Question

Of 116 randomly selected adults, 34 were found to have high blood pressure. Find a 98% comfidence interval for the percentage of adults that have high blood pressure. O A. 25.3% to 33.3 % O B. 19.5 % to 39.1 % O C. 22.3% to 36.2 % O D. 21.0 % to 37.6 %

Answer 1

Solution :

Given that,

n = 116

x = 34

Point estimate = sample proportion = $\hat p$ = x / n = 34 / 116 = 0.293

1 - $\hat p$ = 1 - 0.293 = 0.707

At 98% confidence level the z is,

$\alpha$ = 1 - 98%

$\alpha$ = 1 - 0.98 = 0.02

$\alpha$/2 = 0.01

Z_$\alpha$/2 = Z 0.01 = 2.326

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

= 2.326 * ($\sqrt$((0.293* 0.707) / 116)

= 0.098

A 95% confidence interval for population proportion p is ,

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

0.293 - 0.098 < p < 0.293 + 0.098

0.195 < p < 0.391

19.5% < p < 39.1%

correct option is = B.