Question

What sample size would be required to estimate the true proportion of American female business executives who prefer the title "Ms.," with an error of ±0.025 and 98 percent confidence?

Answer 1

A sample size of 2166 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)`.

The margin of error of a confidence interval is:

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

98% confidence level

So
`\alpha = 0.02`, z is the value of Z that has a pvalue of
`1 - (0.02)/(2) = 0.99`, so
`Z = 2.327`.

Solving

We need to find n when M = 0.025.

We dont know the proportion, so we use
`\pi = 0.5`, which is when we are going to need the largest sample size for this estimate.

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

`0.025 = 2.327\sqrt{(0.5*0.5)/(n)}`

`0.025√(n) = 2.327*0.5`

`√(n) = (2.327*0.5)/(0.025)`

`(√(n))^(2) = ((2.327*0.5)/(0.025))^(2)`

`n = 2166`

A sample size of 2166 is needed.