Question

1-Prop Conf. Int. 07 or 94 adults selected randomly from one town, 67 have health insurance. Find a 90% confidence interval for the true proportion of all adults in the town who have health insurance. Use Interval Notation with decimal rounded to the thousandths, Answer

Answer 1

Solution :

Given that,

Point estimate = sample proportion = $\hat p$ = x / n = 67 / 94 = 0.713

Z_$\alpha$/2 = 1.645

Margin of error = E = Z_{$\alpha$ / 2} * $\sqrt$ (($\hat p$ * (1 - $\hat p$ )) / n)

= 1.645 * ($\sqrt$((0.713 * 0.287) / 94)

= 0.077

A 90% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.713 - 0.077 < p < 0.713 + 0.077

0.636 < p < 0.790

(0.636 , 0.790)