Question

A supervisor records the repair cost for 17 randomly selected TV's. A sample mean of $94.57 and standard deviation of $20.65 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the TV's. 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 3 decimal places.

Step 2 of 2: Construct the 80% confidence interval. Round your answer to two decimal places.

Answer 1

Step 1 of 2:

This is a two tailed t-test, where level of significance, = 0.2 and degrees of freedom, df = 17 - 1 = 16

critical value,

Step 2 of 2:

Margin of error, M =

Lower Limit = - M = 94.57 - 6.70 = 87.87

Upper Limit = + M = 94.57 + 6.70 = 101.27

So 80% confidence Interval = [87.87, 101.27] (Ans)

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