Question

A manager records the repair cost for 4 randomly selected TVs. A sample mean of $88.46 and standard deviation of $17.20 are subsequently computed. Determine the 99% confidence interval for the mean repair cost for the TVs. Assume the population is approximately normal.a. Find the critical value that should be used in constructing the confidence interval. Round your answer to three decimal places.b. Construct the 99% confidence interval. Round your answer to two decimal places.

Answer 1

Answer:

a) The critical value is $z = 2.575$.

b) The 99% confidence interval for the mean repair cost for the TVs is ($66.31, $110.61).

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:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$. This is our critical value

So it is z with a pvalue of $1-0.005 = 0.995$, so $z = 2.575$

a. Find the critical value that should be used in constructing the confidence interval.

The critical value is $z = 2.575$.

b. Construct the 99% confidence interval. Round your answer to two decimal places.

Now, find 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, in this problem

$M = 2.575*\frac{17.20}{\sqrt{4}} = 22.15$

The lower end of the interval is the mean subtracted by M. So it is 88.46 - 22.15 = $66.31

The upper end of the interval is the mean added to M. So it is 88.46 - 22.15 = $110.61

The 99% confidence interval for the mean repair cost for the TVs is ($66.31, $110.61).