A manager records the repair cost for 4 randomly selected stereos. A sample mean of $82.64 and standard deviation of $14.32 are subsequently

Question

A manager records the repair cost for 4 randomly selected stereos. A sample mean of $82.64 and standard deviation of $14.32 are subsequently computed. Determine the 90% confidence interval for the mean repair cost for the stereos. Assume the population is approximately normal. Step 2 of 2 : Construct the 90% confidence interval. Round your answer to two decimal places.

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Lydia 1 week 2021-09-15T22:26:37+00:00 1 Answer 0

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    2021-09-15T22:28:03+00:00

    Answer:

    The 90% confidence interval for the mean repair cost for the stereos is between $70.86 and $94.42.

    Step-by-step explanation:

    We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

    \alpha = \frac{1-0.9}{2} = 0.05

    Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

    So it is z with a pvalue of 1-0.05 = 0.95, so z = 1.645

    Now, find M as such

    M = z*\frac{\sigma}{\sqrt{n}}

    In which \sigma is the standard deviation of the population and n is the size of the sample.

    M = 1.645*\frac{14.32}{\sqrt{4}} = 11.78

    The lower end of the interval is the sample mean subtracted by M. So it is 82.64 – 11.78 = $70.86

    The upper end of the interval is the sample mean added to M. So it is 82.64 + 11.78 = $94.42.

    The 90% confidence interval for the mean repair cost for the stereos is between $70.86 and $94.42.

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