Question

Consider a sampling distribution with p equals 0.12 and samples of size n each. Using the appropriate​ formulas, find the mean and the standard deviation of the sampling distribution of the sample proportion.
(a) For a random sample of size n equals 5000.
(b) For a random sample of size n equals 1000.
(c) For a random sample of size n equals 500.

Answer 1

Answer with explanation:

Consider a sampling distribution with p equals 0.12 and samples of size n each.

Then, formula for mean = p

Formula for standard deviation= $\sqrt{\dfrac{p(1-p)}{n}}$

(a) For a random sample of size n equals 5000.

Mean = 0.12

Standard deviation= $\sqrt{\dfrac{0.12(1-0.12)}{5000}}$

$\\\\0.00459565011723\approx0.0046$

(b) For a random sample of size n equals 1000.

Mean = 0.12

Standard deviation= $\sqrt{\dfrac{0.12(1-0.12)}{1000}}$

$\\\\=0.0102761860629\approx0.0103$

(c) For a random sample of size n equals 500.

Mean = 0.12

Standard deviation= $\sqrt{\dfrac{0.12(1-0.12)}{500}}$

$\\\\=0.0145327216997\approx0.0145$