Question

he National Center for Health Statistics interviewed 5409 adult smokers in 2015, and 2666 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.) The 95% confidence interval is ( , ).

Answer 1

Answer:

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

(0.47956 ,0.50620)

Step-by-step explanation:

Step(i):-

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

Sample size 'n' = 5409

Given that 2666 of them said they had tried to quit smoking during the past year.

proportion

             $P = \frac{x}{n} = \frac{2666}{5409} = 0.49288$

             Q = 1 - P = 1 - 0.49288 = 0.50712

Step(ii):-

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

$({P - Z_{0.05} \sqrt{\frac{PQ}{n} } , P + Z_{0.05} \sqrt{\frac{PQ}{n} )$

$({0.49288 - 1.96 \sqrt{\frac{0.49288 X 0.50712}{5409} } , 0.49288 + 1.96\sqrt{\frac{0.49288 X 0.50712}{5409} )$

(0.49288 - 0.01332 , 0.49288+0.01332)

(0.47956 , 0.50620 )

Final answer:-

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

(0.47956 ,0.50620)