Question

A researcher records the repair cost for 27 randomly selected refrigerators. A sample mean of $60.52 and standard deviation of $23.29 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the refrigerators. 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.

Answer 1

The critical value is
`T_c = 1.7056`

The 90% confidence interval for the mean repair cost for the refrigerators is ($52.875, $68.165).

Explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 27 - 1 = 26

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of
`1 - (1 - 0.9)/(2) = 0.95`. So we have T = 1.7056, which means that the critical value is
`T_c = 1.7056`

The margin of error is:

`M = T(s)/(√(n)) = 1.7056(23.29)/(√(27)) = 7.645`

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 60.52 - 7.645 = $52.875.

The upper end of the interval is the sample mean added to M. So it is 60.52 + 7.645 = $68.165.

The 90% confidence interval for the mean repair cost for the refrigerators is ($52.875, $68.165).