.00 12.50 12.50 6.25 0.00 18.75 18.75 100.00

Cumulative Percentage 6.25 31.25 43.75 56.25 62.50 62.50 81.25 100.00

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DETAILED SOLUTION TO SELECTED TEXT PROBLEMS 2.2 a. Social science majors = 97 + 132 = 229 Male social science majors = 97 % of social science majors who are male: f     100 n  97     100  229   0.4236  100  42.36% b. Business majors = 156 + 139 = 295 Female business majors = 139 Proportion of business majors who are female: f  n 139  295  0.47 c. For humanities: Number of males = 114 Number of females = 83 Ratio of males to females:

2.3 Homicide rate: Number of homicides = 13 Population = 211,732

(

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WHAT CAN I DO IN CLASS? 1. Pass out a sheet of paper in class and have students write down their gender and the academic school year in which they are currently enrolled. Display the results for the entire class and have students decide what type of distribution would best depict the results. For example: Gender Female Female Male Other Male Other Male Female Male Female Female Other Female Female Other Female Female Male Female Male

2.

Academic Year 1 2 1 3 2 1 1 1 2 3 3 4 2 1 1 1 2 2 1 3

Provide students with a column of intervals and have them write out the real limits and midpoints for each interval. This will help familiarize them with statistical definitions.

Class Interval 17–18 15–16

Frequency 7 15

Real Limits 16.5–18.5 14.5–16.5

Therefore, when we subtract the real lower limit from the real upper limit, we get the actual width of the interval, which in this case is 2.

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WHAT CAN I DO ONLINE? 1. Using the Statistics Canada website (http://www.statcan.gc.ca), have students search for the summary table ―Average weekly earnings by industry.‖ Using this table, students can create a bar chart comparing Canada’s average weekly earnings across two or more industries. To simplify this exercise, have students round the numbers to the nearest whole number.

2. Find examples of news stories online and show how statistics are used to strengthen those stories. For example see, http://business.financialpost.com/news/economy/feeling-the-brexit-bite-these-arethe-canadian-sectors-hurt-hardest-by-u-k-vote-to-leave-the-eu.

CHAPTER 3 MEASURES OF CENTRAL TENDENCY AND DISPERSION SUMMARY This chapter presents three measures of central tendency—mode (the most common score), median (the middle score), and mean (the average score), and three related measures of dispersion—index of qualitative variation, range/interquartile range, and variance/standard deviation. An additional graph, the box plot, is also introduced.

The chapter emphasizes the mean and the standard deviation as the most important and commonly used measures. It discusses the algebraic and mathematical characteristics of the mean, especially the point that the mean is affected by every score in a distribution and, therefore, is pulled in the direction of extreme scores relative to the median. This characteristic also serves to introduce the variable distribution characteristics of symmetry and skew. This theme of the ―sensitivity‖ of the mean is also picked up in several of the end-of-chapter problems.

The material on the standard deviation begins by stating some characteristics that a reasonable measure of dispersion should have. The purpose here is to demonstrate that statistics do not just ―happen‖—they are developed to fulfill specific purposes and are designed according to an explicit logic. Beginning students often have trouble interpreting the standard deviation. Section 3.1 and especially Section 3.5 contain demonstrations of several ways to interpret this statistic.

This chapter stresses the importance of level of measurement in selecting a measure of central tendency and dispersion. It closes with a section on how to select an appropriate measure for a given purpose and a given level of measurement.

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NEW TO THE FIFTH CANADIAN EDITION A new statistic, coefficient of variation (CV), has been added to the chapter. The CV provides an easily-to-calculate (CV = s/ X ) measure of dispersion. The application of the CV to real data is presented in Section 3.8 (Interpreting Statistics: The Central Tendency and Dispersion of Income in Canada). Most tables, graphs, and application boxes have been updated.

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ANSWERS TO TEXT PROBLEMS 3.1 ―Region of birth‖ is a nominal-level variable, ―support for stricter gun laws‖ and ―opinion of food‖ are ordinal, and ―expenses‖ and ―number of movies‖ are intervalratio. The mode, the most common score, is the only measure of central tendency available for nominal-level variables. For the ordinal-level variables, do not forget to array the scores from high to low before locating the median. There are 10 lower-year students (n is even), so the median will be the score halfway between the scores of the two middle cases. There are 11 upper-year students (n is odd), so the median will be the score of the middle case. To find the mean for the interval-ratio variables, add the scores and divide by the number of cases.

Variable Region of birth: Gun laws: Expenses: Movies: Food:

Lower Year Mode = Atlantic Median = 3 Mean = 48.50 Mean = 5.80 Median = 6

Upper Year Mode = Atlantic Median = 5 Mean = 63.00 Mean = 5.18 Median = 4

3.2 Complex A B C D

IQV 0.89 0.99 0.71 0.74

Complex B has the highest index of qualitative variation and is the most heterogeneous.

Complex C is the least heterogeneous.

3.3 Variable Gender Social class

Number of years in the party Education Marital status Number of children

Level of Measurement Nominal Ordinal

I-R

Measure of Central Tendency Mode = male Median = ―medium‖ (the middle case is in this category) Mean = 26.15

Ordinal Nominal I-R

Median = high school Mode = married Mean = 2.39

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3.4

Mean Median

Canadian Provinces 2000 2015 48,161.54 76,130.31 47,300.00 69,995.00

U.S. States 1999 2017 42,235.15 60,152.92 40,058.00 57,051.00

For both Canadian provinces and U.S. states, the mean and median increase over the time period. Income is positively skewed—the mean is larger than the median at both times and in both countries.

3.5

Mean Median Range Std Dev.

Canadian Provinces 2000 2015 48,161.54 76,130.31 47,300.00 69,995.00 23,400.00 58,341.00 6,753.64 16,999.07

U.S. States 1999 2017 42,235.15 60,152.92 40,058.00 57,051.00 22,548.00 35,103.00 6,635.91 10,697.71

While incomes increased over this period, so too did the amount of dispersion. In Canada, the standard deviation and range increase over this period, indicating growing variability in the distribution of income. The pattern is similar for the U.S. states.

3.6 Age Happiness # Partners Gender

s = 17.05 R=2 R=8 IQV = 0.89

3.7

The median is 27.50 and the mean is 26.90. Since Q1 is 15 (the median of the lower half of the data) and Q3 is 35 (the median of the upper half of the data), the interquartile range is 35 – 15, or 20. The standard deviation is 12.28.

3.8

Mean = 1.31, median = 0.72. The higher value for the mean indicates a positive skew. Note the very high score of Denmark.

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3.9

Standard deviation = 1.53; CV = 1.17

3.10

2001: Mean = 54.25 2021: Mean = 60.30

3.11

Mean = 40.25, Median = 44.50. The lower value for the mean indicates a negative skew or a few very low scores. For this small group of nations, the skew is caused by the score of Mexico (10), which is much lower than the scores of the other seven nations (which are grouped between 37 and 51).

3.12

Pretest: Mean = 9.33, s = 3.50 Post-test: Mean = 12.93, s = 4.40

3.13

To find the median, the scores for both groups first must be ranked from high to low. Both groups have 25 cases (n = odd), so the median is the score of the 13th case. For first-year students, the median is 35, and for final-year students, the median is 30. The mean score for first-year students is 31.72. For final-year students, the mean is 28.60.

3.14

a. Median = 14.50, Mean = 17.78 b. Q1=6, Q3= 26, IQR=20, s = 13.03 c. Because the score for Montreal (53) is much higher than the other scores (the next highest score is 33), removing this score would reduce the mean; the median remains unaffected. d. The standard deviation of the scores without Montreal would be lower in value; the interquartile range remains unaffected.

3.15

Answers may vary.

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DETAILED SOLUTIONS TO SELECTED TEXT PROBLEM 3.13 a. The scores, in order, are: Case 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 Total =

Score 50 47 45 45 43 42 42 40 40 38 38 37 35 33 32 30 30 26 25 22 12 11 10 10 10 793

Median= 35

Mode = 10

By inspection, the score of 10 occurs most frequently (three times), so the mode is 10.

To find the median with an odd number of scores (n = 25), find the (n+1)/2, or 13th, case.

The score of the 13th case is 35.

To find the mean:

X =

793  31.72 25

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WHAT CAN I DO IN CLASS? 1. Provide three lists of numbers and have students calculate the three measures of central tendency for each: median, mode, and mean. Make sure students understand how level of measurement affects each measure, for example, the fact that only ―mode‖ can be used for nominal level data. Example for median: 10, 10, 8, 7, 5, 4, 2 Here we can see that the median is 7.

2. Provide students with a list of scores (no more than five) and have them calculate the standard deviation. For example, use: 10, 20, 30, 40, 50.

