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 .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)?

Answer 1

Answer)

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)