Question

A random survey of 495 adults found that 59 had dietes. Which of the following is a 98% confidence interval for the population proportion of adults with diabetes?

Answer 1

{0.0853,0.1531}

Explanation:

`\displaystyle CI=\hat{p}\pm z\sqrt{\frac{\hat{p}\bigr(1-\hat{p}\bigr)}{n}}`

`\displaystyle CI_(98\%)=(59)/(495)\pm2.326\sqrt{((59)/(495)\bigr(1-(59)/(495)\bigr))/(495)}\approx\{0.0853,0.1531\}`