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 .30.

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.
B. Assume that the study uses your sample size recommendation in part (a) and finds 520 smokers. What is the point estimate of the proportion of smokers in the population (to 4 decimals)?
C. What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)? ( , )

Answer 1

Answer:

A. 1,653 people 18 years old or older

B. 31.45%

C. [1653-0.0482*s, 1653+0.0482*s] where s is the standard deviation of the sample

Step-by-step explanation:

A.

We are going to assume simple random sampling on an extremely large population.

Then we have that the sample size n should be

$n=\frac{Z^2PQ}{E^2}$

where  

Z= z-score corresponding to the level of confidence, in this case 95% two tails (1.96)

P=Proportion of smokers (0.3)

Q=Proportion of non smokers (0.7)

E=error (0.02)

so  

$n=\frac{3.1486*0.3*0.7}{0.02^2}=1653$

and the size of the sample should be 1,653 people 18 years old or older

B.

If we find 520 smokers out of 1,653 people in the sample, the proportion is

520/1653 = 0.3145 = 31.45%

C.

The 95 % confidence interval would be

$[1653-1.96*\frac{s}{\sqrt{1653}},1653-1.96*\frac{s}{\sqrt{1653}}]=[1653-0.0482s,1653+0.0482s]$

where s is the standard deviation of the sample.