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