Question

in a random sample of 17 people, the mean commute time to work was 32.2 minutes and the standard deviation was 7.3 minutes. assume the population is normally distributed and use a t - distribution to construct a 99% 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: Find the degrees of freedom and critical value

The degrees of freedom \(df=n - 1\), where \(n = 17\), so \(df=17-1 = 16\).
For a \(99\%\) confidence interval, the significance level \(\alpha=1 - 0.99=0.01\), and \(\frac{\alpha}{2}=0.005\).
Using the t - distribution table or a calculator, the critical value \(t_{\frac{\alpha}{2}}\) with \(df = 16\) is \(t_{0.005,16}=2.921\).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\), where \(s = 7.3\), \(n = 17\).
\(E=2.921\times\frac{7.3}{\sqrt{17}}\)
First, calculate \(\sqrt{17}\approx4.123\), then \(\frac{7.3}{4.123}\approx1.77\)
\(E=2.921\times1.77\approx5.2\)

Step3: Calculate the confidence interval

The formula for the confidence interval is \(\bar{x}-E<\mu<\bar{x}+E\), where \(\bar{x}=32.2\)
\(\bar{x}-E=32.2 - 5.2=27.0\)
\(\bar{x}+E=32.2+5.2 = 37.4\)

Answer:

The confidence interval for the population mean \(\mu\) is \((27.0,37.4)\)