Question

A supervisor records the repair cost for 17 randomly selected TVs. A sample mean of $76.76 and standard deviation of $24.02 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

Answer 1

To determine the 99% confidence interval for the mean repair cost for the TVs, you can use the formula: CI = sample mean ± (z * (sample standard deviation / √n)), where z is the z-score corresponding to the desired confidence level.

Step-by-step explanation:

To determine the 99% confidence interval for the mean repair cost for the TVs, we can use the formula: CI = sample mean ± (z * (sample standard deviation / √n)), where z is the z-score corresponding to the desired confidence level. Since we want a 99% confidence interval, the z-score is 2.576.

Plugging in the values from the problem, we get:

CI = 76.76 ± (2.576 * (24.02 / √17))

Calculating this, we find that the confidence interval is approximately $69.46 to $84.06.