Questions: In a random sample of five people, the mean driving distance to work was 18.3 miles and the standard deviation was 6.8 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.

Solution Steps

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

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

where:

• \( Z = 2.5758 \) (Z-score for 99% confidence level),
• \( \sigma = 6.8 \) (standard deviation),
• \( n = 5 \) (sample size).

Substituting the values:

\[ E = 2.5758 \times \frac{6.8}{\sqrt{5}} \approx 7.8332 \]

Thus, the margin of error is:

\[ \text{Margin of Error} = 7.8332 \text{ miles} \]

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

\[ \bar{x} \pm t \times \frac{s}{\sqrt{n}} \]

where:

• \( \bar{x} = 18.3 \) (sample mean),
• \( t \approx 4.6 \) (t-score for 99% confidence level with \( n-1 = 4 \) degrees of freedom),
• \( s = 6.8 \) (sample standard deviation),
• \( n = 5 \) (sample size).

Calculating the confidence interval:

\[ \text{Confidence Interval} = 18.3 \pm 4.6 \times \frac{6.8}{\sqrt{5}} \]

Calculating the lower and upper bounds:

\[ \text{Lower Bound} = 18.3 - 7.8332 \approx 10.4668 \] \[ \text{Upper Bound} = 18.3 + 7.8332 \approx 26.1332 \]

Thus, the 99% confidence interval is:

\[ \text{Confidence Interval} = (10.4668, 26.1332) \text{ miles} \]

We are 99% confident that the true population mean driving distance to work is between:

\[ \boxed{(10.47, 26.13)} \text{ miles} \]

Final Answer