WHAT CAN I DO ONLINE? 1. Have students look up the local housing prices and differentiate between the median price of a house in their city and any other city. For example, Vancouver has a median price of $1,220,000 and Toronto has a median price of $1,045,000.

Use https://www.crea.ca/housing-market-stats/national-price-map/ to compare the prices between major Canadian cities.

2. Have students find examples of positively skewed distributions and negatively skewed distributions, for example, the yearly income of MBA graduates (positively skewed) and the number of job applications completed by MBA graduates (negatively skewed). Use http://stats.stackexchange.com/questions/89179/real-life-examples-ofdistributions-with-negative-skewness.

CHAPTER 4 THE NORMAL CURVE SUMMARY This chapter introduces the normal curve. In combination with the mean and the standard deviation, the normal curve is used to describe the area above, below, and between scores. Numerous problems are provided at the end of the chapter for homework and/or in-class demonstrations of computing Z scores and raw scores, working with percentiles, and finding areas.

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Chapter 4 is intended to bridge the course from descriptive to inferential statistics.

Section 4.6 is particularly important in this regard because it is the major treatment of the concept of probability. Should you need it, the companion website provides an expanded discussion on probability to supplement this material.

Students should acquire two main ideas from Section 4.6: (1) probabilities can be estimated using the properties of the normal curve (even though the application is not particularly realistic or commonly used), and (2) it is extremely likely that randomly selected scores from a normally distributed variable will be close in value to the mean of the distribution. In Part 2, these ideas are linked to the concept of the sampling distribution and used to estimate the probability of a sample outcome from the theoretical distribution of all possible sample outcomes (i.e., the sampling distribution).

NEW TO THE FIFTH CANADIAN EDITION Section 4.6 on probability has been expanded to discuss and compare probability distributions of discrete and continuous variables.

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ANSWERS TO TEXT PROBLEMS 4.1 Xi

Z score

5 6 7 8 9 11 12 14 15 16 18

–1.67 –1.33 –1.00 –0.67 –0.33 0.33 0.67 1.33 1.67 2.00 2.67

% Area Above 95.25 90.82 84.13 74.86 62.93 37.07 25.14 9.18 4.75 2.28 0.38

% Area Below 4.75 9.18 15.87 25.14 37.07 62.93 74.86 90.82 95.25 97.72 99.62

4.2 Z score 1.50 –1.00 –1.25

0.86 –0.63 0.26 1.21 –0.02 0.17 –1.02

% Area Above 6.68 84.13 89.44 19.49 73.57 39.74 11.31 50.80 43.25 84.61

% Area Below 93.32 (see below for detailed solution) 15.87 (see below for detailed solution) 10.56 80.51 26.43 60.26 88.69 49.20 56.75 15.39

4.3 a. b. c. d. e. f. g. h. i. j.

Z scores 0.10 & 1.10 0.60 & 1.10 0.60 0.90 0.60 & –0.40 0.10 & –0.40 0.10 0.30 0.60 1.10

Area (%) 32.45 13.86 27.43 18.41 38.11 19.52 53.98 61.79 72.57 86.43

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4.4 a. 22.34% (see below for detailed solution) d. 86.64% e. 1.39% f. 5.74%

b. 61.79% g. 19.34%

c. 61.79% h. 34.62%

4.5 Xi

Z Score

60 57 55 67 70 72 78 82 90 95

−2.00 −2.50 −2.83 −0.83 −0.33 20.00 21.00 21.67 23.00 23.83

Number of Retirees Above 195 199 199 159 126 100 32 10 1 1

Number of Retirees Below 5 1 1 41 74 100 168 190 199 199

Note: Number of retirees (a discrete variable) has been rounded off to the nearest whole number.

4.6 a. 5.05% f. 41.69% k. 60.64%

b. 23.27% g. 70.67% l. 88.10%

c. 57.14% h. 29.07%

d. 86.21% i. 2.81%

e. 6.85% j. 15.87%

4.7 a. b. c. d. e. f.

Z score −2.20 1.80 −0.20 & 1.80 0.80 & 2.80 −1.20 0.80

4.8 a. Xi = 27 days

Area (%) 1.39 96.41 54.34 20.93 88.49 21.19

b. Xi = 24 days

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c. Xi = 21 days

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4.9 a. b. c. d. e.

Z Scores −1.00 & 1.50 0.25 & 1.50 1.50 0.25 & −2.25 −1.00 & −2.25

Area 0.7745 0.3345 0.0668 0.5865 0.1465

4.10 With a Z score of –0.39 (corresponding as closely as possible to an area between the mean and Z of 0.1500); a. Xi = 980.50 b. Xi = 96.10 4.11 With a Z score of 0.00 (corresponding to an area between the mean and Z of 0.0000) a. Xi = 1000.00 b. Xi = 100.00 4.12 a. 0.1112 e. 0.6711

b. 0.4129 f. 0.3694

c. 0.2177 g. 0.1174

d. 0.0375

4.13 The probability of finding a score greater than 80, when the mean is 65 and the standard deviation is 15, is 0.1587.

4.14

0.0359

4.15

Yes. The raw score of 110 translates into a Z score of +2.88. 99.80% of the area lies below this score, so this individual was in the top 1% on this test.

4.16

Social worker: Z = +0.53 Employment counsellor: Z = +0.67

4.17

For the first event, the probability is 0.0919, and for the second, the probability is 0.0655. The first event is more likely.

4.18

GPA 3.20 3.21 3.25 3.30 3.35

Z score 1.27 1.30 1.42 1.58 1.73

Area Below (%) 89.80 90.32 92.22 94.29 95.82

Only the 3.20 GPA is not in the top 10%.

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DETAILED SOLUTIONS TO SELECTED TEXT PROBLEMS 4.2 For Xi = 650,

Z

Xi  X 650  500   1.50 s 100

The area between the score and the mean is 43.32%, so the area below = 50.00% + 43.32% = 93.32%.

The ―area beyond‖ (or area above a positive Z score) is 6.68%.

For Xi = 400,

Z

Xi  X 400  500   1.00 s 100

The area between the score and the mean is 34.13%, so the area above = 50.00% + 34.13% = 84.13%.

The ―area beyond‖ (or area below a negative Z score) is 15.87.

4.4 a. For Xi = 40,

Z

Xi  X 40  50   1.00 s 10

Z

Xi  X 47  50   0.30 s 10

For Xi = 47,

Areas between scores and mean: For Z = –1.00, area = 0.3413 For Z = –0.30, area = 0.1179 Area between scores: (0.3413) – (0.1179) = 0.2234

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WHAT CAN I DO IN CLASS? 1. Provide students with several different Z scores and have them draw the area depicted by these scores using a Z table. Go over the answers in class. For example, find the area between a positive Z score (Z = 0.45) and a negative Z score (Z = –2.13).

2. Hand out a drawing of a theoretical normal curve and have students fill out the percentages and standard deviations. Ensure they understand the symmetrical nature of the curve.

WHAT CAN I DO ONLINE? 1. Using the Khan Academy website, ask students to practice calculating z-scores: https://www.khanacademy.org/math/ap-statistics/density-curves-normaldistribution-ap/measuring-position/e/z_scores_1 2. Ask students to use http://www.mathcracker.com/normal_probability.php to compute a normal probability distribution.

CHAPTER 5 INTRODUCTION TO INFERENTIAL STATISTICS Sampling and the Sampling Distribution SUMMARY This chapter begins with a brief overview of probability sampling techniques, but the primary goal is to familiarize students with the concept of the sampling distribution and

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its relationship to samples and populations. This concept is, of course, both difficult for students and crucial for an understanding of inferential statistics. The chapter is designed to provide a more tangible understanding of this concept by demonstrating, in Section 5.4, the construction and final shape of a sampling distribution. There is also an exercise at the end of the chapter for constructing a sampling distribution.

NEW TO THE FIFTH CANADIAN EDITION Section 5.3 demonstrates the Central Limit Theorem in ―action,‖ or how the sampling distribution changes with sample size. A new section, 5.5, on the sampling distribution of proportions has been added. It explains and shows how the normal sampling distribution can be extended to nominal- or ordinal-level variables using proportions instead of means.

ANSWERS TO TEXT PROBLEMS 5.1 a–d. Answers may vary.

WHAT CAN I DO IN CLASS? 1. Provide students with a list representing examples of populations and samples and have them label the examples as either representing a population or a sample. For example, all registered voters in the GTA, number of voters who voted: Liberal, Conservative, NDP, or other.

2. Have students list consequences of not having a representative random sample.

For example, use the rule of EPSEM.

