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Will improving customer service result in higher stock prices for the companies providing the better service? "When a company's satisfaction score has improved over the prior year's results and is above the national average (75.7), studies show its shares have a good chance of outperforming the broad stock market in the long run."

The following satisfaction scores of two companies for the 4th quarters of two previous years were obtained from the American Customer Satisfaction Index. Assume that the scores are based on a poll of 60 customers from each company. Because the polling has been done for several years, the standard deviation can be assumed to equal 6 points in each case.

\begin{tabular}{|l|c|c|}
\hline \multicolumn{1}{|c|}{ Company } & Year 1 & Year 2 \\
\hline Rite Aid & 73 & 76 \\
\hline J.C. Penney & 77 & 78 \\
\hline
\end{tabular}

a. To see if Rite Aid's scores increased from Year 1 to Year 2, what are the two populations being studied?

b. What about these populations is being measured? Define the population means using the correct symbols.

c. Formulate the null and alternative hypotheses and test whether Rite Aid's mean satisfaction score increased in Year 2 over Year 1. At [tex]\alpha = 0.05[/tex], what is your conclusion? (Using the formulas, show your work by writing out the solution including a rejection region and clearly state your conclusion in the context of the problem.)

d. Construct a [tex]95\%[/tex] confidence interval to estimate the difference in the population means and explain what the results tell you. (Show your work using the formulas.)

e. Do parts a, b, c, and d for J.C. Penney. However, use your calculator to find a p-value for the test and a confidence interval. (You don't have to use the formulas.)



Answer :

GinnyAnswer
### 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.

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