Question

The National Center for Health Statistics interviewed 5409 adult smokers in 2015, and 2626 of them said they had tried to quit smoking during the past year. Consider this to be a random sample. Find a 95% confidence interval for the proportion of smokers who have tried to quit within the past year. (Round the final answers to three decimal places.)

Answer 1

95% of the confidence interval for the proportion of smokers who have tried to quit within the past year

(0.47228 , 0.49868)

Explanation:

Step(i):-

Given that the National Center for Health Statistics interviewed 5409 adult smokers in 2015

and 2626 of them said they had tried to quit smoking during the past year

proportion

`P = (x)/(n) = (2626)/(5409) =0.48548`

Q = 1-P = 1 - 0.48548 = 0.5146

Step(ii):-

95% of the confidence interval for the proportion of smokers who have tried to quit within the past year

`(P - Z_(\alpha ) \sqrt{(PQ)/(n) } , P+ Z_(\alpha ) \sqrt{(PQ)/(n) } )`

`(0.48548 - 1.96 \sqrt{(0.48548 X 0.51452)/(5409) } , 0.48548+ 1.96\sqrt{(0.48548 X 0.51452)/(5409) } )`

( 0.48548 - 0.0132 , 0.48548 +0.0132)

(0.47228 , 0.49868)

Final answer:-

95% of the confidence interval for the proportion of smokers who have tried to quit within the past year

(0.47228 , 0.49868)