/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Quadro Corporation has two super... [FREE SOLUTION] | 91Ó°ÊÓ

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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.

Short Answer

Expert verified
The 98% confidence interval for the difference between the mean satisfaction indexes for all customers for the two supermarkets is ranging between -17.95 to 16.95. After performing the hypothesis test using the 1% significance level, it has been conclusively shown that we fail to reject the null hypothesis. Hence, we do not have sufficient evidence to say that the supermarkets have different customer satisfaction indexes.

Step by step solution

01

Constructing the Confidence Interval

To construct a confidence interval for the difference between the means of the satisfaction index of the customers from the two supermarkets, we first calculate the mean difference which is \(7.6 - 8.1 = -0.5. \) Now, since the population standard deviation is assumed to be the same for both supermarkets, we calculate the combined standard deviation as \(\sqrt{(75^2/380) + (.59^2/370)} =7.49\). We look for the critical value for 98% confidence level, which is equal to 2.33 (from Z-Table). Hence, the confidence interval is given by \(mean \pm critical \space value * combined \space standard \space deviation\), which will be \(-0.5 \pm (2.33 * 7.49)\)
02

Hypothesis Testing

Next, we need to test whether the mean satisfaction indexes for all customers for the two supermarkets are different. For that we follow a Hypothesis testing approach:a) Set the null and alternative hypotheses. Here, the null hypothesis \((H_0)\) assumes that the mean satisfaction is the same for both supermarkets, \(H_0: \mu_1 = \mu_2\). The alternative hypothesis \((H_A or H_1)\) assumes that the mean satisfaction is not the same, \(H_1: \mu_1 \neq \mu_2\). b) Compute the test statistic which is calculated as: \(Z = (Mean \space difference - 0) / combined \space standard \space deviation\)c) If this test statistic > critical value, we reject the null hypothesis, which is determined by the significance level. Here, the test statistic \(-0.5/7.49= -0.0668\). The critical area for the z-score under a 1% significance level, from Z-table is ±2.57. As -0.0668 is outside this critical region, we fail to reject the null hypothesis.
03

Conclusion

After the analysis, we concluded that the confidence interval for the difference between the mean satisfaction indices is \(-0.5 \pm 17.45\), which implies that the true mean difference of satisfaction index might be in range (-17.95 to 16.95) with 98% confidence. And after executing the hypothesis testing, we have seen that there's not enough evidence to reject the null hypothesis. Therefore, it is not possible to claim that the supermarkets have different customer satisfaction indexes based on the data provided at the 1% significance level.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Confidence Interval
A confidence interval gives us a range of values within which we expect the true difference in means between two groups to lie. In the case of Quadro Corporation's supermarkets, we constructed a 98% confidence interval to estimate the difference between the mean satisfaction indexes from both stores.

To calculate this confidence interval, we found the mean difference of the two satisfaction scores, which was \(-0.5\) (Supermarket I having a lower satisfaction mean than Supermarket II). We calculated a combined standard deviation using both sample sizes and standard deviations, which was approximately \(7.49\). The critical value determined from the Z-table for a 98% confidence level was \(2.33\).

With these values, the confidence interval was determined using the formula: \[ ext{mean difference} \pm ext{critical value} \times ext{combined standard deviation}\] which results in: \(-0.5 \pm (2.33 \times 7.49)\).

This interval shows that we are 98% confident that the true difference in customer satisfaction between the two supermarkets is between approximately \(-17.95\) and \(16.95\), suggesting that the difference might not be substantial.
Central Limit Theorem
The Central Limit Theorem (CLT) is a key concept in statistics that allows us to make inferences about a population based on sample data. It states that when the sample size is large enough, the distribution of the sample mean will be roughly normally distributed, regardless of the distribution of the population from which the sample was taken.

In our exercise, we are dealing with large sample sizes from both supermarkets (380 and 370), which means we can safely apply the Central Limit Theorem. This application allows us to assume that the sample means (customer satisfaction indices in our case) will be approximately normally distributed. This normal approximation is critical for calculating the confidence interval and performing hypothesis testing using z-scores.

Thanks to the CLT, even if the individual satisfaction ratings aren't normally distributed, we don't need to worry. As long as the samples are large, our statistical procedures remain valid.
Customer Satisfaction
Customer satisfaction is a measure of how well a company's products or services meet or exceed customer expectations. For Quadro Corporation, understanding customer satisfaction at their supermarkets is crucial. It's represented here by a satisfaction index, rating service on a scale from \(1\) to \(10\). This helps the company quantify satisfaction into a measurable outcome.

In the context of this problem, the quality control department noted the mean satisfaction index at both supermarkets. Supermarket I had a mean satisfaction index of \(7.6\), while Supermarket II had \(8.1\). These indices provide a snapshot of how satisfied customers are with their shopping experience.

Measuring customer satisfaction can highlight strengths and weaknesses in service. It's a valuable tool for strategizing improvements and maintaining a competitive edge. Whether or not the supermarkets statistically show a difference, continued efforts to understand and improve satisfaction are beneficial.
Z-score
A z-score is a statistical measure that tells us how many standard deviations an element is from the mean. It is utilized in the context of our problem to determine if the observed difference in satisfaction indices is statistically significant.

We calculate the z-score by taking the difference between the means and dividing by the combined standard deviation. Here, the z-score was calculated as: \(Z = \frac{\text{Mean difference}}{\text{Combined standard deviation}} = \frac{-0.5}{7.49} = -0.0668\).

The significance of the z-score is assessed against a critical value from the z-table, determined by the chosen significance level. For a \(1\%\) significance level, our critical value was \(\pm 2.57\). Our calculated z-score, \(-0.0668\), was much smaller in absolute terms and outside the critical region, which means we did not find sufficient evidence to reject the null hypothesis.

Understanding the z-score helps us in hypothesis testing to decide if a supposed effect (here, the difference in satisfaction) is likely due to chance or if it might represent a real difference.

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Most popular questions from this chapter

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