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Repair Costs: Refrigerators In a random sample of 60 refrigerators, the mean repair cost was %150. Assume the population standard deviation is %15.

Repair Costs: Refrigerators In a random sample of 60 refrigerators, the mean repair cost was %150.00. Assume the population standard deviation is %15.50. Construct a 99% confidence interval for the population mean repair cost. Interpret the results.

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We are given data that Sample mean(xbar) = 150 population standard deviation(  σ  ) = 15.5 sample size(n) = 60 Z... View the full answer

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Other Answers

for 99 % confidence interval area outside tailed portion = 0.01/2 = 0.005 also for 0.995... View the full answer

 1 4 4 . 8 4 ≤ μ ≤ 1 5 5 . 1 6  We have... View the full answer

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