Question

120 people are polled and 90 of them say they support labeling foods that contain GMO’s. Build the 90% confidence interval for the true proportion of people that support labeling foods that contain GMO’s.

Answer 1

Solution :

Given that,

n = 120

x = 90

Point estimate = sample proportion = = x / n = 90/120=0.75

1 -   =0.25

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.75*0.25) /120 )

E = 0.065

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.75-0.065 < p < 0.75+0.065

0.685< p < 0.815