Question

The percentage of households that include at least one frequent gamer is 58%. A gaming magazine is interested in studying this further to see how it impacts their magazine advertisements. For what sample size, n, will the sampling distribution of sample proportions have a standard deviation of 0.02

Answer 1

Answer:

For a sample size of n = 609.

Step-by-step explanation:

Central limit theorem for proportions:

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:

We have that p = 0.58.

We have to find n for which s = 0.02. So

$s = \sqrt{\frac{p(1-p)}{n}}$

$0.02 = \sqrt{\frac{0.58*0.42}{n}}$

$0.02\sqrt{n} = \sqrt{0.58*0.42}$

$\sqrt{n} = \frac{\sqrt{0.58*0.42}}{0.02}$

$(\sqrt{n})^{2} = (\frac{\sqrt{0.58*0.42}}{0.02})^{2}$

$n = 609$

For a sample size of n = 609.