The monthly starting salaries of students who receive an MBA degree have a population standard deviation of $120. What sample size should be selected to obtain a .95 probability of estimating the population mean monthly income within a margin of $20? [sample size]
Accepted Solution
A:
Answer:138Step-by-step explanation:Population standard deviation = [tex]\sigma = 120[/tex].95 probability of estimating the population mean monthly income within a margin of $20So, Significance level = 1-0.95 = 0.05α =0.05Margin error = 20[tex]ME =Z \times \frac{\sigma}{\sqrt{n}}[/tex]Z at 0.05 = 1.96[tex]20 =1.96 \times \frac{120}{\sqrt{n}}[/tex][tex]\sqrt{n} =1.96 \times \frac{120}{20}[/tex][tex]n =(1.96 \times \frac{120}{20})^2[/tex][tex]n =138.2976[/tex]So, n = 138Hence sample size should be 138 selected to obtain a .95 probability of estimating the population mean monthly income within a margin of $20