Quadro Corporation has two supermarket stores in a city. The company’s quality control department wanted to check if the customers are equally satisfied with the service provided at these two stores. A sample of 380 customers selected from Supermarket I produced a mean satisfaction index of 7.6 (on a scale of 1 to 10, 1 being the lowest and 10 being the highest) with a standard deviation of .75. Another sample of 370 customers selected from Supermarket II produced a mean satisfaction index of 8.1 with a standard deviation of .59. Assume that the customer satisfaction index for each supermarket has unknown but same population standard deviation.

a. Construct a 98% confidence interval for the difference between the mean satisfaction indexes for all customers for the two supermarkets.

b. Test at the 1% significance level whether the mean satisfaction indexes for all customers for the two supermarkets are different.

To tackle the problem of comparing customer satisfaction between Quadro Corporation's two supermarkets, we will first construct a confidence interval for the difference in mean satisfaction indexes and then perform a hypothesis test to determine if there is a significant difference between the two means.

Constructing the Confidence Interval

To create a 98% confidence interval for the difference between the mean satisfaction indexes of the two supermarkets, we will use the following formula:

Confidence Interval = (M1 - M2) ± Z * SE

Where:

• M1 = Mean satisfaction index for Supermarket I = 7.6
• M2 = Mean satisfaction index for Supermarket II = 8.1
• Z = Z-value for 98% confidence level (approximately 2.33)
• SE = Standard Error of the difference in means

First, we need to calculate the Standard Error (SE) using the formula:

SE = √((σ1²/n1) + (σ2²/n2))

Here, σ1 and σ2 are the standard deviations, and n1 and n2 are the sample sizes:

• σ1 = 0.75, n1 = 380
• σ2 = 0.59, n2 = 370

Now, substituting the values:

SE = √((0.75²/380) + (0.59²/370))

SE = √((0.5625/380) + (0.3481/370))

SE = √(0.001480 + 0.000941) = √0.002421 ≈ 0.0492

Next, we calculate the difference in means:

M1 - M2 = 7.6 - 8.1 = -0.5

Now we can construct the confidence interval:

Confidence Interval = -0.5 ± 2.33 * 0.0492

Confidence Interval = -0.5 ± 0.1144

This results in:

• Lower limit: -0.5 - 0.1144 = -0.6144
• Upper limit: -0.5 + 0.1144 = -0.3856

Thus, the 98% confidence interval for the difference in mean satisfaction indexes is approximately (-0.6144, -0.3856).

Hypothesis Testing

Next, we will test the hypothesis to see if there is a significant difference between the mean satisfaction indexes at a 1% significance level.

We set up our hypotheses as follows:

• Null Hypothesis (H0): μ1 - μ2 = 0 (no difference in means)
• Alternative Hypothesis (H1): μ1 - μ2 ≠ 0 (there is a difference in means)

We will use the Z-test for this hypothesis test, calculated as:

Z = (M1 - M2) / SE

Substituting the values we have:

Z = (-0.5) / 0.0492 ≈ -10.16

Now, we compare the calculated Z-value with the critical Z-value for a two-tailed test at the 1% significance level. The critical Z-values are approximately ±2.576.

Since -10.16 is much less than -2.576, we reject the null hypothesis.

Summary of Findings

In summary, the 98% confidence interval for the difference in mean satisfaction indexes is approximately (-0.6144, -0.3856), which does not include zero. This indicates that there is a statistically significant difference in customer satisfaction between the two supermarkets. Furthermore, the hypothesis test at the 1% significance level also leads us to reject the null hypothesis, confirming that the mean satisfaction indexes for the two supermarkets are indeed different.