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 1 of 2 : Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.

Answer 1

Answer:

CI = (70.861 , 94.418)

Step-by-step explanation:

In order to determine the 90% confidence interval you use the following formula (for a population approximately normal):

$CI=(\overline{x}-Z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\overline{x}+Z_{\alpha/2}\frac{\sigma}{\sqrt{n}})$    (1)

$\overline{x}$: mean = 82.64

σ: standard deviation = 14.32

n: sample = 4

α: tail area = 1 - 0.9 = 0.1

Z_α/2 = Z_0.05: Z factor =  1.645

You replace these values and you obtain:

$Z_{0.05}(\frac{14.32}{\sqrt{4}})=(1.645)(\frac{14.32}{\sqrt{4}})=11.778$

The confidence interval will be:

$CI=(82.64-11.778,82.64+11.778)=(70.861,94.418)$

The 90% confidence interval is (70.861 , 94.418)