Question

What sample size is needed if the research firm's goal is to estimate the current proportion of homes with a stay-at-home parent in which the father is the stay-at-home parent with a margin of error of 0.03? Use a 95% confidence level. (Round your answer up to the nearest whole number.)

Answer 1

We need a sample size of 1068.

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)}`

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`.

In this problem:

We need a sample size of n.

n is found when
`M = 0.03`

We do not know the exact proportion, so we use
`\pi = 0.5`, which is when we are are going to need the largest sample size. So

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

`0.03 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.03√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.03)`

`(√(n))^(2) = ((1.96*0.5)/(0.03))^(2)`

`n = 1067.11`

Rounding up

We need a sample size of 1068.