Question

A study of 4,430 U.S. adults found that 2,913 were obese or overweight. The 95% confidence interval for the proportion of adults in U.S. who are obese or overweight is _____________.

(0.646, 0.669)

(0.644, 0.672)

(0.639, 0.676)

(0.650, 0.665)

Answer 1

Answer:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We ar confident that the true proportion of people in US who are obes is between 0.644 and 0.672

And the best option is given by:

(0.644, 0.672)

Step-by-step explanation:

For this case we can begin calculating the proportion of people overweight with this formula:

$\hat p =\frac{2913}{4430}= 0.658$

The confidence level is 95% , and the significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

Replacing the info given we got:

$0.658 - 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.644$

$0.658 + 1.96\sqrt{\frac{0.658(1-0.658)}{4430}}=0.672$

We ar confident that the true proportion of people in US who are obes is between 0.644 and 0.672

And the best option is given by:

(0.644, 0.672)