In a random sample of 23 people, the mean commute time to work was 34.4 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a $80 \%$ confidence interval for the population mean $\mu$. What is the margin of error of $\mu$ ? Interpret the results.
The confidence interval for the population mean $\mu$ is (Round to one decimal place as needed.)

Step 1 ：We are given a random sample of 23 people, with a mean commute time to work of 34.4 minutes and a standard deviation of 7.2 minutes. We are asked to construct an 80% confidence interval for the population mean \(\mu\) and find the margin of error.

Step 2 ：The margin of error for a confidence interval can be calculated using the formula: Margin of Error = t * (s/\(\sqrt{n}\)), where t is the t-score, s is the standard deviation of the sample, and n is the size of the sample.

Step 3 ：In this case, we have: n = 23 (sample size), s = 7.2 (standard deviation).

Step 4 ：The confidence level is 80%, so the significance level (\(\alpha\)) is 1 - 0.80 = 0.20. Since we are constructing a two-tailed test, the alpha level will be divided by 2, which gives 0.10. We will use this to find the t-score.

Step 5 ：Using a t-distribution table or a calculator with 22 degrees of freedom (n-1), we find that the t-score is approximately 1.321.

Step 6 ：Substituting these values into the margin of error formula, we get: Margin of Error = 1.321 * (7.2/\(\sqrt{23}\)) = 1.98 minutes.

Step 7 ：This means that we are 80% confident that the true population mean commute time falls within 1.98 minutes of our sample mean of 34.4 minutes.

Step 8 ：Final Answer: The margin of error of \(\mu\) is approximately \(\boxed{1.98}\) minutes.