## 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

Question

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

## Answers ( )

Answer: (0.644,0.672)Step-by-step explanation:Let p be the proportion of the adults in U.S. who are obese or overweight.

Confidence interval for p will be :

, where is sample proportion , n is sample size , and z is the critical z value as per confidence level.

As per given, we have

n= 4430

Critical z-value for 95% confidence level is 1.96.

Then, the 95% confidence interval for the proportion of adults in U.S. who are obese or overweight would be:

Hence, the required 95% confidence interval would be (0.644,0.672).