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A study of 4,430 U.S. adults found that 2,913 were obese or overweight. What is the 95% confidence interval for the proportion of U.S. adults who are obese or overweight?

Answer :

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).

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