Question

Mr. Good Wrench advertises that a customer will have to wait no more than 30 minutes for an oil change. A sample of 26 oil changes had a standard deviation of 4.8 minutes. Use this information to calculate a 90% confidence interval for the population standard deviation waiting time for an oil change.

Answer 1

The 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).

Explanation:

The (1 - α)% confidence interval for the population standard deviation is:

`CI=\sqrt{((n-1)s^(2))/(\chi^(2)_(\alpha/2, (n-1)))}\leq \sigma\leq \sqrt{((n-1)s^(2))/(\chi^(2)_(1-\alpha/2, (n-1)))}`

The information provided is:

n = 26

s = 4.8 minutes

Confidence level = 90%

Compute the critical values of Chi-square as follows:

`\chi^(2)_(\alpha/2, (n-1))=\chi^(2)_(0.10/2, (26-1))=\chi^(2)_(0.05, 25)=37.652`

`\chi^(2)_(1-\alpha/2, (n-1))=\chi^(2)_(1-0.10/2, (26-1))=\chi^(2)_(0.95, 25)=14.611`

*Use a Chi-square table.

Compute the 90% confidence interval for the population standard deviation waiting time for an oil change as follows:

`CI=\sqrt{((n-1)s^(2))/(\chi^(2)_(\alpha/2, (n-1)))}\leq \sigma\leq \sqrt{((n-1)s^(2))/(\chi^(2)_(1-\alpha/2, (n-1)))}`

`=\sqrt{((26-1)* 4.8^(2))/(37.652)}\leq \sigma\leq \sqrt{((26-1)* 4.8^(2))/(14.611)}\\\\=3.9113\leq \sigma\leq 6.2787\\\\\approx 3.9 \leq \sigma\leq6.3`

Thus, the 90% confidence interval for the population standard deviation waiting time for an oil change is (3.9, 6.3).