Question

A news station would like to conduct an exit poll to determine the likelihood that a highly debated amendment will receive enough support to pass. The news station plans to construct a 90 percent confidence interval to estimate the proportion of voters supporting the amendment. Whatg expressions would give the smallest sample size that will result in a margin of error of no more than 3 percentage points?

Answer 1

Answer:

The expression is $n = (\frac{1.645*0.5}{0.03})^2$

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 $Z = 1.645$.

What expression would give the smallest sample size that will result in a margin of error of no more than 3 percentage points?

We have to find n for which M = 0.03.

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

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

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

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

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

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

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

The expression is $n = (\frac{1.645*0.5}{0.03})^2$