Question

A political candidate has asked his/her assistant to conduct a poll to determine the percentage of people in the community that supports him/her. If the candidate wants a 3% margin of error at a 90% confidence level, what size of sample is needed

Answer 1

Answer:

A sample size of 752 is needed.

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 z-score that has a p-value of $1 - \frac{\alpha}{2}$.

The margin of error is of:

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

90% confidence level

So $\alpha = 0.1$, z is the value of Z that has a p-value of $1 - \frac{0.1}{2} = 0.95$, so $Z = 1.645$.

If the candidate wants a 3% margin of error at a 90% confidence level, what size of sample is needed?

We have no estimate of the proportion, so we use $\pi = 0.5$.

The sample size is n for which M = 0.03. So

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

$0.03 = 1.645\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.645*0.5$

$\sqrt{n} = \frac{1.645*0.5}{0.03}$

$(\sqrt{n})^2 = (\frac{1.645*0.5}{0.03})^2$

$n = 751.67$

Rounding up:

A sample size of 752 is needed.