Question

A mayoral candidate in a large metropolitan area has hired you to take a poll to determine the proportion of registered voters who plan to vote for him. He would like you to report a 95% confidence interval with a margin of error no more than 0.04. Whiat is the smallest sample size that will produce an interval with these specifications?

Answer 1

Answer:

The smallest sample size that will produce an interval with these specifications is 601.

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:

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

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 $Z = 1.96$.

He would like you to report a 95% confidence interval with a margin of error no more than 0.04. What is the smallest sample size that will produce an interval with these specifications?

We have to find n for which M = 0.04.

We dont know the true proportion, so we use $\pi = 0.5$, which is when the smallest sample size needed will have it's largest value.

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

$0.04 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.04\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.04}$

$(\sqrt{n})^2 = (\frac{1.96*0.5}{0.04})^2$

$n = 600.25$

Rounding up

The smallest sample size that will produce an interval with these specifications is 601.