In a random sample of 50 refrigerators, the mean repair cost was $13, 100 and th

Question

In a random sample of 50 refrigerators, the mean repair cost was $13, 100 and the population standard deviation is $17.60. Construct a 95% confidence for the population mean repair cost the results. Construct a 95% confidence interval for the population mean repair cost. The 95% confidence interval is. (Round to two decimal places as needed.) Interpret you results. Choose the correct answer below. A. With 95% confidence, it can be said that the confidence interval contains the true mean repair cost. B. With 95% confidence, it can be said that the confidence interval contains the sample mean repair cost. C. The confidence interval contains 95% of the mean repair costs.

Explanation / Answer

given random sample n= 50

mean = 131

standard deviation =17.60

standard error = (SE) = s /n = 17.60/50 = 2.489

95% confidence interval z score = 1.96

95% CI = mean ± 1.96 × SE

   = 131 ± 1.96*2.49

= 135.89 to 126.12

option b

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