Question

stat: sample of 780 adults in the U.S.A., it was found that 80 of those had a pinworm infestation. You want to find the 99% confidence interval for the proportion of all U.S. adults with pinworm, In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that 10% of the people in this sample had pinworm but the margin of error for the population estimate was too large. They want an estimate that is in error by no more than 2.5 percentage points at the 95% confidence level.What is the minimum sample size required to obtain this type of accuracy?

Answer 1

Answer: 554

Step-by-step explanation:

If prior population proportion is known, then the formula to find the sample size is given by :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2$

As per given description, we have

p= 0.1

E=0.025

Critical z-value for 95% confidence : $z_{\alpha/2}=1.96$

Then,

$n=0.1(1-0.1)(\dfrac{1.96}{0.025})^2=553.1904\approx554$

Hence, the minimum sample size required = 554.