Question

The American Heart Association is about to conduct an anti-smoking campaign and wants to know the fraction of Americans over 20 who smoke. Step 2 of 2 : Suppose a sample of 966 Americans over 20 is drawn. Of these people, 783 don't smoke. Using the data, construct the 98% confidence interval for the population proportion of Americans over 20 who smoke. Round your answers to three decimal places.

Answer 1

Solution :

Given that,

n = 966

x = 183

Point estimate = sample proportion = = x / n = 183 /966 = 0.189

1 - = 1- 0.189 = 0.811

Z_/2 = 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 * (0.189*(1-0.189) /966 )

= 0.029

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.189 - 0.029 < p <0.189 + 0.029

0.160< p < 0.219

The 98% confidence interval for the population proportion p is : 0.160 , 0.219