Questions: In a random sample of 8 people, the mean commute time to work was 34.5 minutes and the standard deviation was 7.3 minutes. A 90% confidence interval using the t-distribution was calculated to be (29.6,39.4) After researching commute times to work, it was found that the population standard deviation is 9.6 minutes. Find the margin of error and construct a 90% confidence interval using the standard normal distribution with the appropriate calculations for a standard deviation that is known Compare the results. The margin of error of μ is (Round to two decimal places as needed.)

Solution Steps

To find the margin of error \( E \) for the population mean with a known standard deviation, we use the formula:

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

Where:

• \( Z \) is the Z-score corresponding to the desired confidence level (for \( 90\% \), \( Z \approx 1.64 \)),
• \( \sigma = 9.6 \) is the population standard deviation,
• \( n = 8 \) is the sample size.

Substituting the values:

\[ E = 1.64 \cdot \frac{9.6}{\sqrt{8}} \approx 5.58 \]

Thus, the margin of error is:

\[ \text{Margin of Error} = 5.58 \]

The confidence interval for the mean can be calculated using the formula:

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

Where:

• \( \bar{x} = 34.5 \) is the sample mean,
• \( E = 5.58 \) is the margin of error.

Calculating the confidence interval:

\[ 34.5 \pm 5.58 \]

This results in:

\[ (34.5 - 5.58, 34.5 + 5.58) = (28.92, 40.08) \]

The confidence interval calculated using the standard normal distribution is:

\[ (28.92, 40.08) \]

The given confidence interval using the \( t \)-distribution was:

\[ (29.6, 39.4) \]

Final Answer