Question

in a random sample of 25 people, the mean commute time to work was 32.6 minutes and the standard deviation was 7.1 minutes. assume the population is normally distributed and use a t - distribution to construct a 95% 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 degree of freedom and t - value

Degree of freedom \(df=n - 1=25-1 = 24\). For a 95% confidence interval, \(\alpha=1 - 0.95=0.05\), and \(\frac{\alpha}{2}=0.025\). Using the t - distribution table or a calculator, \(t_{\frac{\alpha}{2},df}=t_{0.025,24}=2.064\).

Step2: Calculate the margin of error

The formula for the margin of error \(E=t_{\frac{\alpha}{2}}\frac{s}{\sqrt{n}}\). Given \(s = 7.1\), \(n = 25\), and \(t_{\frac{\alpha}{2}}=2.064\). Then \(E=2.064\times\frac{7.1}{\sqrt{25}}=2.064\times\frac{7.1}{5}=2.064\times1.42 = 2.93088\approx2.9\).

Step3: Calculate the confidence interval

The confidence interval for the population mean \(\mu\) is \(\bar{x}-E<\mu<\bar{x}+E\). Given \(\bar{x}=32.6\), \(E = 2.9\). So \(32.6-2.9<\mu<32.6 + 2.9\), which is \(29.7<\mu<35.5\).

Answer:

\((29.7,35.5)\)