Question

A manager records the repair cost for 6 randomly selected TVs. A sample mean of $67.87 and standard deviation of $28.03 are subsequently computed. Determine the 95% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Step 2 of 2 : Construct the 95%95% confidence interval. Round your answer to two decimal places.

Answer 1

The critical value is approximately 2.571, and the 95% confidence interval for the mean repair cost is [$40.11, $95.63].

Step-by-step explanation:

Step 1 of 2: To find the critical value, we first need to determine the degrees of freedom for the sample. Since we have 6 TVs, the degrees of freedom are 6 - 1 = 5.

Using a t-distribution table or a calculator, the critical value for a 95% confidence interval with 5 degrees of freedom is approximately 2.571.

Step 2 of 2: To construct the 95% confidence interval, we use the formula:

CI = sample mean ± (critical value * standard deviation / sqrt(sample size)).

Plugging in the given values, we get: CI = $67.87 ± (2.571 * $28.03 / sqrt(6)).

=[$40.11, $95.63]

the lower and upper bounds of the confidence interval.