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A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19...

A student records the repair cost for 13 randomly selected TVs. A sample mean of $72.19 and standard deviation of $15.88 are subsequently computed. Determine the 98% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.

1) Find the critical value that should be used in constructing the confidence interval. Round you answer to three decimal places.

2) Construct the 98% confidence interval. Round your answer to two decimal places.

Solutions

Expert Solution

Answer a)

The number of degrees of freedom are df = 13 - 1 = 12, and the significance level is α = 0.02.

Based on the provided information, the critical t-value for α = 0.02 and df = 12 is tc​ = 2.681 (Obtained using t-value calculator)

Answer b)

The 98% confidence interval is (60.38, 84.00)

milcah answered 1 year ago
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