## A supervisor records the repair cost for 17 randomly selected stereos. A sample mean of $66.34 and standard deviation of $15.22 are subseque

Question

A supervisor records the repair cost for 17 randomly selected stereos. A sample mean of $66.34 and standard deviation of $15.22 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the stereos. 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.

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Math
6 hours
2021-09-14T23:43:59+00:00
2021-09-14T23:43:59+00:00 1 Answer
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## Answers ( )

Answer:Critical value: T = 1.337

The 80% confidence interval for the mean repair cost for the stereos is between $45.991 and $86.689

Step-by-step explanation:We have the standard deviation for the sample. So we use the t-distribution to solve this question.

The

first stepto solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. Sodf = 17 – 1 = 16

80% confidence intervalNow, we have to find a value of T, which is found looking at the t table, with 16 degrees of freedom(y-axis) and a confidence level of . So we have T = 1.337, which is the critical value

The margin of error is:M = T*s = 1.337*15.22 = 20.349

In which s is the standard deviation of the sample.

The

lower endof the interval is the sample mean subtracted by M. So it is 66.34 – 20.349 = $45.991The

upper endof the interval is the sample mean added to M. So it is 66.34 + 20.349 = $86.689The 80% confidence interval for the mean repair cost for the stereos is between $45.991 and $86.689