Question

a political candidate has asked you to conduct a poll to determine what percentage of people support her.
if the candidate wants a 4% margin of error at a 90% confidence level, what sample size is needed?
use technology to find the z - score and do not round between steps. make sure you use the correct rounding rule for samples size, and give your answer in whole people.
question help: message instructor

Explanation:

Step1: Find the z - score

For a 90% confidence level, the significance level \(\alpha=1 - 0.90=0.10\), and \(\alpha/2=0.05\). Using technology (e.g., a standard normal table or a calculator with a normal distribution function), the \(z\) - score \(z_{\alpha/2}=z_{0.05}\approx1.645\).

Step2: Assume \(p = 0.5\) (most conservative estimate for sample size in proportion problems)

The formula for the margin of error \(E\) in a proportion is \(E = z_{\alpha/2}\sqrt{\frac{p(1 - p)}{n}}\), and we want to solve for \(n\). Rearranging the formula for \(n\) gives \(n=\frac{z_{\alpha/2}^{2}p(1 - p)}{E^{2}}\).
Substitute \(z_{\alpha/2}=1.645\), \(p = 0.5\), and \(E = 0.04\) into the formula:

$$
LATEXBLOCK0
$$

First, calculate \((1.645)^{2}=2.706025\). Then \(2.706025\times0.25 = 0.67650625\).

$$ n=\frac{0.67650625}{0.0016}=422.81640625 $$

Step3: Round up the sample size

Since the sample size \(n\) must be a whole number and we cannot have a fraction of a person, we round up \(422.81640625\) to \(423\).

Answer:

\(423\)