Question

A newspaper article about the results of a poll states: "In theory, the results of such a poll, in 99 cases out of 100 should differ by no more than 2 percentage points in either direction from what would have been obtained by interviewing all voters in the United States." Find the sample size suggested by this statement.

Answer 1

Answer:

The sample size suggested by this statement is of at least 4145.

Step-by-step explanation:

This statement states that the 99% confidence interval has a margin of error of 2 percentage points.

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:

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

99% confidence level:

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

Find the sample size suggested by this statement.

The sample size is at least n.

n is found when M = 0.02.

We don't know the true proportion, so we use $\pi = 0.5$, which is when the largest sample size will be needed.

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

$0.02 = 2.575\sqrt{\frac{0.5*0.5}{n}}$

$0.02\sqrt{n} = 2.575*0.5$

$\sqrt{n} = \frac{2.575*0.5}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.575*0.5}{0.02})^{2}$

$n = 4144.14$

Rounding up

The sample size suggested by this statement is of at least 4145.