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

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

Step-by-step explanation:

We can begin founding the estimated proportion of people overweight:

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

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:  

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

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