The 2003 Statistical Abstract of the United States reported

The 2003 Statistical Abstract of the United States reported the percentage of people 18 years of age and older who smoke. Suppose that a study designed to collect new data on smokers and nonsmokers uses a preliminary estimate of the proportion who smoke of .30.   

What is the 95% confidence interval for the proportion of smokers in the population (to 4 decimals)?

Thanks!

Solution

Answer to the question)

p = 0.30

n = 520

SE = sqrt(p*(1-p)/n)

SE = sqrt(0.3*0.7/520) = 0.0201

.

Formula of confidence interval

p^ - z*SE , p^ + z* SE

.

we got z = 1.96 for 95% confidence level

.

On pluggign the values we get

0.30 - 1.96 *0.0201 , 0.30 +1.96 * 0.0201

0.2606 , 0.3394