Question

In the past, 21% of all homes with a stay-at-home parent, the father is the stay-at-home parent. An independent research firm has been charged with conducting a sample survey to obtain more current information. (a) 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 0.09? Use a 95% confidence level. (Round your answer up to the nearest whole number.)

Answer 1

A sample size of 79 is needed.

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)`.

For this problem, we have that:

`\pi = 0.21`

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`.

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 0.09?

A sample size of n is needed.

n is found when M = 0.09. So

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

`0.09 = 1.96\sqrt{(0.21*0.79)/(n)}`

`0.09√(n) = 1.96√(0.21*0.79)`

`√(n) = (1.96√(0.21*0.79))/(0.09)`

`(√(n))^(2) = ((1.96√(0.21*0.79))/(0.09))^(2)`

`n = 78.68`

Rounding up to the nearest whole number.

A sample size of 79 is needed.