Question

Obesity is defined as a body mass index (BMI)of30kg/m 2 or more. A 95% confidence interval for the percentage of U.S. adults aged 20 years and over who were obese was found to be 22.4 to 23.5%. What was the sample size?

The answer should be 318, but I don't know how to work it out.

Answer 1

The formula for CI is,

So, lower limit = 0.224 = $p-z*\sqrt{\frac{p(1-p)}{n}}$

and upper limit = 0.235 = $p+z*\sqrt{\frac{p(1-p)}{n}}$

Now, z score for 95% confidence level is, 1.96

and lower limit + upper limit = 2p = 0.224+0.235 = 0.459

=> p = 0.2295

Putting p and z in lower limit we get,

0.224 = $0.2295-1.96*\sqrt{\frac{0.2295(1-0.2295)}{n}}$

=> (0.224-0.2295)/1.96 = $-\sqrt{\frac{0.2295(1-0.2295)}{n}}$

=> 0.0028 = $\sqrt{\frac{0.1768}{n}}$

=> n= 0.1768/0.0028²=