Question

ga political candidate has asked you to conduct a poll to determine what percentage of people support her. if the candidate only wants a 8% margin of error at a 95% cnofidence level, what size of sample is needed

Answer 1

Answer: 151

Step-by-step explanation:

if prior population proportion is unknown , then the formula is used to find the sample size :

$n=0.25(\frac{z_{\alpha/2}}{E})^2$

, where $z_{\alpha/2}$ = Two tailed critical value for significance level of $\alpha.$

E = Margin of error.

Given : margin of error = 8%= .08

For 95% confidence level , two tailed critical value = 1.96

Now, the required sample size :

$n=0.25(\frac{1.96}{0.08})^2\\\\=0.25(24.5)^2\\\\=150.0625\approx151$

Hence, the size of the sample needed = 151.