Question

According to data from the 2010 United States Census, 43.1% of Americans over the age of 65 were male.

Suppose Maria, a researcher, takes a random sample of 60 Americans over the age of 65 and finds that 18 are male.
Let p represent the sample proportion of Americans over the age of 65 that were male.

What are the mean and standard deviation of the sampling distribution of p?

Answer 1

Answer:
`\mu_{\hat{p}}=0.431\\\\ \sigma_{\hat{p}}=0.064`

Explanation:

The mean and standard deviation of the sampling distribution of p is given by :-

`\mu_{\hat{p}}=p\\\\ \sigma_{\hat{p}}=\sqrt{(p(1-p))/(n)}`

, where p= population proportion.

n= sample size.

Let p represent the sample proportion of Americans over the age of 65 that were male.

Given : The proportion of Americans over the age of 65 were male.

p= 43.1%=0.431

sample size : n= 60

Then, the mean and standard deviation of the sampling distribution of p will be :-

`\mu_{\hat{p}}=p =0.431\\\\ \sigma_{\hat{p}}=\sqrt{(p(1-p))/(n)}=\sqrt{(0.431(1-0.431))/(60)}\\\\=√(0.0041)\approx0.064`

Hence, the mean and standard deviation of the sampling distribution of p :

`\mu_{\hat{p}}=0.431\\\\ \sigma_{\hat{p}}=0.064`