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WHAT CAN I DO ONLINE? 1. Have students find examples of outcomes where the sample size was not large enough for the inference to the population (e.g., studies that make claims based on only a few cases). Use http://www.nature.com/nrn/journal/v14/n5/full/nrn3475.html.

2. Instruct students to go to the Statistics Canada website and write a brief summary (one to two paragraphs) of StatsCan’s sampling techniques. Use the sampling design for the Canadian Labour Force Survey (https://www150.statcan.gc.ca/n1/pub/71-526-x/2017001/chapt2-eng.htm) as an example.

CHAPTER 6 ESTIMATION PROCEDURES FOR SAMPLE MEANS AND PROPORTIONS SUMMARY Estimation procedures, as presented in this chapter, provide a more straightforward launch into inferential statistics than hypothesis testing. Students are more familiar with estimation, at least as consumers, and the logic involved is somewhat simpler. Once this material is covered, students will understand the concept of the sampling distribution, alpha levels, and several other pieces of the logic that will be introduced and developed in the chapters on hypothesis testing.

Thus, the overall goal of this chapter is to teach students how to estimate population values and to build more groundwork for hypothesis testing. This chapter builds on what was already presented about the standard normal curve and Z scores in Chapters 4 and 5 and introduces the student’s t distribution that is needed when the population standard deviation is unknown. The chapter shows to construct interval estimates for sample means and sample proportions when either the population standard deviation is known (using Z) or unknown (using t). There is a full discussion of the relationships among interval width, sample size, and confidence level, as well as discussion on how to determine appropriate sample size.

NEW TO THE FIFTH CANADIAN EDITION Most tables, figures, and application boxes have been updated, including polling data from the 2021 federal election. A new formula, Formula 6.4, has been added. In previous editions, the confidence interval for the proportion was constructed with Formula 6.3, which assumes knowledge of the population proportion Pu. Without this knowledge, Pu was set at the most conservative possible value, 0.50. However, this approach unnecessarily increases margin of error in the confidence interval. Formula 6.4, on the

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other hand, uses the sample proportion Ps as an estimate of Pu. Formula 6.4 is also the formula typically used in other statistics textbooks.

On a related matter, previous editions of the textbook used the ―n ≥ 100‖ criterion to invoke the Central Limit Theorem for sample proportions. This edition uses the more customary assumption that both (n)(Pu) and n(1 – Pu), are ≥ 15. Because the value is Pu is almost always unknown, the assumption becomes (n)(Ps) and n(1 – Ps) ≥15.

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ANSWERS TO TEXT PROBLEMS 6.1 a. 5.2  0.11 b. 100  0.71 c. 20  0.40 d. 1,020  5.42 e. 7.3  0.23 f. 33  0.80

6.2 a. 0.14  0.07 b. 0.37  0.03 c. 0.79  0.07 d. 0.43  0.03 e. 0.40  0.04 f. 0.63  0.05

6.3 Confidence Level 95%

0.05

Area Beyond Z 0.0250

94%

0.06

0.0300

1.88 or 1.89

92%

0.08

0.0400

1.75 or 1.76

97%

0.03

0.0150

2.17

98%

0.02

0.0100

2.33

99.9%

0.001

0.0005

3.27

Alpha

Z score 1.96

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6.4 Sample Size

Confidence Level (%)

Alpha

t Score

t Score Needed to Construct Confidence Interval

5

95

0.05

2.571

2.776

6

99

0.01

3.707

4.032

15

95

0.05

2.131

2.145

15

99

0.01

2.947

2.977

30

99

0.01

2.750

2.756

31

95

0.05

2.042

2.042

35

95

0.05

2.042

2.042

39

95

0.05

2.042

2.042

61

90

0.10

1.671

1.671

6.5 a. 2.30  0.04 b. 2.10  0.01, 0.78  0.07 c. 6.00  0.37 6.6 2.37  0.80

6.7 a. 178.23  1.97. The estimate is that students spent between $176.26 and $180.20 per book on average. b. 1.5  0.04. The estimate is that students visited the clinic between 1.46 and 1.54 times on the average. c. 2.8  0.13 d. 3.5  0.19 6.8 0.10  0.04

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6.9 0.14  0.05 6.10 0.30  0.07

6.11 a. Ps = (823/1,496) = 0.55 Confidence interval: 0.55  0.02 Between 53% and 57% of the population agrees with the statement. b. Ps = (650/1,496) = 0.43 Confidence interval: 0.43  0.02 c. Ps = (375/1,496) = 0.25 Confidence interval: 0.25  0.02 d. Ps = (1,023/1,496) = 0.68 Confidence interval: 0.68  0.02 e. Ps = (800/1,496) = 0.53 Confidence interval: 0.53  0.02

6.12

0.53  0.10

6.13 Alpha ( a ) 0.10 0.05 0.01 0.001

Confidence Level (%) 90 95 99 99.9

Confidence Interval 100  7.38 100  8.77 100  11.55 100  14.73

6.14

Sample A (n = 100): 0.40  0.10 Sample B (n = 1,000): 0.40  0.03 Sample C (n = 10,000): 0.40  0.01

6.15

The confidence interval is 0.51  0.05. The estimate would be that between 46% and 56% of the population prefer candidate A. The population parameter ( Pu) is equally likely to be anywhere in the interval (i.e., it is just as likely to be 46% as it is to be 56%), so a winner cannot be predicted.

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6.16 Nation United States Japan Brazil Nigeria China

Confidence Interval 0.36  0.02 0.32  0.02 0.35  0.02 0.55  0.02 0.16  0.01

6.17

The confidence interval is 0.23  0.07. At the 95% confidence level, the estimate would be that between 256 (16%) and 480 (30%) of the 1,600 incoming students would be extremely interested.

The estimated numbers are found by multiplying n (1,600) by the upper (0.30) and lower (0.16) limits of the interval.

6.18

a. b. c. d. e. f.

34.60 + 1.76 68 + 2.11 164.80 + 13.31 76 + 1.88 0.35 + 0.09 0.56 + 0.05

6.19 a. 43.87  0.49 b. 2.86  0.08 c. 1.81  0.06 d. 0.29  0.02 e. 0.18  0.01 f. 0.36  0.02 g. 0.81  0.01

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DETAILED SOLUTION TO SELECTED TEXT PROBLEM 6.9

c.i. = 0.14 ± (1.96)(0.026) c.i. = 0.14 + 0.05

WHAT CAN I DO IN CLASS? 1. According to a random sample of 325 university students, the mean for the number of hours per week spent doing housework was 18.1, with a standard deviation of 12. Calculate a 99% confidence interval for the population mean.

Make sure to explain what this confidence interval means, using a complete sentence.

2. If you used a 95% confidence interval instead of a 99% confidence interval, what would occur to the size of the interval, the precision of the interval, and the accuracy of the interval you calculated? Explain why.

WHAT CAN I DO ONLINE? 1. Using https://bc.ctvnews.ca/51-of-canadians-support-return-of-capitalpunishment-for-murder-convictions-poll-suggests-1.4837434, ask students to create a 95% confidence interval for the proportion of the population that favours the death penalty.

2. Statistics Canada reports years of education attained by Canadians in many of its surveys. Answer the following hypothetical question: Student A creates a 95% confidence interval for mean years of education, using a random sample of 200 Canadian residents. Student B creates a 95% confidence interval for mean years of education, using a random sample of 4,000 Canadian residents. Which confidence interval is likely to be smaller?

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CHAPTER 7 HYPOTHESIS TESTING WITH NOMINAL AND ORDINAL VARIABLES Chi Square SUMMARY The chapter centres on the logic of hypothesis testing and computational routines for the chi square test of independence. The structure of bivariate tables is explained in Section 7.6, and the logic of chi square is summarized in Section 7.7. (Section 7.12 presents the Yates correction for small samples.) Section 7.10 explains the distinction between statistical significance and importance. This distinction is stressed, in part, as a way of making the transition to measures of association, which are more direct indicators of the ―importance‖ (regarding relationship characteristics such as strength) of a relationship between two variables.

NEW TO THE FIFTH CANADIAN EDITION Sections 7.3 and 7.8 now include visual aids (see Figures 7.1 and 7.2) to help students better understand the sampling distribution for purposes of hypothesis testing. For example, Figure 7.1 illustrates differences and similarities between the rejection area, alpha level, test/obtained statistic, and p value. Most data and examples have also been updated.

ANSWERS TO TEXT PROBLEMS 7.1 a. 1.11 b. 0.00 c. 1.52 d. 1.46

7.2 The critical value for a chi square test with five degrees of freedom at the 0.01 level is 15.086.

7.3 The critical value for a chi square test at the 0.01 level, with three degrees of freedom is 11.345, so the researcher could not reject the null hypothesis at that level, but could reject it at the 0.10 level, which would indicate the probability of error is 10% rather than 1%.

