Question

confidence interval of months to months has been found for the mean duration of​ imprisonment, ​, of political prisoners of a certain country with chronic PTSD.a. Determine the margin of​ error, E.b. Explain the meaning of E in this context in terms of the accuracy of the estimate.c. Find the sample size required to have a margin of error of months and a ​% confidence level.​ (Use ​months.)d. Find a ​% confidence interval for the mean duration of​ imprisonment, ​, if a sample of the size determined in part​ (c) has a mean of months.

Answer 1

Complete Question

A 95% confidence interval of 19.3 months to 47.5 months has been found for the mean duration of? imprisonment, ??,of political prisoners of a certain country with chronic PTSD.

a. Determine the margin of error, E.

b. Explain the meaning of E in this context in terms of the accuracy of the estimate.

c. Find the sample size required to have a margin of error of 13 months and a 99% confidence level.? (Use 38 months. for standard deviation )

d. Find a 99% confidence interval for the mean duration of? imprisonment, ??, if a sample of the size determined in part? (c) has a mean of 36.5 months.

Answer:

a

`E =  14.1`

b

In this context E tell us that the true mean will lie within E = 14.1 of the sample mean

c

`n  =57`

d

`23.514 < \mu < 49.486`

Explanation:

Considering question a

From the question we are told that

The upper limit is U = 47.5 months

The lower limit is L = 19.3 months

Generally the margin of error is mathematically represented as

`E = (U - L )/(2)`

=>
`E = ( 47.5 - 19.3 )/(2)`

=>
`E = 14.1`

Considering question b

In this context E tell us that the true mean will lie within E = 14.1 of the sample mean

Considering question c

Generally the sample size is mathematically represented as

`n = [ \frac{ Z_{(\alpha )/(2) * \sigma }}{ E} ]^2`

Here E is given as E = 13

Given that the confidence level is 99% then the level of significance is

`\alpha = (100 - 99 )\%`

=>
`\alpha = 0.01`

From the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } = Z_{(0.01 )/(2) } = 2.58`

So

`n = [ (2.58 *  38)/(13)]^2`

=>
`n =57`

Considering question d

From the question

The sample mean is
`\= x = 36.5`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } * (\sigma )/(n)`

=>
`E = 2.58 * (38 )/(57)`

=>
`E = 12.986`

Generally the 99% confidence interval for mean distribution is mathematically represented as

`36.5 - 12.986 < \mu < 36.5 + 12.986`

=>
`23.514 < \mu < 49.486`