Question

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]

Answer 1

Answer:

138

Step-by-step explanation:

Population standard deviation = $\sigma = 120$

.95 probability of estimating the population mean monthly income within a margin of $20

So, Significance level = 1-0.95 = 0.05

α =0.05

Margin error = 20

$ME =Z \times \frac{\sigma}{\sqrt{n}}$

Z at 0.05 = 1.96

$20 =1.96 \times \frac{120}{\sqrt{n}}$

$\sqrt{n} =1.96 \times \frac{120}{20}$

$n =(1.96 \times \frac{120}{20})^2$

$n =138.2976$

So, n = 138

Hence sample size should be 138 selected to obtain a .95 probability of estimating the population mean monthly income within a margin of $20