Question

in a random sample of ten people, the mean driving distance to work was 20.9 miles and the standard deviation was 5.6 miles. assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 90% confidence interval for the population mean μ. interpret the results. identify the margin of error. 3.2 miles (round to one decimal place as needed.) construct a 90% confidence interval for the population mean. (round to one decimal place as needed.)

Explanation:

Step1: Recall Confidence Interval Formula

The formula for a confidence interval using the t - distribution is \(\bar{x}\pm E\), where \(E = t_{\alpha/2}\frac{s}{\sqrt{n}}\) is the margin of error, \(\bar{x}\) is the sample mean, \(s\) is the sample standard deviation, \(n\) is the sample size, and \(t_{\alpha/2}\) is the critical t - value.

We know that \(\bar{x}=20.9\), \(s = 5.6\), \(n = 10\), and for a 90% confidence interval, \(\alpha=1 - 0.90=0.10\), so \(\alpha/2 = 0.05\). The degrees of freedom \(df=n - 1=10 - 1 = 9\). Looking up the t - value in the t - distribution table, \(t_{0.05,9}\approx1.833\).

Step2: Calculate Margin of Error (already given as 3.2, let's verify)

\(E=t_{\alpha/2}\frac{s}{\sqrt{n}}=1.833\times\frac{5.6}{\sqrt{10}}\)
\(\sqrt{10}\approx3.1623\), \(\frac{5.6}{3.1623}\approx1.771\)
\(1.833\times1.771\approx3.2\) (matches the given margin of error)

Step3: Construct Confidence Interval

The confidence interval is \(\bar{x}-E\) to \(\bar{x}+E\)
\(\bar{x}-E=20.9 - 3.2 = 17.7\)
\(\bar{x}+E=20.9+3.2 = 24.1\)

Answer:

The 90% confidence interval for the population mean \(\mu\) is \((17.7, 24.1)\) miles. This means we are 90% confident that the true population mean driving distance to work lies between 17.7 miles and 24.1 miles. The margin of error is 3.2 miles.