Question

We conduct a simulation to mimic randomly sampling from a population with of college graduates. In the population 62% had student loans. Each sample has 50 graduates in it. According to theory, what will be the standard deviation of the distribution of sample proportions?

Enter a number in decimal form. For example, enter 0.50, not 50 or 50%.
Round your answer to three decimal places, e.g. 0.125, not 0.1254

Answer 1

Answer: 0.069

========================================

Work Shown:

p = 0.62 = given population proportion

n = 50 = given sample size

`\sigma_{\hat{p}}` = standard deviation of the sample proportions
`\hat{p}`

`\sigma_{\hat{p}}` = standard error of
`\hat{p}`

`\sigma_{\hat{p}} = \sqrt{(p(1-p))/(n)}\\\\\sigma_{\hat{p}} = \sqrt{(0.62(1-0.62))/(50)}\\\\\sigma_{\hat{p}} = \sqrt{(0.62*0.38)/(50)}\\\\\sigma_{\hat{p}} = \sqrt{(0.2356)/(50)}\\\\\sigma_{\hat{p}} = √(0.004712)\\\\\sigma_{\hat{p}} \approx 0.06864400920692\\\\\sigma_{\hat{p}} \approx 0.069\\\\`

This value tells us how spread out the distribution of sample proportions
`\hat{p}` will be.