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7.4 ² (obtained) = 2.00

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7.5 Computational Table for Problem 7.5 (1) fo 21 29 14 36 n = 100

(2) fe 17.50 32.50 17.50 32.50 n = 100.0

(3) fo – fe 3.50 −3.50 −3.50 3.50 0

(4) (5) 2 (fo – fe) (fo – fe)2/fe 12.25 0.70 12.25 0.38 12.25 0.70 12.25 0.38 2  (obtained) = 2.16

With 1 degree of freedom and alpha set at 0.05, the critical region will begin at 3.841.

The obtained chi square of 2.15 does not fall within this area, so the null hypothesis cannot be rejected. There is not a statistically significant relationship between unionization and salary.

7.6 ² (obtained) = 3.33 7.7 The obtained chi square is 5.12, which is significant (df = 1, alpha = 0.05).

7.8 ² (obtained) = 0.03 7.9 ² (obtained) = 6.67 7.10 ² (obtained) = 0.30 7.11 ² (obtained) = 12.59 7.12 ² (obtained) = 25.73 7.13 ² (obtained) = 19.33

7.14 Answers may vary.

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7.15 Problem

Chi Square 25.19 1.80 5.23 28.43 14.17

a. b. c. d. e.

Significant at  = 0.05?

Yes No No Yes Yes

DETAILED SOLUTION TO SELECTED TEXT PROBLEM 7.5 Expected frequencies (fe): For top left cell: For top right cell:

(50)(35) 1750   17.50 100 100 (50)(65) 3250 fe    32.50 100 100

fe 

fe 

(50)(35) 1750   17.50 100 100

For bottom right cell: fe 

(50)(65) 3250   32.50 100 100

For bottom left cell:

Step 4. Computing the test statistic fo 21 29 14 36 n = 100

fe 17.50 32.50 17.50 32.50 n = 100.00

fo – fe 3.50 -3.50 -3.50 3.50 0.00

(fo – fe)2 12.25 12.25 12.25 12.25

(fo – fe)2 / fe 0.70 0.38 0.70 0.38 2 χ (obtained) = 2.16

χ2(obtained) = 2.16

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WHAT CAN I DO IN CLASS? 1. Review each formula for the calculation of chi square in detail. It is helpful if statistical terms such as sampling distribution, sample data, alpha level, degrees of freedom, and so forth, are clearly identified and repeated so that students can comprehend statistics with little anxiety. For example, alpha simply means the percentage in the curve beyond the chi square (critical) value.

WHAT CAN I DO ONLINE? 1. There are many research articles online that use the chi square test. Using an online database of publications, have students find articles using this test. For example, see https://journals.plos.org/plosone/article?id=10.1371/journal.pone.0250223 2. Instruct students to find a news story and create their own question to test using chi square and the five-step method. For example, see the following article https://www.cbc.ca/news/canada/covid-restaurants-changing-model-1.6302810

CHAPTER 8 MEASURES OF ASSOCIATION FOR VARIABLES MEASURED AT THE NOMINAL LEVEL SUMMARY This chapter introduces the concept of association between variables in the context of bivariate tables. To make the transition from statistical significance to statistical association, the chapter begins by explaining in detail how measures of association are used to describe and analyze the importance of a relationship as opposed to the statistical significance of a relationship. Also, the distinction between association and causation is made here.

Next, the definition (Section 8.2) and three characteristics of a bivariate association (Section 8.3) are provided. The three characteristics are used throughout Part 3 to provide continuity across the various measures of association.

Section 8.3 also presents procedures for calculating percentages for bivariate tables and interpreting the results vis-à-vis the three characteristics of bivariate association.

Bivariate tables in the text are always organized with the column variable increasing in

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value from left to right, and the row variable increasing in value from top to bottom.

Although there is no ―right way‖ to set up tables, using this convention makes the tables comparable to output from statistics packages like SPSS. The chapter also introduces measures of association as a more precise indicator of the strength of a relationship. It considers two chi square–based measures of association, phi and Cramer’s V, and lambda. The emphasis is on lambda and the logic of proportional reduction in error. Lambda is presented as the preferred measure of association for nominal variables because of the directness of interpretation for values between zero and one.

NEW TO THE FIFTH CANADIAN EDITION Tables, graphs, and application boxes have been updated throughout the chapter. The SPSS exercises have been updated to reflect the new survey data used in the textbook.

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ANSWERS TO TEXT PROBLEMS Tables display conditional distributions (or column percentages).

8.1 Sense of Isolation Low High Totals

Living Arrangement Housing Integrated Development Neighbourhood 80.00 20.00 20.00 80.00 100.00 100.00

The conditional distributions change, so there is a relationship between the variables. The change from column to column is very large and the maximum difference is 60 percent (80.00 – 20.00). Using Table 8.5 as a guideline, we can say that this relationship is strong. From inspection of the percentages, we can see that sense of isolation is much lower for seniors living in a housing development for retirees compared to an ageintegrated neighbourhood.

8.2 a. Support for raising fees by age:

Support For Against Totals

Under 45 44.44 55.56 100.00

Age 45 or older 40.00 60.00 100.00

Maximum difference = 4.44

b. Support for raising fees by discipline:

Support For Against Totals

Discipline Social Sciences Science & & Arts Business 30.00 48.15 70.00 51.85 100.00 100.00

Maximum difference = 18.15

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c. Support for raising fees by tenured status:

Support For Against Totals

Status Tenured Non-Tenured 45.45 28.57 54.55 71.43 100.00 100.00

Maximum difference = 16.88

8.3 2019 Election Voted Didn’t Vote 87.31 11.44 12.69 88.56 100.00 100.00

2021 Election Voted Didn’t vote Totals

The maximum difference for this table is (87.31 − 11.44) or 75.87. This is a very strong relationship. People are very consistent in their voting habits.

8.4 Age More Social Occasions?

Yes No Totals

55 to 64 64.72 35.28 100.00

65 or older 62.92 37.08 100.00

f 0.59 0.04 0.18 0.16 0.75 0.02

l

8.5 Problem 8.1 8.2 a. b. c.

8.3 8.4

0.55 0.00 0.00 0.00 0.71 0.00

8.6 Answers will vary.

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8.7 a.

Political Ideology Supports SameSex Marriage?

Favour Oppose Totals

Liberal 59.42 40.58 100.00

Moderate 39.39 60.61 100.00

Conservative 26.88 73.12 100.00

The maximum difference of 32.54 indicates a strong relationship. Liberals support samesex marriage, conservatives are opposed, and moderates are intermediate.

8.8

a. Cramer's V = 0.27; lambda = 0.14 b. Cramer's V = 0.16; lambda = 0 c. Cramer's V= 0.17; lambda = 0 d. Cramer's V = 0.08; lambda = 0 e. Cramer's V = 0.20; lambda = 0.12

WHAT CAN I DO IN CLASS? 1. Review definitions of independent and dependent variables and how they are represented in statistics (x = independent, y = dependent). There is always confusion regarding these two definitions. Also, review the difference between association and statistical significance.

2. Give students Problem 8.8a. in this chapter as a practice problem in class. Go over the answer in detail, explaining each step.

WHAT CAN I DO ONLINE? 1. Using the 2021 Canadian federal election, have students pick a candidate and identify any possible association between the candidate and public opinion, and the ultimate outcome of the election. For example, see https://en.wikipedia.org/wiki/Opinion_polling_for_the_2021_Canadian_federal_e lection 2. Have students find news stories where associations between variables are listed, including the strength of those associations. For example, see https://www.theglobeandmail.com/opinion/article-covid-19s-demographicfallout-has-begun-we-have-fewer-babies-fewer/

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MEASURES OF ASSOCIATION FOR VARIABLES MEASURED AT THE ORDINAL LEVEL SUMMARY This chapter presents gamma, Kendall’s tau-b and tau-c, and Somers’ d as measures of association for ordinal-level variables organized in tables (―collapsed‖ ordinal variables), and Spearman’s rho and rho-squared for ordinal-level variables with many scores.

Considerable attention is given to the logic of pairs, which underlies the computation of gamma, Kendall’s tau-b and tau-c, and Somers’ d. This chapter also presents the procedures for testing the significance of gamma and Spearman’s rho.

NEW TO THE FIFTH CANADIAN EDITION Tables and boxes have been updated throughout the chapter. The SPSS exercises have been revised to reflect the new Canadian survey data used in the textbook.

ANSWERS TO TEXT PROBLEMS 9.1 a. G = 0.71 b. G = 0.69 c. G = –0.88 These relationships are strong. Facility in English/French and income increase with length of residence. Use the percentages to help interpret the direction of a relationship.

