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Sagot :
### Rite Aid Analysis
#### a. Populations
The two populations being studied are:
1. The customer satisfaction scores for Rite Aid in Year 1.
2. The customer satisfaction scores for Rite Aid in Year 2.
#### b. Measurement
We are measuring the mean satisfaction scores for these populations. Define the population means using the following symbols:
- μY1: Mean satisfaction score for Year 1.
- μY2: Mean satisfaction score for Year 2.
#### c. Hypotheses Testing
We want to test whether the mean satisfaction score for Rite Aid increased from Year 1 to Year 2.
- Null hypothesis (H0): μY2 ≤ μY1 (Rite Aid's satisfaction did not increase)
- Alternative hypothesis (H1): μY2 > μY1 (Rite Aid's satisfaction increased)
##### Test Statistic and Conclusion
The test statistic for comparing the two means (assuming we use a Z-test):
- Mean difference: [tex]\( \overline{X_2} - \overline{X_1} = 76 - 73 = 3 \)[/tex]
- Standard error (SE): [tex]\( \sigma / \sqrt{n} = 6 / \sqrt{60} \approx 0.7746 \)[/tex]
- Z-statistic: [tex]\( \text{Z} = \frac{\text{Mean Difference}}{\text{SE}} \approx \frac{3}{0.7746} \approx 3.873 \)[/tex]
The critical value for a one-tailed test at α = 0.05 is approximately 1.645.
- Since 3.873 > 1.645, we reject the null hypothesis.
Conclusion: There is sufficient evidence to conclude that Rite Aid's satisfaction score increased from Year 1 to Year 2.
#### d. Confidence Interval
Construct a 95% confidence interval to estimate the difference in the population means.
- Critical value (Z-critical): 1.645 for a one-tailed test at α = 0.05.
- Confidence interval for the difference in means:
[tex]\[ \text{CI} = \left( (\overline{X_2} - \overline{X_1}) - Z_{\alpha} \cdot \text{SE}, (\overline{X_2} - \overline{X_1}) + Z_{\alpha} \cdot \text{SE} \right) = (3 - 1.645 \cdot 0.7746, 3 + 1.645 \cdot 0.7746) \approx (1.726, 4.274) \][/tex]
Interpretation:
The 95% confidence interval for the difference in satisfaction scores is (1.726, 4.274), suggesting the difference is significant and positively different from zero. This confirms that customer satisfaction at Rite Aid likely improved.
### J.C. Penney Analysis
#### a. Populations
The two populations being studied are:
1. The customer satisfaction scores for J.C. Penney in Year 1.
2. The customer satisfaction scores for J.C. Penney in Year 2.
#### b. Measurement
We are measuring the mean satisfaction scores for these populations. Define the population means using the following symbols:
- μY1: Mean satisfaction score for Year 1.
- μY2: Mean satisfaction score for Year 2.
#### c. Hypotheses Testing
We want to test whether the mean satisfaction score for J.C. Penney increased from Year 1 to Year 2.
- Null hypothesis (H0): μY2 ≤ μY1 (J.C. Penney's satisfaction did not increase)
- Alternative hypothesis (H1): μY2 > μY1 (J.C. Penney's satisfaction increased)
##### Test Statistic and P-value
The test statistic for comparing the two means (assuming we use a t-test):
- Mean difference: [tex]\( \overline{X_2} - \overline{X_1} = 78 - 77 = 1 \)[/tex]
- Standard error (SE): [tex]\( \sigma / \sqrt{n} = 6 / \sqrt{60} \approx 0.7746 \)[/tex]
- t-statistic: [tex]\( \text{t} = \frac{\text{Mean Difference}}{\text{SE}} \approx \frac{1}{0.7746} \approx 1.291 \)[/tex]
Using a t-distribution with 59 degrees of freedom:
- P-value: Using a calculator, we find the p-value ≈ 0.1009.
Since the p-value (0.1009) > α (0.05), we fail to reject the null hypothesis.
Conclusion: There is insufficient evidence to conclude that J.C. Penney's satisfaction score significantly increased from Year 1 to Year 2.
#### d. Confidence Interval
Construct a 95% confidence interval to estimate the difference in the population means.
- Critical value (t-critical): Approximately 1.671 for 59 degrees of freedom and α = 0.05.
- Confidence interval for the difference in means:
[tex]\[ \text{CI} = \left( (\overline{X_2} - \overline{X_1}) - t_{\alpha} \cdot \text{SE}, (\overline{X_2} - \overline{X_1}) + t_{\alpha} \cdot \text{SE} \right) = (1 - 1.671 \cdot 0.7746, 1 + 1.671 \cdot 0.7746) \approx (-0.2944, 2.2944) \][/tex]
Interpretation:
The 95% confidence interval for the difference in satisfaction scores is (-0.2944, 2.2944), which contains zero. This implies the difference is not statistically significant, supporting the conclusion that there is insufficient evidence to suggest a significant increase in customer satisfaction for J.C. Penney.
