Question

A researcher wishes to estimate with 95% confidence, the proportion of the people who own a home computer. A previous study shows that 40% of the interviewed had a computer at home. The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary?

Answer 1

The minimum sample size necessary is 2305.

Explanation:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

For this problem, we have that:

`\pi = 0.4`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The researcher wishes to be accurate within 2% of the true proportion. Find the minimum sample size necessary?

We need a sample size of n.

n is found when M = 0.02. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.02 = 1.96\sqrt{(0.4*0.6)/(n)}`

`0.02√(n) = 1.96*√(0.4*0.6)`

`√(n) = (1.96*√(0.4*0.6))/(0.02)`

`(√(n))^(2) = ((1.96*√(0.4*0.6))/(0.02))^(2)`

`n = 2304.96`

Rounding up

The minimum sample size necessary is 2305.