Question

A mayoral candidate in a large metropolitan area has hired you to take a poll to determine the proportion of registered voters who plan to vote for him. He would like you to report a 95% confidence interval with a margin of error no more than 0.04. Whiat is the smallest sample size that will produce an interval with these specifications?

Answer 1

The smallest sample size that will produce an interval with these specifications is 601.

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`.

He would like you to report a 95% confidence interval with a margin of error no more than 0.04. What is the smallest sample size that will produce an interval with these specifications?

We have to find n for which M = 0.04.

We dont know the true proportion, so we use
`\pi = 0.5`, which is when the smallest sample size needed will have it's largest value.

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

`0.04 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.04√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.04)`

`(√(n))^2 = ((1.96*0.5)/(0.04))^2`

`n = 600.25`

Rounding up

The smallest sample size that will produce an interval with these specifications is 601.