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.

Answer 1

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

Explanation:

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

`\alpha = (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*(\sigma)/(√(n))`

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

`M = 1.645*(14.32)/(√(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.