Question

Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union decides to sample 348 of its members at random and monitor the working time of each of them for 1 month. At the end of the month, the total number of hours absent from work is recorded for each employee. If the mean and standard deviation of the sample are x = 7.5 hours and s = 3.5 hours, find a 99% confidence interval for the true mean number of hours a union member is absent per month. Round to the nearest thousandth.

Answer 1

Answer:

The 99% confidence interval for the true mean number of hours a union member is absent per month is (0 hours, 16.5125 hours).

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 $1-\alpha$.

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

Now, find M as such

$M = z*s$

In which s is the standard deviation of the sample.

So

$M = 2.575*3.5 = 9.0125$

The lower end of the interval is the mean subtracted by M. So it is 7.5 - 9.0125 = -...

There is not a negative number of hours. So the lower end of the interval is 0 hours.

The upper end of the interval is the mean added to M. So it is 7.5 + 9.0125 = 16.5125 hours.

The 99% confidence interval for the true mean number of hours a union member is absent per month is (0 hours, 16.5125 hours).