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:

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

             [tex]P = \frac{x}{n} = \frac{2666}{5409} = 0.49288[/tex]

             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

[tex]({P - Z_{0.05} \sqrt{\frac{PQ}{n} } , P + Z_{0.05} \sqrt{\frac{PQ}{n} )[/tex]

[tex]({0.49288 - 1.96 \sqrt{\frac{0.49288 X 0.50712}{5409} } , 0.49288 + 1.96\sqrt{\frac{0.49288 X 0.50712}{5409} )[/tex]

(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)

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