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 4000. b. For a random sample of size n equals 1000. c. For a random sample of size n equals 500.

Answer 1

Answer:

a) Mean 0.12, standard deviation 0.0051

b) Mean 0.12, standard deviation 0.0103

c) Mean 0.12, standard deviation 0.0145

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For proportions, the mean is $\mu = p$, and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$

In this problem, we have that:

$p = 0.12$

So

a. For a random sample of size n equals 4000.

Mean 0.12, standard deviation $s = \sqrt{\frac{0.12*0.88}{4000}} = 0.0051$

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

Mean 0.12, standard deviation $s = \sqrt{\frac{0.12*0.88}{1000}} = 0.0103$

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

Mean 0.12, standard deviation $s = \sqrt{\frac{0.12*0.88}{500}} = 0.0145$