Question

A consumer affairs investigator records the repair cost for 4 randomly selected washers. A sample mean of $52.63 and standard deviation of $22.01 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the washers. Assume the population is approximately normal. Step 2 of 2 : Construct the 80% confidence interval. Round your answer to two decimal places.

Answer 1

The 80% confidence interval for the mean repair cost for the washers is between $34.60 and $70.66.

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 = 4 - 1 = 3

80% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 3 degrees of freedom(y-axis) and a confidence level of
. So we have T = 1.638

The margin of error is:

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 52.63 - 18.03 = $34.60

The upper end of the interval is the sample mean added to M. So it is 52.63 + 18.03 = $70.66.

The 80% confidence interval for the mean repair cost for the washers is between $34.60 and $70.66.