Question

A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 2 percentage points with 99​% confidence if ​(a) he uses a previous estimate of 25​%? ​(b) he does not use any prior​ estimates?

Answer 1

Answer:

a) We need a sample size of at least 3109.

b) We need a sample size of at least 4145.

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:

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

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 $Z = 2.575$.

(a) he uses a previous estimate of 25​%?

we need a sample of size at least n.

n is found when $M = 0.02, \pi = 0.25$. So

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

$0.02 = 2.575\sqrt{\frac{0.25*0.75}{n}}$

$0.02\sqrt{n} = 2.575\sqrt{0.25*0.75}$

$\sqrt{n} = \frac{2.575\sqrt{0.25*0.75}}{0.02}$

$(\sqrt{n})^{2} = (\frac{2.575\sqrt{0.25*0.75}}{0.02})^{2}$

$n = 3108.1$

We need a sample size of at least 3109.

(b) he does not use any prior​ estimates?

When we do not use any prior estimate, we use $\pi = 0.5$

So

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

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

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

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

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

$n = 4144.1$

Rounding up

We need a sample size of at least 4145.