14. In a random sample of 35 refrigerators, the mean repair cost was $138.00 and

Question

14. In a random sample of 35 refrigerators, the mean repair cost was $138.00 and the population standard deviation is $18.50. Construct a 95% confidence interval for the population mean repair cost. Interpret the results. Construct a 95% confidence interval for the population mean repair cost. The 95% confidence interval is ( Interpret you results. Choose the correct answer below. 0 A. The confidence interval contains 95% of the mean repair costs. ). (Round to two decimal places as needed.) With 95% confidence, it can be said that the confidence interval contains the sample mean repair cost. With 95% confidence, it can be said that the confidence interval contains the true mean repair cost. 0 B. ° C.

Explanation / Answer

n = 35 x' = 138.00 s = 18.50 z = 1.96

LCL = x' - zs / n = 138.00 - 1.96 * 18.50 / 35 = 131.8709.

UCL = x' + zs / n = 138.00 + 1.96 * 18.50 / 35 = 144.1291.

The 95% confidence interval is 131.87, 144.13.

Interpretation:

C. With 95% confidence, it can be said that the confidence interval contains the true mean repair cost.

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