In the first table, 80% of recent immigrants were ―Low‖ in English/French facility, while 60% of long-term immigrants were ―High.‖ In this relationship, low scores on one variable are associated with low scores on the other, and scores increase together (as one increases, the other increases), so this is a positive relationship. In contrast, contact with country of origin decreases with length of residence (–0.88). Most recent immigrants have higher levels of contact, and most long-term immigrants have lower levels.

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9.2 Problem 7.7 7.9 7.12

Gamma 0.58 –0.41 0.50

9.3 G = 0.88 This is a very strong positive relationship.

9.4 Gamma = 0.40 Kendall’s tau-c = 0.23 Somers’ dyx = 0.17 9.5 G = 0.27 Be careful interpreting the direction of this relationship. The positive sign of gamma means that cases tend to fall along the diagonal from upper left to lower right. In this case, families with middle socio-economic status are more associated with organized sports and lower socio-economic status families with non-organized sports. Computing percentages will help you identify the direction of the relationship.

9.6 a. G = 0.24 b. G = 0.31 c. G = 0.16

9.7 G = 0.22, Z (obtained) = 0.92 9.8 Gamma = –0.17 Kendall’s tau-b = –0.12 Somers’ dyx = –0.11 9.9 G = –0.14

9.10 G = 0.08 Z (obtained) = 1.26 9.11 rs = –0.46, t (obtained) = –1.55

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9.12 Spearman’s rho = 0.78

9.13 rs = 0.10 For these nations, there is a weak relationship between diversity and inequality.

9.14 Spearman’s rho = 0.62

Z (obtained) = 3.35

9.15 a. G = –0.14 b. G = –0.17 c. G = –0.14 d. G = 0.39 e. G = –0.13 Income has moderate negative relationships with the first three and the last dependent variables. Be careful in interpreting direction for these tables and remember that a negative gamma means that cases tend to be clustered along the diagonal from lower left to upper right. Computing percentages will clarify direction. For example, for the first table, low income is associated with opposition to same-sex marriage (62% of the people in this column said ―Oppose‖), and high income is associated with support (47% of the people in this column said ―Favour,‖ and this is the highest percentage of support across the three income groups).

Income has a strong positive relationship with support for traditional gender roles. Note the way in which the dependent variable is coded: ―Favour‖ means support for traditional gender roles. A positive relationship means that cases tend to fall along the diagonal from upper left to lower right. In this case, favouring is greater for low income and declines as income increases. Is this truly a ―positive‖ relationship? As always, percentages will help clarify the direction of the relationship.

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DETAILED SOLUTIONS TO SELECTED TEXT PROBLEMS 9.7 a. ns: 11 (10 + 9 + 9 + 9) 6 (9 + 9) 9 (9 + 9) 10 (9)

nd:

7 (10 + 9 + 9 + 5) 6 (9 + 5) 9 (9 + 5) 10 (5)

= = =

11 (37) = 407 6 (18) = 108 9 (18) = 162 = 90 767

= = =

7 (33) = 6 (14) = 9 (14) = =

231 84 126 50 491

G = (767–491) / (767+491) = (276/1258) = 0.22

9.7 b. Step 4. Computing the test statistic Z (obtained )  G

ns  nd n(1 – G 2 )

 0.22

767 + 491 75(1 – 0.222 )

1258 71.37 = (0.22)(4.20) = 0.92  0.22

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WHAT CAN I DO IN CLASS? 1. Review which measure (gamma, Kendall’s tau-b and tau-c, Somers’ d, or Spearman’s rho) should be used to measure the association for two collapsed ordinal variables under different circumstances. For example, when there are tied pairs of cases such as income: Education Low High

José Noah 2.

Income Low Low

Have students work on problem 9.7 in class and go over the solution in detail, making sure there is no confusion in the definition of the many statistical terms used in this example.

WHAT CAN I DO ONLINE? 1. Find recent news stories online and have students pick one with ordinal variables that can be analyzed using the methods in this chapter. For example, see https://www.statcan.gc.ca/sites/default/files/scvsc.pdf.

2. Have students find a published article that examines depression as a dependent variable and identify the variables that the author(s) of the paper uses as the independent variables. Use https://www.nature.com/articles/s41598-021-92963w.

CHAPTER 10 HYPOTHESIS TESTING WITH MEANS AND PROPORTIONS The One-Sample Case SUMMARY This chapter applies the logic of hypothesis testing that was introduced in Chapter 7, in the context of testing for the significance of single sample means and proportions against a population value.

The chapter begins by considering the kinds of situations in which one-sample tests of hypothesis would be appropriate. It then presents an example problem, in informal language, before rephrasing the example in the terms of the five-step hypothesis testing in Section 10.2. Figure 10.1 provides an overview of the process in the one sample case.

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The section following the example problem covers tests for means computed with a known population standard deviation, means computed with an unknown standard deviation (using the t distribution introduced in Chapter 6), and proportions computed from a large sample. Section 10.3 discusses the use of one-tailed versus two-tailed tests and considers the implications of using a one-tailed or two-tailed test for the risk of Type I and Type II errors.

NEW TO THE FIFTH CANADIAN EDITION Figures, tables, and boxes have been extensively updated and new end of chapter problems added.

ANSWERS TO TEXT PROBLEMS 10.1 a.

Alpha 0.05 0.10 0.06 0.01 0.02

Form One-tailed Two-tailed Two-tailed One-tailed Two-tailed

Z (Critical) +/–1.65 +/–1.65 +/–1.88 +/–2.33 +/–2.33

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b.

Alpha 0.10 0.02 0.01 0.01 0.05

Form Two-tailed Two-tailed Two-tailed One-tailed One-tailed

n 31 24 121 31 61

t(Critical) +/–1.697 +/–2.500 +/–2.617 +/–2.457 +/–1.671

c.

1. Z (obtained) = −3.77 2. t (obtained) = −2.21 3. t (obtained) = −5.75 4. Z (obtained) = +0.65 5. Z (obtained) = −0.77

10.2 t (obtained) = +10.16 (see below for detailed solution)

10.3 a. t (obtained) = −40.54 b. t (obtained) = +29.09

10.4 Z (obtained) = +6.36

10.5 t (obtained) = -1.77

10.6 Z (obtained) = +2.84

10.7 a. t (obtained) = −13.66 b. t (obtained) = +25.51

10.8 a. t (obtained) = +0.67 b. t (obtained) = −7.09 c. Math: t (obtained) = +8.75

Reading: t (obtained) = +8.75

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10.9 t (obtained) =-13.06 10.10 t (obtained) = –37.86

10.11 Z (obtained) = +3.06 (see below for detailed solution) 10.12 Z (obtained) = –1.22 10.13 Z (obtained) = −1.48

10.14 Robberies: Ps = 0.40, Z (obtained) = +1.51 Aggravated assaults, Ps = 0.45, Z (obtained) = +0.81 10.15 a. Z (obtained) = –0.74 b. Z (obtained) = +2.19 c. Z (obtained) = −8.55 d. Z (obtained) = −18.07 e. Z (obtained) = +2.09 f. Z (obtained) = −53.33

10.16 Z (obtained) = +1.73 10.17 t (obtained) = −1.14

10.18 Answers may vary.

10.19 a. Z(obtained) = −1.28 b. Z(obtained) = 3.57 c. t(obtained) = 34.96 d. t(obtained) = −19.11 e. t(obtained) = 35.03 f. t(obtained) = −139.83

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DETAILED SOLUTION TO SELECTED TEXT PROBLEMS 10.2 Step 4. Computing the test statistic

 = 3.30 = 3.80 σ= 0.53 n = 117 Z (obtained) = ( Z (obtained) = (

)

√ √

Z (obtained) = (

) )

Z (obtained) = ( ) Z (obtained) = +10.20

10.11 Step 4. Computing the test statistic Pu = 0.10

Ps = 0.14 n = 527

  Ps  Pμ  Z (obtained) =   Pμ 1  Pμ  / n       .14  .10   Z (obtained) =   .10 .90  / 527   .04   Z (obtained) =   .00017   .04  Z (obtained) =    .013  Z (obtained) = + 3.06

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WHAT CAN I DO IN CLASS? 1. A random sample of 500 new immigrants to Canada had an average of 16.0 years of education with a standard deviation of 1.2. Is this significantly different from the national average of 13.4 years? (The sample size is for classroom purposes only as we know that 2.15 million recent immigrants were identified in the 2016 Canadian Census and that 42% had a university education.)

