Mathematics, 03.04.2020 22:02 jwbri
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.
Answers: 2
![Interval= \hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\\\\\\\=0.4873\pm 1.96\times\sqrt{\frac{0.4873(1-0.4873)}{5409}}\\\\\\\\=0.4873\pm0.0133\\\\\\=[0.4740,0.5006]](/tpl/images/0580/6243/88a4a.png)
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The National Center for Health Statistics interviewed 5409 adults smokers in 2015, and 2636 of them...
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