Question

# The Centers for Disease Control reported the percentage of people 18 years of age and older...

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 .35.

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)?

A)

Given p = 0.35

Margin of error = 0.02

We know that margin of error (MOE)= Z*(√P*(1-P)/√N)

We know p and we need to find the N

To find N, first we need to estimate the value of z

First we need to divide 95% by 100 and then subtract it from 1

1-(95/100) = 0.05

Now we need to divide it by 2

=0.025

From z table, 1.96 corresponds to 0.025

Therefore, z = 1.96

0.02 = 1.96*√0.35*(1-0.35)/√n

√n = (1.96/0.02)*√0.35*(1-0.35)

N = (1.96/0.02)^2*{0.35*(1-0.35}

N = 2185

b)

Point estimate = 520/2185

= 0.23798627002

= 0.2380

C)

Margin of error = z*√p*(1-p)/√n

1.96*√0.2380*(1-0.2380)/√2185)

= 0.01785649904

Confidence interval is given by

P-MOE, P+MOE

0.238-0.01786, 0.238+0.1786

= 0.22014350095 to 0.25585649904

= (0.2201 to 0.2559)