Question

If n = 200 and (p-hat) = 0.45, construct a 90% confidence interval.

Give your answers to three decimals

__ < p < __

Answer 1

0.392 < p < 0.508

Explanation:

`\displaystyle CI=\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\\\CI_(90\%)=0.45\pm 1.645\sqrt{(0.45(1-0.45))/(200)}\\\\CI_(90\%)\approx0.45\pm0.058\\\\CI_(90\%)=\{0.392,0.508\}`

Thus, we are 90% confident that the true population proportion
`\hat{p}` is between 0.392 and 0.508, assuming conditions are met.