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 :
[tex]\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p)}}{n}}[/tex], where [tex]\hat{p}[/tex] 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
[tex]\hat{p}=\frac{2913}{4430}\approx0.658[/tex]
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:
[tex]0.658\pm (1.96)\sqrt{\frac{0.658(1-0.658)}{4430}}\\\\=0.658\pm1.96\times0.0071272851865\\\\=0.658\pm0.0139694789655\\\\\approx(0.658-0.014,0.658+0.014)\\\\=(0.644,0.672)[/tex]
Hence, the required 95% confidence interval would be (0.644,0.672).
