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 ( , ).

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

             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

(0.49288 - 0.01332 , 0.49288+0.01332)

(0.47956 , 0.50620 )

Final -

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

(0.47956 ,0.50620)

y=3n + 1

step-by-step explanation:

this is wrong because it says the second number is one less and i added one instead