SOLUTION: In a random sample of 60 refrigerators, the mean repair cost was $150.00 and the standard deviation was $15.50. Construct a 99% confidence interval for the mean repair cost.

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Question 1188352: In a random sample of 60 refrigerators, the mean repair cost was $150.00 and the
standard deviation was $15.50. Construct a 99% confidence interval for the mean
repair cost.
Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
150+-+2.576+%2A+%2815.50%2Fsqrt%2860%29%29 = 150+-+2.576+%2A+%2815.50%2F7.7460%29 = 150+-+2.576+%2A+%282.0010%29 = 150+-+5.1546 = 144.8454 = $144.85

150+%2B+2.576+%2A+%2815.50%2Fsqrt%2860%29%29 = 150+-+2.576+%2A+%2815.50%2F7.7460%29 = 150+-+2.576+%2A+%282.0010%29 = 150+%2B+5.1546 = 155.1546 = $155.15