Question

in a random sample of 27 people, the mean commute time to work was 34.4 minutes and the standard deviation was 7.2 minutes. assume the population is normally distributed and use a t - distribution to construct a 98% confidence interval for the population mean μ. what is the margin of error of μ? interpret the results.
the confidence interval for the population mean μ is (□,□). (round to one decimal place as needed.)

Explanation:

Step1: Determine the degrees of freedom and the critical value

The degrees of freedom \(df=n - 1\), where \(n = 27\), so \(df=27-1 = 26\).
For a \(98\%\) confidence interval, \(\alpha=1 - 0.98=0.02\), and \(\frac{\alpha}{2}=0.01\).
Using a t - distribution table or a calculator, the critical value \(t_{\frac{\alpha}{2}}\) with \(df = 26\) is \(t_{0.01}\approx 2.479\).

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a t - distribution is \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 7.2\) and \(n = 27\).

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Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=34.4\) and \(E\approx3.4\).

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Answer:

The confidence interval for the population mean \(\mu\) is \((31.0,37.8)\)