Question

The National Center for Health Statistics interviewed 5409 adults smokers in 2015, and 2636 of them said they had tried to quit smoking during the past year. Consider this to be a random sample. a) Find a 95% confidence interval for the proportion of smokers who have tried to quit within the past year.

Answer 1

0.4740<p<0.5006

Explanation:

-Given
`n=5409, \ x=2636 , \ CI=0.95`

#we calculate the proportion of trial quitters;

`\hat p=(2636)/(5409)\\\\=0.4873`

For a confidence level of 95%:

`z_(\alpha/2)=z_(0.025)\\\\=1.96`

The confidence interval is calculated as follows:

`Interval= \hat p\pm z_(\alpha/2)\sqrt{(\hat p(1-\hat p))/(n)}\\\\\\\\=0.4873\pm 1.96*\sqrt{(0.4873(1-0.4873))/(5409)}\\\\\\\\=0.4873\pm0.0133\\\\\\=[0.4740,0.5006]`

Hence, the 95% confidence interval is 0.4740<p<0.5006