Question

A candidate for a US Representative seat from Indiana hires a polling firm to gauge her percentage of support among voters in her district.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)
b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Answer 1

a) The minimum sample size is 601.

b) The minimum sample size is 2401.

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)}`

For this problem, we have that:

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

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`.

a. If a 95% confidence interval with a margin of error of no more than 0.04 is desired, give a close estimate of the minimum sample size that will guarantee that the desired margin of error is achieved. (Remember to round up any result, if necessary.)

This is n for which M = 0.04. So

`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 minimum sample size is 601.

b. If a 95% confidence interval with a margin of error of no more than 0.02 is desired, give a close estimate of the minimum sample size necessary to achieve the desired margin of error.

Now we want n for which M = 0.02. So

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

`0.02 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.02√(n) = 1.96*0.5`

`√(n) = (1.96*0.5)/(0.02)`

`(√(n))^2 = ((1.96*0.5)/(0.02))^2`

`n = 2401`

The minimum sample size is 2401.