Question

Consider a sampling distribution with p equals 0.11 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 250.

Answer 1

Answer:

a) Mean 0.11 and standard deviation 0.0044.

b) Mean 0.11 and standard deviation 0.0099.

c) Mean 0.11 and standard deviation 0.0198

Step-by-step explanation:

Central Limit Theorem:

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

In this question:

$p = 0.11$

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

Mean:

$\mu = p = 0.11$

Standard deviation:

$s = \sqrt{\frac{0.11*0.89}{5000}} = 0.0044$

Mean 0.11 and standard deviation 0.0044.

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

Mean:

$\mu = p = 0.11$

Standard deviation:

$s = \sqrt{\frac{0.11*0.89}{1000}} = 0.0044$

Mean 0.11 and standard deviation 0.0099.

c. For a random sample of size n equals 250.

Mean:

$\mu = p = 0.11$

Standard deviation:

$s = \sqrt{\frac{0.11*0.89}{250}} = 0.0198$

Mean 0.11 and standard deviation 0.0198