Question

According to a study of political​ prisoners, the mean duration of imprisonment for 30 prisoners with chronic​ post-traumatic stress disorder​ (PTSD) was 30.6 months. Assuming that sigmaequals42 ​months, determine a 95​% confidence interval for the mean duration of​ imprisonment, mu​, of all political prisoners with chronic PTSD. Interpret your answer in words.

Answer 1

The 95​% confidence interval for the mean duration of​ imprisonment, mu​, of all political prisoners with chronic PTSD is between 15.57 months and 45.63 months.

This means that we are 95% sure that the true mean mean duration of​ imprisonment, mu​, of all political prisoners with chronic PTSD is between 15.57 months and 45.63 months.

Explanation:

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

`\alpha = (1-0.95)/(2) = 0.025`

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
`1-0.025 = 0.975`, so
`z = 1.96`

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 = 1.96(42)/(√(30)) = 15.03`

The lower end of the interval is the sample mean subtracted by M. So it is 30.6 - 15.03 = 15.57 months

The upper end of the interval is the sample mean added to M. So it is 30.6 + 15.03 = 45.63 months

The 95​% confidence interval for the mean duration of​ imprisonment, mu​, of all political prisoners with chronic PTSD is between 15.57 months and 45.63 months.

This means that we are 95% sure that the true mean mean duration of​ imprisonment, mu​, of all political prisoners with chronic PTSD is between 15.57 months and 45.63 months.