To solve this problem, let's go through each part carefully.

Part (a)

Since we are dealing with a sample and estimating the population mean with a sample size of 15 (which is less than 30), we should use the t-distribution to compute the confidence interval. The t-distribution is appropriate here because the sample size is small and the population standard deviation is unknown.

Answer for (a): t-distribution

Part (b)

To find a 90% confidence interval for the population mean commute miles:

1. Calculate the sample mean ($\bar{x}$) and the sample standard deviation ($s$) from the data provided: $\text{Data} = [25, 20, 22, 11, 21, 21, 17, 28, 14, 8, 12, 11, 6, 8, 28]$

2. Find the sample mean ($\bar{x}$): $\bar{x} = \frac{\sum x_i}{n}$

3. Calculate the sample standard deviation ($s$).

4. Determine the t-score for a 90% confidence level with $n - 1 = 14$ degrees of freedom.

5. Compute the margin of error (ME): $ME = t \cdot \frac{s}{\sqrt{n}}$

6. Calculate the confidence interval: $\text{Confidence Interval} = \left(\bar{x} - ME, \bar{x} + ME\right)$

I'll calculate these values now.### Part (b) Solution

1. Sample Mean ($\bar{x}$) = 16.8 miles
2. Sample Standard Deviation ($s$) = 7.370 miles (rounded to three decimal places)
3. t-score for a 90% confidence level with 14 degrees of freedom = 1.761
4. Margin of Error (ME) is calculated using this t-score and standard deviation.

Thus, the 90% confidence interval for the population mean commute distance is approximately: $\text{Confidence Interval} = (13.448, 20.152)$ rounded to three decimal places.

Answer for (b): Between 13.448 and 20.152 miles

Part (c)

A 90% confidence interval means that if we were to take many samples and create a confidence interval for each sample, about 90% of these intervals would contain the true population mean, and about 10% would not contain it.

Answer for (c): 90% will contain the true mean, and 10% will not contain the true mean.

Would you like further details on any of these steps or clarifications?

Related Questions

1. What changes if we use a 95% confidence interval instead of 90%?
2. How is the t-distribution different from the normal distribution in this context?
3. Why is it essential to use a t-distribution instead of a z-distribution for small sample sizes?
4. How would the confidence interval change if the sample size were larger?
5. What assumptions are made when constructing this confidence interval?

Tip

When sample sizes are small, always consider whether the population standard deviation is known. If it's unknown, use the t-distribution.