Question

In a prior sample of U.S. adults, the Center for Disease Control (CDC), found that 8% 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. Enter your answers as whole numbers.
(a) What is the minimum sample size required to obtain this type of accuracy? Use the prior sample proportion in your calculation. The minimum sample size is 1 U.S. adults.


(b) What is the minimum sample size required to obtain this type of accuracy when you assume no prior knowledge of the sample proportion? The minimum sample size is 2 U.S. adults.

Answer 1

Answer: a) 453  b) 1537

Step-by-step explanation:

As per given , we have

Margin of error : E= 0.025

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

a) The prior estimate of population proportion : p=0.08

Required sample size :-

$n=p(1-p)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.08(1-0.08)(\dfrac{1.96}{0.025})^2\\\\=452.386816\approx453$

The minimum sample size is 453 U.S. adults.

b) Since the prior estimate of population proportion is not available , so we take p= 0.5

Required sample size :-

$n=0.5(1-0.5)(\dfrac{z_{\alpha/2}}{E})^2\\\\=0.25(\dfrac{1.96}{0.025})^2\\\\=1536.64\approx1537$

The minimum sample size is 1537 U.S. adults.