Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 50 who smoke. Step 2 of 2 : Suppose a sample of 2851 Americans over 50 is drawn. Of these people, 1911 don't smoke. Using the data, construct the 98% confidence interval for the population proportion of Americans over 50 who smoke. Round your answers to three decimal places.

Answer 1

Confidence interval for the population proportion :p

Sample Size : n : Number of Americans over 50 in the sample = 2851

Number of people don't smoke of the sample : 1911

Number of people who smoke of the sample : x= 2851 - 1911=940

Sample proportion of Americans over 50 who smoke = x/n = 940/2851 = 0.3297

Confidence Level	98%
= (100-98)/100=2/100 = 0.02	0.02
/2 =0.02/2=0.01	0.01
Z/2= Z0.01	2.3263

98% confidence interval for the population proportion of Americans over 50 who smoke

98% confidence interval for the population proportion of Americans over 50 who smoke = (0.3092, 0.3502)