Question

he United States Marine Corps is reviewing its orders for uniforms because it has a surplus of uniforms for tall men recruits and a shortage for shorter men recruits. Its review involves data for 772 men recruits between the ages of 18 to 24. That sample group has a mean height of 69.7 inches with a population standard deviation of 2.8 inches. Construct a 99% confidence interval for the mean height of all men recruits between the ages 18 and 24.

Answer 1

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

Explanation:

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

`\alpha = (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 the margin of error M as such

`M = z*(\sigma)/(√(n))`

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

`M = 2.575(2.8)/(√(772)) = 0.26`

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

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

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