A consumer affairs investigator records the repair cost for 23 randomly selected refrigerators. A sample mean of $88.05 and standard deviation of $10.52 are subsequently computed. Determine the 80%...










A consumer affairs investigator records the repair cost for 23 randomly selected refrigerators. A sample mean of $88.05 and standard deviation of $10.52 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the refrigerators. 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 three decimal places.

















Jun 10, 2022
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