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

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 are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672

Explanation:

We can begin founding the estimated proportion of people overweight:

`\hat p =(2913)/(4430)= 0.658`

We need to find a critical value for the confidence level using the normal standard distribution. We know that 95% is the confidence level, then 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 true population proportion of interest is given by the following formula:

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

Replacing the data provided 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 are 95% confident that the true proportion of adults overweight is between 0.644 and 0.672