A consumer affairs investigator records the repair cost for 4 randomly selected washers. A sample mean of $64.26 and standard deviation of $27.77 are subsequently computed. Determine the 80% confidence interval for the mean repair cost for the washers. Assume the population is approximately normal. Step 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Critical value: z = 1.28

The 80% confidence interval for the mean repair cost for the washers is between $46.487 and $82.033.

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 1-0.1 = 0.9, so z = 1.28

Now, find the margin of error M as such

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

So

The lower end of the interval is the mean subtracted by M. So 64.26 - 17.773 = $46.487.

The upper end of the interval is M added to the mean. So 64.26 + 17.773 = $82.033.

The 80% confidence interval for the mean repair cost for the washers is between $46.487 and $82.033.

my answer is$ 12

step-by-step explanation:

because 15 -2=12

the answer is 150.6

step-by-step explanation:

you multiply .02 by 30 and get .6.

and then you multiply 5.00 by 30 and get 150.

hope this !