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1. In a random sample of 15 CD players brought in form repair, the average repair cost was $80 and the sample standard deviation was $14. Construct a 90% confidence interval for. Assume the repair costs are normally distributed and there are no outliers.

Question

1. In a random sample of 15 CD players brought in form repair, the average repair cost was $80 and the sample standard deviation was $14. Construct a 90% confidence interval for. Assume the repair costs are normally distributed and there are no outliers.

 

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Step 1

It is given that the average repair cost (mean) is $80, sample standard deviation s is $14, the sample size (n ) is 15 and the level of confidence is 90%.

 

Step 2

The formula for finding the 90% confidence interval for mean is as follows:

Confi dence Interval = Mean t Confidence
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Confi dence Interval = Mean t Confidence

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Step 3

The confidence is 6.367 using Excel formula “=CONFIDENCE.T((1–0.90),14,15)”.

Substitut...

Confidence Interval = 80 t6.367
(80 6.367,80-6.367)
(73.633,86.367)
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Confidence Interval = 80 t6.367 (80 6.367,80-6.367) (73.633,86.367)

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