Mathematics, 07.10.2020 22:01 tressasill

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.

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Answer from: cvbnkifdfg

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.

E = 14.1

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

n =57

$23.514 < \mu < 49.486$

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

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

$n = [ \frac{2.58 * 38}{13}]^2$

=> 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

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

=> $23.514 < \mu < 49.486$

Answer from: Quest

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