There is an average of 22.455 employees at 22 office furniture dealers in a major metropolitan area, with a standard deviation of 18.52. Construct a 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers.

The 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers is (12.288, 32.622).

Step-by-step explanation:

We have that to find our level, that is the subtraction of 1 by the confidence interval divided by 2. So:

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

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

The lower end of the interval is the mean subtracted by M. So it is 22.455 - 10.167 = 12.288

The upper end of the interval is the mean added to M. So it is 22.455 + 10.167 = 32.622

The 99 percent confidence interval for the true mean number of full-time employees at office furniture dealers is (12.288, 32.622).

the answer is 0.5 mpm

step-by-step explanation:

the line goes half way through the square each time (which is 0.5 or half of course) so that would mean its 0.5 mpm.

23

step-by-step explanation: