Question

The Centers for Disease Control reported the percentage of people 18 years of age and older who smoke (CDC website, December 14, 2014). Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .31. a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (to the nearest whole number)? Use 95% confidence.

Answer 1

Answer:   2056

Step-by-step explanation:

The formula for margin of error for population proportion  :-

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

Given : Significance level : $\alpha = 1-0.95=0.05$

Critical value : $z_{0.025}=1.96$      

The proportion of people smoke : p=0.31.

Margin of error : E= 0.02

Substitute all the value in the above formula, we get

$0.02=1.96\times\sqrt{\dfrac{0.31(0.69)}{n}}\\\\\Rightarrow\0.0102=\sqrt{\dfrac{0.2139}{n}}$

Squaring both sides , we get

$0.00010404=\dfrac{0.2139}{n}\\\\\Rightarrow\ n=\dfrac{0.2139}{0.00010404}=2055.9400230\approx2056$

Hence, the required sample size = 2056