Respuesta :
Answer:
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.
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 the margin of error 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}{\sqrt{772}} = 0.26[/tex]
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.
