Answer:
95% of the confidence interval for the proportion of smokers who have tried to quit within the past year
(0.47228 , 0.49868)
Step-by-step 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
[tex]P = \frac{x}{n} = \frac{2626}{5409} =0.48548[/tex]
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
[tex](P - Z_{\alpha } \sqrt{\frac{PQ}{n} } , P+ Z_{\alpha } \sqrt{\frac{PQ}{n} } )[/tex]
[tex](0.48548 - 1.96 \sqrt{\frac{0.48548 X 0.51452}{5409} } , 0.48548+ 1.96\sqrt{\frac{0.48548 X 0.51452}{5409} } )[/tex]
( 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)
