Question

if a random sample of five people, the mean driving distance to work was 19.2 miles and the standard deviation was 4.7 miles. assume the population is normally distributed and use the t - distribution to find the margin of error and construct a 80% confidence interval for the population mean μ. interpret the results.
identify the margin of error
5.21 miles
(round to one decimal place as needed.)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 5\). So \(df=5-1 = 4\).

Step2: Find the critical value \(t_{\alpha/2}\)

For a \(89\%\) confidence interval, \(\alpha=1 - 0.89=0.11\), and \(\alpha/2=0.055\). Using a t - distribution table or calculator, \(t_{0.055,4}\approx1.895\).

Step3: Calculate the margin of error \(E\)

The formula for the margin of error when using the t - distribution is \(E=t_{\alpha/2}\frac{s}{\sqrt{n}}\), where \(s = 4.7\) and \(n = 5\).

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Answer:

The margin of error \(E\approx4.0\) (rounded to one decimal place)