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

`0.658 - 1.96\sqrt{(0.658(1-0.658))/(4430)}=0.644`

`0.658 + 1.96\sqrt{(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)

Explanation:

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

`\hat p =(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{(\hat p (1-\hat p))/(n)}`

Replacing the info given we got:

`0.658 - 1.96\sqrt{(0.658(1-0.658))/(4430)}=0.644`

`0.658 + 1.96\sqrt{(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)