Question

Consider a population proportion p=0.12. [You may find it useful to reference the z table.] a. What is the expected value and the stonderd error of the sampling distribution of the sample proportion with n= 20 and n= 50. (Round the standard error to 4 decimal places.)

Answer 1

The expected value is 0.12, standard error given that n = 20 = 0.0727

The expected value is 0.12, standard error given that n = 50 = 0.0460

How to solve for the standard error

The population proportion p=0.12

we have the expected value as well as the standard error as what we have to compute here

when we have n = 20

`\sqrt{(p(1-p))/(n) }`

`\sqrt{(0.12(1 - 0.12))/(20) }`

= 0.0727

Then we have n = 50

using proportion as 0.12

`\sqrt{(0.12(1 - 0.12))/(50) }`

Then n = 0.0460

Hence the standard error given that n = 20 = 0.0727

The standard error given that n = 50 = 0.0460