Question

11. 140 people are polled and 60 of them say they support labeling foods that contain GMO’s. Build the 95% confidence interval for the true proportion of people that support labeling foods that contain GMO’s.

Answer 1

Solution :

Given that,

n = 140

x = 60

Point estimate = sample proportion = = x / n = 60/140=0.429

1 -   = 1- 0.429 =0.571

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

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

= 1.96 (((0.429*0.571) / 140)

E = 0.082

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.429-0.082 < p <0.429+0.082

(0.347 , 0.511)