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Chapter 8: Problem 8

The National Quality Research Center at the University of Michigan provides a quarterly measure of consumer opinions about products and services (The Wall Street Journal, February 18,2003 ). A survey of 10 restaurants in the Fast Food/Pizza group showed a sample mean customer satisfaction index of \(71 .\) Past data indicate that the population standard deviation of the index has been relatively stable with \(\sigma=5\) a. What assumption should the researcher be willing to make if a margin of error is desired? b. Using \(95 \%\) confidence, what is the margin of error? c. What is the margin of error if \(99 \%\) confidence is desired?

Short Answer

Expert verified
The researcher should assume that the customer satisfaction index scores are normally distributed. The margin of error for a 95% confidence interval is 3.09, while for a 99% confidence interval, it is 4.079.

Step by step solution

01

Assumption for Margin of Error

The researcher should be willing to assume that the customer satisfaction index scores for the population are normally distributed to use the given population standard deviation \( (\sigma=5) \) and use a t-distribution or z-distribution for calculating the margin of error.
02

Margin of Error for 95% Confidence Interval

To calculate the margin of error for 95% confidence interval, we will use the formula: \[Margin\_of\_Error = z_{\alpha/2} * \frac{\sigma}{\sqrt{n}} \] Where \(z_{\alpha/2}\) is the z-score corresponding to the desired level of confidence, \(\sigma\) is the population standard deviation, and \(n\) is the sample size. Since we are given a 95% confidence level, we must find the \(z_{\alpha/2}\) score for 95%. This corresponds to the z-score of 1.96: \(z_{\alpha/2} = 1.96\) Now, we have everything to calculate the margin of error: \(Margin\_of\_Error = 1.96 * \frac{5}{\sqrt{10}}\) \(Margin\_of\_Error = 3.09\) The margin of error for a 95% confidence interval is 3.09.
03

Margin of Error for 99% Confidence Interval

To calculate the margin of error for 99% confidence interval, we will use the same formula: \[Margin\_of\_Error = z_{\alpha/2} * \frac{\sigma}{\sqrt{n}} \] This time, we are given a 99% confidence level, so we must find the \(z_{\alpha/2}\) score for 99%. This corresponds to the z-score of 2.576: \(z_{\alpha/2} = 2.576\) Now, we have everything to calculate the margin of error: \(Margin\_of\_Error = 2.576 * \frac{5}{\sqrt{10}}\) \(Margin\_of\_Error = 4.079\) The margin of error for a 99% confidence interval is 4.079.

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