Question

Data from a representative sample were used to estimate that 31% of all computer users in a recent year had tried to get on a Wi-Fi network that was not their own in order to save money. You decide to conduct a survey to estimate this proportion for the current year. What is the required sample size if you want to estimate this proportion with a margin of error of 0.05

Answer 1

The required sample size is 329.

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)}`

31% of all computer users in a recent year had tried to get on a Wi-Fi network that was not their own in order to save money.

This means that
`\pi = 0.31`

Confidence level is not given, so we assume 95%.

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`.

What is the required sample size if you want to estimate this proportion with a margin of error of 0.05?

We need a sample size of n. n is found when M = 0.05. So

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

`0.05 = 1.96\sqrt{(0.31*0.69)/(n)}`

`0.05√(n) = 1.96√(0.31*0.69)`

`√(n) = (1.96√(0.31*0.69))/(0.05)`

`(√(n))^2 = ((1.96√(0.31*0.69))/(0.05))^2`

`n = 328.6`

Rounding up

The required sample size is 329.