#### a. Populations
The two populations being studied are:
1. The customer satisfaction scores for Rite Aid in Year 1.
2. The customer satisfaction scores for Rite Aid in Year 2.
#### b. Measurement
We are measuring the mean satisfaction scores for these populations. Define the population means using the following symbols:
- μY1: Mean satisfaction score for Year 1.
- μY2: Mean satisfaction score for Year 2.
#### c. Hypotheses Testing
We want to test whether the mean satisfaction score for Rite Aid increased from Year 1 to Year 2.
- Null hypothesis (H0): μY2 ≤ μY1 (Rite Aid's satisfaction did not increase)
- Alternative hypothesis (H1): μY2 > μY1 (Rite Aid's satisfaction increased)
##### Test Statistic and Conclusion
The test statistic for comparing the two means (assuming we use a Z-test):
- Mean difference: [tex]\( \overline{X_2} - \overline{X_1} = 76 - 73 = 3 \)[/tex]
- Standard error (SE): [tex]\( \sigma / \sqrt{n} = 6 / \sqrt{60} \approx 0.7746 \)[/tex]
- Z-statistic: [tex]\( \text{Z} = \frac{\text{Mean Difference}}{\text{SE}} \approx \frac{3}{0.7746} \approx 3.873 \)[/tex]
The critical value for a one-tailed test at α = 0.05 is approximately 1.645.
- Since 3.873 > 1.645, we reject the null hypothesis.
Conclusion: There is sufficient evidence to conclude that Rite Aid's satisfaction score increased from Year 1 to Year 2.
#### d. Confidence Interval
Construct a 95% confidence interval to estimate the difference in the population means.
- Critical value (Z-critical): 1.645 for a one-tailed test at α = 0.05.
- Confidence interval for the difference in means:
[tex]\[ \text{CI} = \left( (\overline{X_2} - \overline{X_1}) - Z_{\alpha} \cdot \text{SE}, (\overline{X_2} - \overline{X_1}) + Z_{\alpha} \cdot \text{SE} \right) = (3 - 1.645 \cdot 0.7746, 3 + 1.645 \cdot 0.7746) \approx (1.726, 4.274) \][/tex]
Interpretation:
The 95% confidence interval for the difference in satisfaction scores is (1.726, 4.274), suggesting the difference is significant and positively different from zero. This confirms that customer satisfaction at Rite Aid likely improved.
### J.C. Penney Analysis
#### a. Populations
The two populations being studied are:
1. The customer satisfaction scores for J.C. Penney in Year 1.
2. The customer satisfaction scores for J.C. Penney in Year 2.
#### b. Measurement
We are measuring the mean satisfaction scores for these populations. Define the population means using the following symbols:
- μY1: Mean satisfaction score for Year 1.
- μY2: Mean satisfaction score for Year 2.
#### c. Hypotheses Testing
We want to test whether the mean satisfaction score for J.C. Penney increased from Year 1 to Year 2.
- Null hypothesis (H0): μY2 ≤ μY1 (J.C. Penney's satisfaction did not increase)
- Alternative hypothesis (H1): μY2 > μY1 (J.C. Penney's satisfaction increased)
##### Test Statistic and P-value
The test statistic for comparing the two means (assuming we use a t-test):
- Mean difference: [tex]\( \overline{X_2} - \overline{X_1} = 78 - 77 = 1 \)[/tex]
- Standard error (SE): [tex]\( \sigma / \sqrt{n} = 6 / \sqrt{60} \approx 0.7746 \)[/tex]
- t-statistic: [tex]\( \text{t} = \frac{\text{Mean Difference}}{\text{SE}} \approx \frac{1}{0.7746} \approx 1.291 \)[/tex]
Using a t-distribution with 59 degrees of freedom:
- P-value: Using a calculator, we find the p-value ≈ 0.1009.
Since the p-value (0.1009) > α (0.05), we fail to reject the null hypothesis.
Conclusion: There is insufficient evidence to conclude that J.C. Penney's satisfaction score significantly increased from Year 1 to Year 2.
#### d. Confidence Interval
Construct a 95% confidence interval to estimate the difference in the population means.
- Critical value (t-critical): Approximately 1.671 for 59 degrees of freedom and α = 0.05.
- Confidence interval for the difference in means:
[tex]\[ \text{CI} = \left( (\overline{X_2} - \overline{X_1}) - t_{\alpha} \cdot \text{SE}, (\overline{X_2} - \overline{X_1}) + t_{\alpha} \cdot \text{SE} \right) = (1 - 1.671 \cdot 0.7746, 1 + 1.671 \cdot 0.7746) \approx (-0.2944, 2.2944) \][/tex]
Interpretation:
The 95% confidence interval for the difference in satisfaction scores is (-0.2944, 2.2944), which contains zero. This implies the difference is not statistically significant, supporting the conclusion that there is insufficient evidence to suggest a significant increase in customer satisfaction for J.C. Penney.
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