Question

U.2.1 According to a study of political prisoners, the mean duration of imprisonment for 38 prisoners with chronic post-traumatic stress disorder (PTSD) was 32 3 months. Assuming that = 39 months, determine a 95% confidence interval for the mean duration of imprisonment of all political prisoners with chronic PTSD. Interpret your answer in words. Click here to view Page 1 of the table of areas under the standard normal curve Click here to view Page 2 of the table of areas under the standard normal curve A 95% confidence interval for the population mean is from months to months (Round to one decimal place as needed)

Answer 1

Solution:

Given,

$\bar x$ = 32.3 ....... Sample mean

$\sigma$   = 39 ........Population standard deviation

n = 38   ....... Sample size

Note that, Population standard deviation($\sigma$) is known..So we use z distribution. Our aim is to construct 95% confidence interval.

$\therefore$ c = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

$\therefore$  $\alpha$/2 = 0.05 $\slash$ 2 = 0.025 and 1- $\alpha$ /2 = 0.975

$\therefore$ = 1.96

a/2

Now , confidence interval for mean($\mu$) is given by:

Margin.of.error< u< Margin.of.error

  

Za/2(a/Vn)<u<Za/2* (a/Vn)

32.3 -1.96*(39/ $\sqrt{}$ 38)   
Κμκ ΤΕ
  32.3 + 1.96*(39/ $\sqrt{}$ 38)

32.3 -  12.4  < $\mu$ < 32.3 + 12.4

19.9  < $\mu$ < 44.7

is the required 95% confidence interval for mean....

95% Confidence Interval for population mean is from 19.9 to 44.7