mulan5446
24.02.2020 •
Mathematics
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.
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Ответ:
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).
Ответ:
One type of proof is a two-column proof. It contains statements and reasons in columns. Another type is a paragraph proof, in which statements and reasons are written in words. A third type is a flowchart proof, which uses a diagram to show the steps of a proof.