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Chapter 8: Problem 5

In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a sample of 49 customers. Assume a population standard deviation of \$5. a. At \(95 \%\) confidence, what is the margin of error? b. If the sample mean is \(\$ 24.80\), what is the \(95 \%\) confidence interval for the population mean?

Short Answer

Expert verified
The margin of error for a 95% confidence level is approximately \$1.40. The 95% confidence interval for the population mean amount spent per customer is approximately between \$23.40 and \$26.20.

Step by step solution

01

Identifying the given values

We are given the following values: Sample size (n) = 49 customers Population standard deviation (σ) = \$5 Sample mean (\(\bar{x}\)) = \$24.80 Confidence level = 95%
02

Calculating the Margin of Error

To calculate the margin of error, we need to determine the value of the Standard Error. The Standard Error is given by the formula SE = \(\frac{\sigma}{\sqrt{n}}\). Then, we will use the Z-score of 1.96 for a 95% confidence level. The formula for margin of error (ME) is: ME = Z * SE First, calculate the Standard Error (SE): SE = \(\frac{\sigma}{\sqrt{n}}\) SE = \(\frac{5}{\sqrt{49}}\) SE ≈ 0.714 Now, calculate the Margin of Error (ME): ME = Z * SE ME = 1.96 * 0.714 ME ≈ 1.40 The margin of error for a 95% confidence level is approximately \$1.40.
03

Constructing the Confidence Interval

Now that we have the margin of error, we can construct the confidence interval for the population mean using the following formula: CI = \(\bar{x} \pm \) ME CI = \$24.80 ± \$1.40 This gives us the 95% confidence interval: CI = (\$23.40, \$26.20) The 95% confidence interval for the population mean amount spent per customer is approximately between \$23.40 and \$26.20.

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Most popular questions from this chapter

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