Question

Of 116 randomly selected adults, 34 were found to have high blood pressure. Find a 98% confidence interval for the percentage of adults that have high blood pressure.

A. 19.5% to 39.1%                                         

B. 21.0% to 37.6%

C. 22.3% to 36.2%                                           

D. 25.3% t0 33.3%

Answer 1

Solution :

Given that,

n = 116

x = 34

Point estimate = sample proportion = = x / n = 34 / 116 = 0.293

1 - = 1 - 0.293 = 0.707

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z _0.01 = 2.326

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 2.326 * (((0.707 * 0.293 ) / 116 )

= 0.098

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.293 - 0.98 < p < 0.293 + 0.098

0.195 < p < 0.391

19.5% , 39.1 %  

Option (A) is correct = 19.5% to 39.1%