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6.1 6 In a random sample of 60 refrigerators, the mean repair cost was $147.00 and the population standard deviation is $19.30. Construct a 90%

6.1 6

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In a random sample of 60 refrigerators, the mean repair cost was $147.00 and the population standard deviation is $19.30. Construct a 90% confidence interval for the population mean repair cost. Interpret the results. Construct a 90% confidence interval for the population mean repair cost. The 90% confidence interval is (Round to two decimal places as needed.)

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