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 .33.

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

Part a

We are given

Estimate for proportion = p = 0.33

q = 1 – p = 1 – 0.33 = 0.67

Margin of error = E = 0.02

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Sample size formula is given as below:

n = p*q*(Z/E)^2

n = 0.33*0.67*(1.96/0.02)^2

n = 2123.444

n = 2124

Required sample size = 2124

Part b

We have

Sample size = n = 2124

Numbers of interest = x = 520

Point estimate = x/n = 520/2124 = 0.244821

Point estimate = P = 0.244821

Part c

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

n = 2124

P = 0.244821

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.244821 ± 1.96* sqrt(0.244821*(1 – 0.244821)/2124)

Confidence Interval = 0.244821 ± 1.96* 0.0093

Confidence Interval = 0.244821 ± 0.0183

Lower limit = 0.244821 - 0.0183 = 0.2265

Upper limit = 0.244821 + 0.0183 = 0.2631

Confidence interval = (0.2265, 0.2631)