Question

Assume that the president’s job approval is at 43%. Assume the cost of conducting a poll is $2.50 per person. Imagine if you were to be tasked with helping to create a poll for next month on the president’s job approval (specifically, on determining the number of people to be sampled). What would be the cost difference for a poll designed to have a margin of error of 1 percentage point vs. that of a poll designed to have a margin of error of 4 percentage points?

Answer 1

The cost difference would be of $22,067.5

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)}`

In this question, we have that:

`\pi = 0.43`

I think there was a small typing mistake, and the confidence level was omitted. I will use a 95% confidence level.

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`.

How many people are needed for a margin of error of 4 percentage points?

This is n when M = 0.04. So

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

`0.04 = 1.96\sqrt{(0.43*0.57)/(n)}`

`0.04√(n) = 1.96√(0.43*0.57)`

`√(n) = (1.96√(0.43*0.57))/(0.04)`

`(√(n))^(2) = ((1.96√(0.43*0.57))/(0.04))^(2)`

`n = 588.5`

Rounding up

For a margin of error of 4 percentage points, 589 people will be sampled.

How many people are needed for a margin of error of 1 percentage point?

This is n when M = 0.01. So

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

`0.01 = 1.96\sqrt{(0.43*0.57)/(n)}`

`0.01√(n) = 1.96√(0.43*0.57)`

`√(n) = (1.96√(0.43*0.57))/(0.01)`

`(√(n))^(2) = ((1.96√(0.43*0.57))/(0.01))^(2)`

`n = 9415.7`

Rounding up

For a margin of error of 1 percentage point, 9416 people will be sampled.

What would be the cost difference for a poll designed to have a margin of error of 1 percentage point vs. that of a poll designed to have a margin of error of 4 percentage points?

Cost per person $2.50.

Margin of error of 0.01: Cost of 9416*2.50 = $23,540

Margin of error of 0.04: Cost of 589*2.50 = $1472.5

Difference:

23540 - 1472.5 = $22,067.5

The cost difference would be of $22,067.5