Question

11. 200 people are polled and 100 of them say they support labeling foods that contain GMO’s. Build the 99% confidence interval for the true proportion of people that support labeling foods that contain GMO’s.

Answer 1

Solution :

Given that,

n = 200

x = 100

Point estimate = sample proportion = = x / n = 100/200=0.5

1 -   =0.5

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576 ( Using z table )

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

= 2.576* (((0.5*0.5) /200 )

E = 0.091

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.5-0.091 < p < 0.5+0.091

0.409< p < 0.591