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

Answer:

The cost difference would be of $22,067.5

Step-by-step 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.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

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 $1 - \frac{0.05}{2} = 0.975$, 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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.04 = 1.96\sqrt{\frac{0.43*0.57}{n}}$

$0.04\sqrt{n} = 1.96\sqrt{0.43*0.57}$

$\sqrt{n} = \frac{1.96\sqrt{0.43*0.57}}{0.04}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.43*0.57}}{0.04})^{2}$

$n = 588.5$

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

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.01 = 1.96\sqrt{\frac{0.43*0.57}{n}}$

$0.01\sqrt{n} = 1.96\sqrt{0.43*0.57}$

$\sqrt{n} = \frac{1.96\sqrt{0.43*0.57}}{0.01}$

$(\sqrt{n})^{2} = (\frac{1.96\sqrt{0.43*0.57}}{0.01})^{2}$

$n = 9415.7$

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