2. Lack of credential recognition for newcomers to Canada has been the principal reason for precarious labour participation among these groups. Many recent Canadian immigrants say that they are overlooked for jobs they are qualified for because they did not acquire their credentials in Canada.

Let’s assume that the average number of non-precarious jobs per decade for all Canadian residents is μ = 4.70 with a standard deviation of σ = 0.80. Further a random sample of 500 new immigrants found that the mean number of nonprecarious jobs/decade is ̅ = 2.7. Is there a statistically significant difference between the number of non-precarious jobs/decade obtained by all Canadian residents and the number obtained by new immigrants? Have students answer this problem using the five-step model.

WHAT CAN I DO ONLINE? 1. To illustrate significant differences in attainment between Canadian immigrants and their Canadian-born counterparts, have students access the Statistics Canada website (www.statcan.gc.ca) and review the articles depicting these disparities.

For example, see https://www150.statcan.gc.ca/n1/pub/11f0019m/11f0019m2019024-eng.htm.

2. Have students search online for a possible scenario where it would be best to use the t statistic and where the Z statistic would be more appropriate. For example, see https://www.statisticshowto.com/probability-and-statistics/hypothesistesting/t-score-vs-z-score.

CHAPTER 11 HYPOTHESIS TESTING WITH MEANS AND PROPORTIONS The Two-Sample Case SUMMARY After considering the kinds of research situations in which two sample tests of significance are appropriate, this chapter presents tests of significance for sample means (for situations where population standard deviations are known and unknown) and

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sample proportions (large samples only). Throughout the chapter, we continue to note the factors that affect the probability of rejecting the null hypothesis, and to distinguish between statistical significance and theoretical and/or practical importance.

NEW TO THE FIFTH CANADIAN EDITION There are two notable changes in the chapter. First, a different (and more precise and commonly used) sample-size requirement for the two-sample proportion Z test is used.

Previous editions of the textbook assumed that the sampling distribution was approximately normal at n ≥ 100. The new sample size requirement is (n1)(Pµ1) and n1(1 – Pµ1) are ≥ 15 and (n2)(Pµ2) and n2(1 – Pµ2) are ≥ 15. Because Pu1 and Pµ2 are typically unknown, Ps1 and Ps2 can be used as substitutes.

Second, Cohen’s d has been added to provide measure of association or more precisely effect size. It also compliments the eta statistic in Chapter 12, as both Cohen’s d and eta measure effect size. Cohen’s d is often used in conjunction with the t test.

Also, most examples and tables have been revised and a new ―You Are the Researcher‖ section (SPSS Demonstration 11.3) has been added to illustrate use of the ―IndependentSamples Proportions‖ Z test function in SPSS.

ANSWERS TO TEXT PROBLEMS 11.1 a.  = 1.39, Z (obtained) = −2.53 b.  = 1.61, Z (obtained) = 2.50

11.2 a.  = 0.10, Z (obtained) = 7.00 b.  = 0.12, Z (obtained) = 2.25 c.  = 0.11, Z (obtained) = −1.91

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11.3 a. Z (obtained) = 1.71 b. Z (obtained) = −2.50

11.4 The degrees of freedom for this problem is 11 + 12 - 2 , or 21. The critical value for a one-tailed test at the 0.01 level with 21 degrees of freedom is 2.518. Therefore, the test statistic was significant.

11.5 a.  = 0.07, Z (obtained) = 12.85 b. Phone calls:  = 0.12, Z (obtained) = –3.33 E-mail messages:  = 0.15, Z (obtained) = 20.00 11.6  = 0.22, Z (obtained) = 10.00 11.7 These are small samples (combined n’s of less than 100), so be sure to use Formulas 11.4 and 11.5in Step 4. a.  = 0.12, t (obtained) = −1.33 b.  = 0.13, t (obtained) = 14.85 11.8  = 0.47, t (obtained) = –4.25, Cohen’s d = -0.51

11.9

(Canada)  = 0.0097, t (obtained) = –30.93 (Nigeria)  = 0.0071, t (obtained) = –154.93 (China)  = 0.0067, t (obtained) = 74.63 (Mexico)  = 0.0107, t (obtained) = –74.77 (Japan)  = 0.0115, t (obtained) = –43.48

The large values for the t scores indicate that the differences are significant at very low alpha levels (i.e., they are extremely unlikely to have been caused by random chance alone). Note that women are significantly happier than men in every nation except China, where men are significantly happier.

11.10 Z (obtained) is 1.10, which is not significant for a two-tailed, 0.05 test. Therefore, there is no significant difference between Bolivian women and Peruvian women with regards to knowledge of modern contraceptives.

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(1  Pu )

11.11 Pu = 0.45, p = 0.06, Z (obtained) = 0.67

11.12 a. Pu = 0.56,  = 0.08, Z (obtained) = –0.75 (see below for detailed solution) b. Pu = 0.13,  = 0.07, Z (obtained) = –0.71

11.13 a. Pu = 0.46, p = 0.06, Z (obtained) = 2.17 b. Pu = 0.80, p = 0.07, Z (obtained) = −1.43 c. Pu = 0.72, p = 0.08, Z (obtained) = 0.75 11.14 Pu = 0.30,  = 0.05, Z (obtained) = 2.00

11.15 a. Z (obtained) = 21.50 b. Z (obtained) = −2.75 c. Z (obtained) = 24.00 d. t (obtained) = 1.89 e. t (obtained) = –5.65 f. t (obtained) = –5.50

DETAILED SOLUTION TO SELECTED TEXT PROBLEM 11.12 a.

Step 4. Computing the test statistic Special Ps1 = 0.53 n1 = 78

P 

Regular Ps2 = 0.59 n2 = 82

n1 PS 1  n2 Ps 2 78.53  82.59   .56 n1  n2 78  82

 X  X  P1  P 

n1  n2  n1n2

N1  N 2 78  82 160  (.56)(.44)  .2464  (.4964)(.1582)  .079 N 1N 2 (78)(82) (6396)

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Z (obtained) 

( Ps1  Ps 2)



p  p



.53  .59 .06   0.76 .079 .079

WHAT CAN I DO IN CLASS? 1. The following table shows sample outcomes for the number of text messages sent per day for iPhone vs. Android phone owners: Text Messages Sample 1 (iPhones) Sample 2 (Android) X 1 = 18.20 X 1 = 22.40 s1 = 2.20 s1 = 1.80 n1 = 170 n1 = 168 Is the difference between iPhones and Android owners significant?

2. Go over with the class when to use a t statistic and when to use a Z statistic and write out the five steps for the above problem.

WHAT CAN I DO ONLINE? 1. Have students find examples of recent news stories that can be analyzed using the five-step model. For example, see https://urldefense.com/v3/__https://www.mhrc.ca/national-pollingcovid__;!!MXVguWEtGgZw!do69xt1eJAuFlok_n_mH3GlLNrSdvvYbDq9Yw8c ZI5UNxNahDbI8n0ONKko4yuTqHg$.

2. From the examples found for the question above, have students choose one and write out a null and research hypothesis as best they can, and fill in the five steps.

CHAPTER 12 HYPOTHESIS TESTING WITH MORE THAN TWO MEANS One-Way Analysis of Variance SUMMARY This chapter introduces one-way analysis of variance (ANOVA). It begins with an example and emphasizes the association between ANOVA and the t test for sample means presented in Chapter 11. This connection with previous material and the continued use of the five-step model should maximize continuity for students.

The myriad computations required by ANOVA are presented in Section 12.3, and the usual computational shortcut, based on SST = SSB + SSW, is presented in Section 12.4.

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Although not particularly difficult (many students won’t believe this), the computations are certainly time consuming, so this might provide an opportunity to mention computers and statistical packages if they have not already been incorporated into the course.

Students should at least be aware that there are ways of easing the computational burden so that more emphasis and energy can be placed on analysis and interpretation.

NEW TO THE FIFTH CANADIAN EDITION New questions have been added to the end-of-chapter problem section and the SPSS exercises revised with the new survey data used in the textbook.

ANSWERS TO TEXT PROBLEMS 12.1 Problem a. b. c.

SST 231.67 455.73 8,362.55

SSB 173.17 78.53 5,053.35

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SSW 58.50 377.20 3,309.20

F Ratio 13.32 1.25 8.14

η2 0.7475 N/A 0.6043

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12.2 a. F (obtained) = 0.81, 2 is not appropriate since F (obtained) < F (critical) b. F (obtained) = 0.17, 2 is not appropriate since F (obtained) < F (critical) c. F (obtained) = 41.64, 2 = 0.8740 d. F (obtained) = 2.32, 2 is not appropriate since F (obtained) < F (critical)

12.3 Problem a. b.

SST 86.28 332.44

SSB 45.78 65.44

SSW 40.50 267.00

F Ratio 8.48 1.84

η2 0.5306 N/A

For problem 12.3a, with alpha = 0.05 and df = 2, 15, the critical F ratio would be 3.68.

