Good luck
A researcher records the repair cost for 51 randomly selected dryers. A sample mean of $85.68 and standard deviation of $27.28 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the dryers. Assume the population is normally distributed. Construct the 99% confidence interval. Round your answer to two decimal places.
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1 solution
x‾‾ ± tα/2sn‾‾√ ≈ 85.68 ± 2.678 27.28/√51
≈ 85.68 ± 10.2299
≈ 75.45 and 95.91