Questions: In a random sample of seven people, the mean driving distance to work was 19.4 miles and the standard deviation was 6.5 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean μ. Interpret the results. Identify the margin of error. □ □ (Round to one decimal place as needed.) Construct a 99% confidence interval for the population mean. □ □ (Round to one decimal place as needed.)

Solution Steps

To find the margin of error \( E \) for the population mean, we use the formula:

\[ E = Z \cdot \frac{\sigma}{\sqrt{n}} \]

where:

• \( Z \) is the Z-score corresponding to the desired confidence level (for \( 99\% \), \( Z \approx 2.6 \)),
• \( \sigma \) is the standard deviation of the sample (6.5 miles),
• \( n \) is the sample size (7).

Substituting the values:

\[ E = 2.6 \cdot \frac{6.5}{\sqrt{7}} \approx 6.3 \]

Thus, the margin of error is:

\[ \text{Margin of Error} = 6.3 \text{ miles} \]

The confidence interval for the population mean \( \mu \) is given by:

\[ \bar{x} \pm E \]

where:

• \( \bar{x} \) is the sample mean (19.4 miles),
• \( E \) is the margin of error (6.3 miles).

Calculating the confidence interval:

\[ \text{Lower Bound} = 19.4 - 6.3 = 13.1 \] \[ \text{Upper Bound} = 19.4 + 6.3 = 25.7 \]

Thus, the 99% confidence interval for the population mean is:

\[ (13.1, 25.7) \text{ miles} \]

Final Answer