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

   $23.514 < \mu < 49.486$

Step-by-step 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

       $E = \frac{U - L }{2}$

 =>  $E = \frac{ 47.5 - 19.3 }{2}$

 =>  $E = 14.1$

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

        $n = [ \frac{ Z_{\frac{\alpha }{2} * \sigma }}{ E} ]^2$

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  $\frac{\alpha }{2}$  is  

      $Z_{\frac{\alpha }{2} } = Z_{\frac{0.01 }{2} } = 2.58$

So

    $n = [ \frac{2.58 * 38}{13}]^2$

=>    $n =57$

Considering question d

From the question

   The sample mean is   $\= x = 36.5$

Generally the margin of error is mathematically represented as  

         $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{n}$

=>      $E = 2.58 * \frac{38 }{57}$

=>      $E = 12.986$    

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

    $36.5 - 12.986 < \mu < 36.5 + 12.986$

=>     $23.514 < \mu < 49.486$