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 .34.

a. How large a sample should be taken to estimate the proportion of smokers in the population with a margin of error of .02 (rounded up to the next 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

Solution =

Given,

E = 0.02

c = 95% = 0.95

or p = 0.34

1 - 1- p = 1 - 0.34 = 0.66

Now,

= 1 - c = 1 - 0.95 = 0.05

/2 = 0.025

= 1.96 (using z table)

The sample size for estimating the proportion is given by

n =

= (1.96)2 * 0.34 * 0.66 / (0.022)

= 2155.1376

= 2156 ..(round to the next whole number)

Answer : Required Sample size is n = 2156

B)  the point estimate of the proportion,

= x/n   = 520 / 2156 = 0.2412

C) Solution:

Given,

n = 2156 ....... Sample size

x = 520 .......no. of successes in the sample

Let denotes the sample proportion.

     = x/n   = 520 / 2156 = 0.2412

Our aim is to construct 95% confidence interval.

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.025 and 1- /2 = 0.975

Search the probability 0.975 in the Z table and see corresponding z value

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.96 * [ 0.2412 *(1 - 0.2412)/2156 ]

= 0.018

Now the confidence interval is given by

( - E)   ( + E)

( 0.2412 - 0.018 )   ( 0.2412 + 0.018 )

0.2232   0.2592  

Required 95% Confidence Interval is ( 0.2232 , 0.2592 )