Question

in a random sample of 23 people, the mean commute time to work was 31.8 minutes and the standard deviation was 7.2 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: Identify the values

We have sample size \( n = 23 \), sample mean \( \bar{x}=31.8 \), sample standard deviation \( s = 7.2 \), and confidence level \( 95\% \) (so \( \alpha=0.05 \)). The degrees of freedom \( df=n - 1=23 - 1 = 22 \).

Step2: Find the t - critical value

For a \( 95\% \) confidence level and \( df = 22 \), the two - tailed t - critical value \( t_{\alpha/2} \) (from t - distribution table or calculator) is approximately \( t_{0.025,22}=2.074 \).

Step3: Calculate the standard error (SE)

The formula for the standard error is \( SE=\frac{s}{\sqrt{n}} \). Substituting the values, we get \( SE=\frac{7.2}{\sqrt{23}}\approx\frac{7.2}{4.796}\approx1.501 \).

Step4: Calculate the margin of error (E)

The formula for the margin of error for a t - distribution is \( E = t_{\alpha/2}\times SE \). Substituting the values, we have \( E=2.074\times1.501\approx3.113 \).

Step5: Calculate the confidence interval

The confidence interval is \( \bar{x}-E<\mu<\bar{x} + E \). Substituting \( \bar{x}=31.8 \) and \( E\approx3.1 \) (rounded to one decimal place), we get \( 31.8-3.1<\mu<31.8 + 3.1 \), which is \( 28.7<\mu<34.9 \).

Answer:

The 95% confidence interval for the population mean \(\mu\) is \(28.7 < \mu < 34.9\) (rounded to one decimal place). The margin of error is approximately \(3.1\) minutes. We can interpret this as: we are 95% confident that the true population mean commute time lies between \(28.7\) minutes and \(34.9\) minutes.