Question

in a random sample of 20 people, the mean commute time to work was 30.7 minutes and the standard deviation was 7.1 minutes. assume the population is normally distributed and use a t - distribution to construct a 90% 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 t - value

The degrees of freedom \(df=n - 1\), where \(n = 20\), so \(df=20-1 = 19\).
For a \(90\%\) confidence interval, \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\).
Looking up in the t - distribution table, \(t_{\frac{\alpha}{2},df}=t_{0.05,19}=1.729\).

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 \(s = 7.1\) and \(n = 20\).

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Step3: Calculate the confidence interval

The confidence interval is \(\bar{x}-E<\mu<\bar{x} + E\), where \(\bar{x}=30.7\).
\(\bar{x}-E=30.7-2.75 = 27.95\approx28.0\)
\(\bar{x}+E=30.7 + 2.75=33.45\approx33.5\)

Answer:

The confidence interval for the population mean \(\mu\) is \((28.0,33.5)\)