And Alternate
The given problem involves testing hypotheses regarding the difference in population means, comparing sample data from two independent groups, and assessing whether there is a statistically significant difference at a specified significance level.
The hypotheses to be tested are:
H0: µ1 = µ2 (there is no difference between the population means)
H1: µ1 ≠ µ2 (there is a difference between the population means)
A sample of 9 observations from one population yielded a sample mean of 22 with a standard deviation of 4.7. A second sample of 9 observations from another population had a sample mean of 26 and a standard deviation of 3.6. The significance level is 0.10. The task is to determine if there is sufficient evidence to conclude a difference exists between the population means.
Additionally, another scenario involves testing whether the population proportion of cars with damage exceeds the park’s claimed probability of 0.65, based on a sample of 20 cars with 18 experiencing damage.
Furthermore, a linear programming problem involves optimizing meal preparation at a restaurant, considering constraints related to time, demand, and profits, and evaluating the impacts of advertising investment, staff reliability, and pricing strategies.
Paper For Above instruction
Introduction
Statistical hypothesis testing is a fundamental aspect of inferential statistics, allowing researchers to make decisions about population parameters based on sample data. This paper addresses multiple hypothesis testing scenarios, including comparing population means and proportions, and discusses how decisions are made under different significance levels. Additionally, it explores the application of linear programming in operational decision-making in a restaurant setting, emphasizing how constraints and strategic decisions influence outcomes.
Comparing Population Means: Independent Samples t-Test
The first part of the analysis involves testing whether two population means differ significantly. The hypotheses are set as H0: µ1 = µ2 and H1: µ1 ≠ µ2. With sample sizes of 9 in each group, the sample means are 22 and 26, with standard deviations of 4.7 and 3.6 respectively. To determine if the difference is statistically significant at the 0.10 level, the pooled variance estimate and t-statistic are calculated.
Decision Rule
The decision rule involves comparing the calculated t-statistic with the critical t-value at the degrees of freedom (df = n1 + n2 - 2 = 16) for a two-tailed test at α=0.10. The critical t-value for df=16 and α=0.10/2=0.05 in each tail is approximately ±1.746.
*If the calculated t exceeds ±1.746, the null hypothesis is rejected; otherwise, it is not rejected.*
Calculation of Pooled Variance
The pooled variance is given by:
\[ S_p^2 = \frac{(n_1 - 1) S_1^2 + (n_2 - 1) S_2^2}{n_1 + n_2 - 2} \]
Calculating:
\[ S_p^2 = \frac{(8)(4.7)^2 + (8)(3.6)^2}{16} \approx \frac{8 \times 22.09 + 8 \times 12.96}{16} = \frac{176.72 + 103.68}{16} = \frac{280.4}{16} \approx 17.525 \]
Thus, the pooled variance estimate is approximately 17.525.
Calculation of Test Statistic
The standard error (SE) for the difference in means is:
\[ SE = \sqrt{S_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} \]
\[ SE = \sqrt{17.525 \left(\frac{1}{9} + \frac{1}{9}\right)} = \sqrt{17.525 \times \frac{2}{9}} \approx \sqrt{17.525 \times 0.222} \approx \sqrt{3.894} \approx 1.973 \]
The t-statistic is then:
\[ t = \frac{\bar{X}_1 - \bar{X}_2}{SE} = \frac{22 - 26}{1.973} \approx -4 / 1.973 \approx -2.026 \]
Since |-2.026| > 1.746, the null hypothesis is rejected, indicating a significant difference at the 0.10 level.
Testing Proportion: Damage in Safari Park Scenario
The second scenario models the probability that the proportion of damaged cars exceeds the claimed probability of 0.65.
Parameter Definition
Let p be the true proportion of cars with damage during a safari drive-through.
Observed Test Statistic
Sample data: 20 cars, with 18 damaged. The sample proportion \( \hat{p} \) is:
\[ \hat{p} = \frac{18}{20} = 0.9 \]
The test statistic for testing proportions is:
\[ Z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \]
Where \( p_0 = 0.65 \). Calculating:
\[ Z = \frac{0.9 - 0.65}{\sqrt{\frac{0.65 \times 0.35}{20}}} = \frac{0.25}{\sqrt{\frac{0.2275}{20}}} = \frac{0.25}{\sqrt{0.011375}} \approx \frac{0.25}{0.1067} \approx 2.345 \]
P-Value Calculation
Since this is a one-sided test \( (p > 0.65) \), the p-value is:
\[ p = P(Z > 2.345) \]
Using Z-tables, the p-value is approximately 0.0096, indicating strong evidence to reject the null hypothesis at the 0.05 significance level.
Distribution Under Null Hypothesis
The test statistic follows a standard normal distribution, N(0,1), under the null hypothesis, suitable for p-value calculation.
Decision and Conclusion
Given the p-value < 0.05 and the decision to reject H0 at the 0.05 level, there is significant evidence that the true proportion exceeds 0.65. The appropriate conclusion is:
**There is sufficient evidence to conclude that the population proportion of cars with damage exceeds the park’s claim.**
Operational Decision-Making Using Linear Programming
The restaurant scenario involves optimizing meal preparation considering constraints like demand, time, and profit margins. The primary goal is to maximize profit subject to these constraints, which are formulated as a linear programming model.
**Objective Function:**
Maximize \( Z = 12x + 16y \), where \( x \) = number of chicken dinners, \( y \) = number of pork dinners.
**Constraints:**
- Time constraint: \( 15x + 30y \leq 1200 \) (20 hours = 1200 minutes)
- Demand constraint: \( x + y \leq 60 \) (maximum meals per night)
- Demand ratio: \( x \geq \frac{3}{2} y \) (at least 1.5 chicken for every pork)
- Minimum pork proportion: \( y \geq 0.10 (x + y) \)
- Non-negativity: \( x, y \geq 0 \)
**Impact of Advertising:**
Increasing maximum capacity from 60 to 70 meals through advertising potentially increases profit, assuming other constraints are met and demand justifies it.
**Staff Reliability:**
A reduction of 5 hours (300 minutes) would decrease available time and thus the maximum feasible number of meals, reducing profit margins.
**Price Increase for Chicken Dinners:**
Raising profit per chicken meal from $12 to $14 increases the overall maximization objective and could lead to increased profit, provided demand and constraints are satisfied.
Conclusion
The hypothesis testing scenarios confirm that there is statistical evidence of a difference in population means and that the proportion of damaged cars exceeds the park’s claim. The decision-making framework in the restaurant scenario demonstrates how operational constraints influence profit strategies and how strategic investments could optimize revenue.
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