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.

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 level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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

So it is z with a pvalue of , so

Now, find M as such

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

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.

the answer is 2/54

step-by-step explanation: