Question

A 95​% confidence interval of 17.6 months to 49.2 months has been found for the mean duration of​ imprisonment, mu​, 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 11 months and a 99​% confidence level.​ (Use sigmaequals45 ​months.) d. Find a 99​% confidence interval for the mean duration of​ imprisonment, mu​, if a sample of the size determined in part​ (c) has a mean of 36.5 months.

Answer 1

Answer:

a) $E = \frac{49.2-17.6}{2}= 15.8$

b) For this case we have 95% of confidence that the true mean would be between $\pm 15.8$ units respect the true mean.

c) $n=(\frac{2.58(45)}{11})^2 =111.39 \approx 112$

So the answer for this case would be n=112

d) $36.5-2.58\frac{45}{\sqrt{112}}=25.53$    

$36.5+2.58\frac{45}{\sqrt{112}}=47.47$

Step-by-step explanation:

Part a

For this case we know that the cinfidence interval for the true mean is given by:

$\bar X \pm E$

Where E represent the margin of error. For this case we have the confidence interval at 95% of confidence and we can estimate the margin of error like this:

$E = \frac{49.2-17.6}{2}= 15.8$

Part b

For this case we have 95% of confidence that the true mean would be between $\pm 15.8$ units respect the true mean.

Part c

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$    (a)

And on this case we have that ME =11 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$   (b)

The critical value for 99% of confidence interval now can be founded using the normal distribution. The critical value would be $z_{\alpha/2}=2.58$, replacing into formula (b) we got:

$n=(\frac{2.58(45)}{11})^2 =111.39 \approx 112$

So the answer for this case would be n=112

Part d

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Replcaing the info given we got:

$36.5-2.58\frac{45}{\sqrt{112}}=25.53$    

$36.5+2.58\frac{45}{\sqrt{112}}=47.47$