We would reject the null hypothesis and conclude that decision making does vary significantly by type of relationship. By inspection of the group means, it seems that the ―cohabitational‖ category accounts for most of the differences.

12.4 SST = 26,181.78 SSB = 10,208.44 SSW = 15,973.33 F (obtained) = 4.79 2 = 0.3899

12.5 SST = 213.61 SSB = 2.11 SSW = 211.50 F (obtained) = 0.07 2 is not appropriate since F (obtained) < F (critical)

12.6 SST = 10,608.31 SSB = 3,342.89 SSW = 7,265.42 F (obtained) = 7.59 2 = 0.3151

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12.7 SST = 429.48 SSB = 124.06 SSW = 305.42 F (obtained) = 5.96 2 = 0.2889

12.8 a. F (obtained) = 3.02 (see below for detailed solution), 2 = 0.18 b. F (obtained) = 8.00 , 2 = 0.372 c. F (obtained) = 3.76, 2 = 0.218 d. F (obtained) = 0.19, 2 is not appropriate since F (obtained) < F (critical) e. F (obtained) = 0.15 , 2 is not appropriate since F (obtained) < F (critical)

12.9 Nation Mexico Canada United States

SST 300.98 156.38 286.38

SSB 154.08 20.08 135.28

SSW 146.90 136.30 151.10

F Ratio 12.59 1.77 10.74

η2 0.5119 N/A 0.4724

At alpha = 0.05 and df = 3, 36, the critical F ratio is 2.92. There is a significant difference in support for medically assisted suicide by class in Mexico and the United States, but not in Canada. The category means for Mexico suggest that the upper social class accounts for most of the differences. For the United States, there is more variation across the category means, and lower social class seems to account for most of the differences.

12.10 a. F (obtained) = 5.13 2 = 0.30 b. F (obtained) = 2.18 , 2 is not appropriate since F (obtained) < F (critical) c. F (obtained) = 1.39, 2 is not appropriate since F (obtained) < F (critical)

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DETAILED SOLUTION TO SELECTED TEXT PROBLEM 12.8 a.

SST   X 2  n X 2 SST = 3,720– (30)(10.47)2 SST = 3,702 - 3,307.5 SST =431.37





SSB   nk Xk  X

2

SSB = 10(8.90 – 10.47)2 + 10(12.70 – 10. 47)2 + 10(9.80 – 10. 47)2 SSB = 10(–1.57)2 + 10(2.23)2 + 10(–0.67)2 SSB = 10(2.46) + 10(4.97) + 10(0.45) SSB = 24.65 + 49.73 + 4.49 SSB = 78.87 SSW = SST – SSB SSW = 431.37– 78.87 SSW = 352.50 dfw = n – k = 30 – 3 = 27 dfb = k – 1 = 3 – 1 = 2

Within estimate (MSW ) 

SSW 352.50   13.06 dfw 27

Between estimate (MSB) 

F

MSB 39.43   3.02 MSW 13.06

2 

SSB 78.87   0.18 SST 431.37

SSB 78.87   39.43 dfb 2

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WHAT CAN I DO IN CLASS? 1. The table below shows the total range (kilometres per full tank of gas) for three types of cars. Calculate the ANOVA using the six steps described in this chapter to examine the effect of type of car on total kilometre range.

Compact

Mid-size

Full-size

643

469

484

655

427

456

702

525

402

X 1 = 666.67

X 2 = 473.67

X 3 = 447.33

Grand Mean = 529.22 SST = 96,324.72 SSB = 8,6052.84 dfw = 6 dfb = 2 MSW = 1,711.98 MSB = 43,026.42

SSW = 10,271.884

F (obtained) = 25.13 F (critical) = 5.14

2 = 0.8934 2. Have the students write out three limitations of ANOVA and explain each in comparison to other methods of analysis.

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WHAT CAN I DO ONLINE? 1. Ask students to find a study that uses one way ANOVA. This will help illustrate its use with empirical data in research. For example, https://ijbnpa.biomedcentral.com/articles/10.1186/1479-5868-11-51.

2. Ask students to find an example of a publication that uses ANOVA with post hoc tests. For example, see https://www.mdpi.com/1660-4601/17/17/6366/htm

CHAPTER 13 HYPOTHESIS TESTING AND MEASURES OF ASSOCIATION FOR VARIABLES MEASURED AT THE INTERVAL-RATIO LEVEL SUMMARY This chapter is a full treatment of regression and correlation. The chapter begins with the construction and interpretation of scattergrams and then proceeds to the least-squares regression line and the regression equation. Pearson’s r is presented in a separate section, followed by a section on the logic and interpretation of the coefficient of determination.

The chapter ends with a section on testing the significance of r. This chapter presents many new concepts, and a single example problem is used throughout to provide continuity and to help students integrate this material.

NEW TO THE FIFTH CANADIAN EDITION A new section (13.7 Regression, Level of Measurement, and Dummy Variables) has been added to explain, with examples, how linear regression can be extended to nominal/ordinal-level independent variables by treating them as ―dummy‖ variables.

Most tables and application boxes have also been updated, as well as the SPSS exercises.

ANSWERS TO TEXT PROBLEMS 13.1

a. Answers may vary. b. to e.

Unemployment Slope (b) 3.00 Y intercept (a) 39.00 Reg. Eq. Y = (39) + (3)X r 0.94

For Turnout (Y) and Education 12.67 -94.73 Y = (-94.73) + (12.67)X 0.98

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Neg. Campaigning -0.90 114.01 Y=(114.01) + (-0.90)X -0.87

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r2 t (obtained)

0.89 4.70

0.97 9.80

0.76 -3.08

f. Answers may vary.

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13.2

a. Answers may vary. b. Son’s Prestige = –13.12 + (1.17) Father’s Prestige Daughter’s Prestige = –6.99 + (1.04) Father’s Prestige c. For X = 72, Y’ = 71.12 For X = 72, Y’ = 67.89 d. r and (r²): Father’s

Son’s 0.83 (0.69)

Daughter’s 0.73 (0.53)

e. Father’s and Son’s: t (obtained) = 3.65 Father’s and Daughter’s: t (obtained) = 2.61 f. Answers may vary.

13.3 Slope (b) –0.37 Y intercept (a) 13.42 r –0.31 r2 0.09

13.4

Correlations

Community Engage.

Performance Score Social Media Posts * p < 0.05

13.5

Community Engage. Performance Score -0.463

Social Media Posts 0.847 * -0.238

a. Answers may vary. b.

Dependent Variables Homicide Violent Property

a b a b a b

Independent Variables Growth Density 65+ 0.78 2.54 6.15 0.15 -0.001 -0.28 829.79 983.51 949.48 9.54 -0.18 -3.28 1611.14 3770.71 8222.25 222.67 -1.13 -325.95

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c. For a growth rate of 4%, the predicted homicide rate would be 1.38. For a population density of 250, the predicted property crime rate would be 3,488.21.

For a city with 15% of its population aged 65+, the predicted rate of violent crime would be 900.28. d.

Dependent Variables Homicide

Independent Variables Growth Density 65+ 0.61 −0.47 −0.69 0.37 0.22 0.48 0.19 −0.30 −0.04 0.04 0.09 0.01 0.64 −0.28 −0.58 0.41 0.08 0.34

r r2 r r2 r r2

Violent Property

e.

Independent Variables

Dependent Variables

Growth

Density

65+

Homicide

t (obtained)

2.04

−1.42

−2.53

Violent

t (obtained)

0.50

−0.83

−0.11

Property

t (obtained)

2.21

−0.78

−1.89

f. Answers may vary

13.6 Fertility Education of Females

13.7

Correlations: r and (r²) Education of Females Maternal Mortality –0.45 (0.21) 0.91 (0.83) –0.52 (0.27)

b = 0.05, a = 53.18, r = 0.39, r2 = 0.16

13.8

Correlations: r and (t obtained) HS Grads Income Rank Expenditures 0.53(2.25) –0.71 (–3.64) HS Grads –0.74 (–3.97)

13.9

The correlation between the variables (r = 0.47) is positive and moderate to weak (r2 = 0.22) in strength.

