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. (round to one decimal place as needed.)

Explanation:

Step1: Identify parameters

Sample size \( n = 10 \), sample mean \( \bar{x}=20.9 \), sample standard deviation \( s = 5.6 \), confidence level \( 90\% \). Degrees of freedom \( df=n - 1=9 \).

Step2: Find t - critical value

For \( 90\% \) confidence and \( df = 9 \), the t - critical value \( t_{\alpha/2} \) (from t - table or calculator) is \( t_{0.05,9}\approx1.833 \).

Step3: Calculate margin of error

The formula for margin of error \( E \) using t - distribution is \( E=t_{\alpha/2}\times\frac{s}{\sqrt{n}} \). Substitute values: \( E = 1.833\times\frac{5.6}{\sqrt{10}} \). Calculate \( \sqrt{10}\approx3.1623 \), \( \frac{5.6}{3.1623}\approx1.771 \), then \( E\approx1.833\times1.771\approx3.2 \) (rounded to one decimal place).

Answer:

3.2