Question

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.

Answer 1

Answer:

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

$\alpha = \frac{1-0.80}{2} = 0.10$

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 1-0.1 = 0.9, so z = 1.28

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

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

So

$M = 1.28*\frac{27.77}{\sqrt{4}} = 17.773$

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.