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DETAILED SOLUTION TO SELECTED TEXT PROBLEM 13.3 Quantities needed: = 96 X = 99 Y = 1010 X² = 1107 Y² = 918 XY N XY  (X )(Y ) b X 2  (X ) 2 Calculating the N slope: (10)(918)  (96)(99) b (10)(1010)  (96)2 9180  9504 b 10100  9216 324 b 884 b  0.37 Calculating the Y-intercept _ _ a = Y – bX a= 9.90 – (–0.37)(9.60) a= 9.90 – (–3.55) a= 13.45 N XY  (X )(Y ) r Calculating Pearson’s r [ N X 2  (X ) 2 ][ N Y 2  (Y ) 2 ]

r r

(10)(918)  (96)(99) [(10)(1010)  (96) 2 ][(10)(1107)  (99) 2 ] 324 (884)(1269)

324 1121796 324 r 1059.15 r  .31 r

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WHAT CAN I DO IN CLASS? 3. Provide students with a list of scores for education (X) and number of chronic health problems (Y) for a sample of five seniors and ask them to calculate the ―least-squares‖ regression line. What can they conclude from these data? For example:

Education 10 12 15 16 20 4.

Number of Health Problems 6 3 5 2 0

Have students do problem 13.3 in class and go over the solution in detail.

WHAT CAN I DO ONLINE? 3. Using Statistics Canada’s interactive scatterplot at http://www12.statcan.gc.ca/census-recensement/2016/dp-pd/dv-vd/scatterplotdiagrammededispersion/index-en.html, ask students to describe the nature of the relationship between population aging (% of population 65+ years and older) and population growth (% change in the population from 2011 to 2016) for provinces and territories.

4. Find an article on Google Scholar (https://scholar.google.ca) that uses linear regression or correlation analysis, and have students discuss the findings.

CHAPTER 14 PARTIAL CORRELATION AND MULTIPLE REGRESSION AND CORRELATION SUMMARY The chapter begins with partial correlation. Next, Section 14.3 discusses multiple regression using partial slopes to predict scores on Y, and then Section 14.4 introduces the concept of standardized partial slopes (beta-weights) as a way of assessing the relative importance of the independents. Section 14.5 introduces the multiple correlation coefficient.

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This chapter is divided into short subsections so that instructors can more conveniently omit material—especially subsections that focus on computation. Further, the chapter uses a single example problem throughout to provide continuity and to keep students focused on the material at hand. This chapter also reexamines the problem introduced in Chapter 13 (number of children and contribution to housework) in the context of a third variable, education.

NEW TO THE FIFTH CANADIAN EDITION A discussion of multicollinearity and the correlation matrix has been added to the chapter. Some tables and application boxes have been updated and the SPSS exercises have been revised.

ANSWERS TO TEXT PROBLEMS 14.1

a. For turnout (Y) and unemployment (X) while controlling for negative advertising (Z), ryx.z = 0.97. The relationship between X and Y is not affected by the control variable Z. b. For turnout (Y) and negative advertising (X) while controlling for unemployment (Z), ryx.z = –0.92. The bivariate relationship is not affected by the control variable. c. Turnout (Y) = 70.25 + (2.09) unemployment (X1) + (-0.43) negative advertising (X2). For unemployment (X1) = 10 and negative advertising (X2) = 75, turnout (Y) = 58.90. d. For unemployment (X1): b*1 = 0.66. For negative advertising (X2): b*2 = –0.41.

Unemployment has a stronger effect on turnout than negative advertising. Note that the independent variables’ effect on turnout is in opposite directions. e. R2 = 0.98

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14.2

a. Z = illness. People who are very sick are more likely to be in a hospital and more likely to die. b. Z = extent of the fire. More serious fires attract more firefighters and result in more damage. c. Z = age. Older men are both more likely to listen to this type of music and more likely to develop cataracts. d. Z = temperature. Higher temperature affects both ice cream sale and drownings.

14.3

a. For strife (Y) and unemployment (X), controlling for urbanization (Z), ryx.z = 0.79. b. For strife (Y) and urbanization (X), controlling for unemployment (Z), ryx.z = 0.20. c. Strife (Y) = (–14.61) + (4.94) unemployment (X1) + (0.16) urbanization (X2).

With unemployment = 10 and urbanization = 90, strife (Y’) would be 49.19. d. For unemployment (X1): b*1 = 0.78. For urbanization (X2): b*2 = 0.12. e. R² = 0.65

14.4

a. Homicide = 1.35 + (0.13) Growth + (−0.001) Density Violent = 945.15 + (4.04) Growth + (−0.16) Density Property = 1691.88 + (218.82) Growth + (−0.11) Density b. For growth = 8 and density = 500 Homicide = 0.87 Violent = 897.47 Property = 3,387.44 c. Homicide = (−0.915) Growth + (-0.594) Density + (−1.419) %65 Violent = (0.665) Growth + (−0.132) Density + (0.587) %65 Property = (0.589) Growth + (−0.037) Density + (−0.041) %65 d. Homicide: R = 0.84, R² = 0.70 Violent: R = 0.37, R² = 0.14 Property: R = 0.64, R² = 0.41

14.5 a. b.

Performance score (Y) = 76.43 + (-0.83) community engagement (X1) + (1.88) social media posts (X2).

For X1= 20 and X2 = 0, Y = 59.83

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c. d.

Zy = (-0.93)Z1 + (0.55)Z2 R 2 = 0.30

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14.6

Zero-order correlations Income # of kids Cohesion –.04 .37 a. Cohesion = 4.31 + (–.00) income + (.65) #kids b. For income = 20,000 and #kids = 6, cohesion = 8.21 c. Cohesion = (–.06) income + (.37) #kids d. R = .37, R² = .14

14.7

a. Zy = (–0.001) HS grads (Z1) + (−0.74) Rank (Z2) b. R² = 0.54

14.8

Zero-order correlations # kids Age Prestige TV hours .01 –.27 –.03 a. TV hours = (–.32) Age + (.14) # kids b. R = .31, R² = .09

DETAILED SOLUTION TO SELECTED TEXT PROBLEM 14.3 a.

Strife Unemployment

Zero order correlations Unemployment Urban .80 .25 .17

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Strife (Y) and unemployment (X), controlling for urban (Z): ryx  (ryz )(rxz ) ryx. z  1  ryz 2 1  rxz 2 .80  (.25)(.17) ryx. z  1  (.25) 2 1  (.17) 2

.80  .04 1  .06 1  .03 .76 ryx. z  .94 .97 .76 ryx. z  (.97)(.98) ryx. z  .76 .95 ryx. z  .80 ryx. z 

b.

Partial slopes:

 sy  ry1  ry 2r12   16.73   .80  (.25)(.17)   .80  .04   .76  b1       6.77        6.77     (6.77)(.78)  5.30 2 2  s1  1  r 12   2.47   1  (.17)   1  .03   .97 

 sy  ry 2  ry1r12   16.73   .25  (.80)(.17)   .25  .14   .11  b 2      1.38       1.38     (1.38)(.11)  .15 2 2  s 2  1  r 12   12.13   1  (.17)  1  .03   .97  

Y-intercept: a  Y  b1 X 1  b2 X 2

a  18.00  (5.30)(4.55)  (.15)(62.7) a  18.00  24.12  9.41 a  15.53 c.

Multiple regression equation: Y = a + b1X1 + b2X2 Y = –15.53 + (5.30) X1 + (.15) X2

d.

Beta-weights:

 s1   2.47  b* 1  b1    5.30    (5.30)(.15)  .80  16.73   sy 

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 s2   12.13  b* 2  b 2    .15    (.15)(.73)  .11  16.73   sy 

e.

Coefficient of multiple determination:

R 2  r 2 y1  r 2 y 2.1(1  r 2 y1) R 2  (.80) 2  (.20)2 (1  (.80)2 ) R 2  .64  (.04)(1  .64) R 2  .64  (.04)(.36) R 2  .64  .01 R 2  .65

WHAT CAN I DO IN CLASS? 5. Have students re-examine the problem introduced in Chapter 13 (Instructor’s Manual) using a second independent variable, income. Provide students with a list of five scores for education (X1), annual income (in 000s) (X2), and number of chronic health problems (Y), and have them calculate the multiple regression equation. What can students conclude from these data? For example:

Education 10 12 15 16 20

6.

Income 4 7 7 3 2

Number of Health Problems 6 3 5 2 0

Give students Problem 14.3 in this chapter to do as a practice problem in class.

Go over the answer in detail, explaining each step.

WHAT CAN I DO ONLINE? 5. Use PubMed Central (https://www.ncbi.nlm.nih.gov/pmc/) to access an article where social class (e.g., income, education, occupation) is used as an independent variable in a multiple linear regression analysis and ask students to discuss the findings.

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6. Find recent news stories online and have students pick one with interval-ratio variables that can be analyzed using the methods in this chapter. For example, see https://www.thestar.com/news/gta/2017/10/19/torontos-tech-talent-being-used-towoo-amazon.html.

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