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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.)

User Smithclay
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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

User Jim Puls
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