Question

A researcher records the repair cost for 8 randomly selected washers. A sample mean of $60.46 and standard deviation of $18.36 are subsequently computed. Determine the 90% 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.

Answer 1

Answer:

The critical value is T = 1.895.

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 6

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 6 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.895, which is the critical value.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.895\frac{18.36}{\sqrt{8}} = 12.301$

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

The lower end of the interval is the sample mean subtracted by M. So it is 60.46 - 12.301 = $48.159

The upper end of the interval is the sample mean added to M. So it is 60.46 + 12.301 = $72.761

The 90% confidence interval for the mean repair cost for the washers is between $48.159 and $72.761