Question #199872

Read and analyze the given problem. Provide a complete solution, Table, and a graph corresponds to the answer.


Based from the result of a survey conducted by a group of researchers, the proportion of females holding executive positions in the government agencies in a certain populated city is normally distributed. In a random sample of 300 female employees in the government agencies, 75 female hold executive positions.

Compute the proportion estimate of the population

Compute for the margin of error of the proportion at 90% CL.

Find 90% and 95% CI for all the female employees in the government agencies holding executive positions.

Interpret the result.


1
Expert's answer
2022-01-07T17:42:03-0500

We have given n=300n = 300

1.) Proportion estimate of the population =75300=p=0.25= \dfrac{75}{300} = p = 0.25


2.) Margin error =z×p(1p)n= z\times \sqrt{\dfrac{p(1-p)}{n}}


z=1.645z = 1.645

Margin error =1.645×0.75×0.25300=0.041= 1.645 \times \sqrt{\dfrac{0.75\times 0.25}{300}} = 0.041


3.) 90% confidence interval


CI=p±zp(1p)nCI = p \pm z\sqrt{\dfrac{p(1-p)}{n}}


CI=0.25±1.645×0.25×0.75300CI = 0.25 \pm 1.645\times \sqrt{\dfrac{0.25\times 0.75}{300}}


CI=0.25±0.041CI = 0.25 \pm 0.041


95% confidence interval


CI=p±zp(1p)nCI = p \pm z\sqrt{\dfrac{p(1-p)}{n}}


CI=0.25±1.96×0.25×0.75300CI = 0.25 \pm 1.96\times \sqrt{\dfrac{0.25\times 0.75}{300}}


CI=0.25±0.049CI = 0.25 \pm 0.049


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