The U.S. Department of Health and Human Services collected sample data for 772 males between the ages of 18 and 24. That sample group has a mean height of 69.7 inches with a standard deviation of 2.8 inches. Find the 99 percent confidence interval for the mean height of all males between the ages of 18 and 24.

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Answer:

The 99 percent confidence interval for the mean height of all males between the ages of 18 and 24 is between 69.44 inches and 69.96 inches.

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.99}{2} = 0.005[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.005 = 0.995[/tex], so [tex]z = 2.575[/tex]

Now, find M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

[tex]M = 2.575*\frac{2.8}{772} = 0.26[/tex]

The lower end of the interval is the mean subtracted by M. So it is 69.7 - 0.26 = 69.44 inches

The upper end of the interval is the mean added to M. So it is 69.7 + 0.26 = 69.96 inches.

The 99 percent confidence interval for the mean height of all males between the ages of 18 and 24 is between 69.44 inches and 69.96 inches.