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

For a sample size of n = 609.

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{(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{(p(1-p))/(n)}`

`0.02 = \sqrt{(0.58*0.42)/(n)}`

`0.02√(n) = √(0.58*0.42)`

`√(n) = (√(0.58*0.42))/(0.02)`

`(√(n))^(2) = ((√(0.58*0.42))/(0.02))^(2)`

`n = 609`

For a sample size of n = 609.