Question

In a statewide opinion poll of a random sample of 300 people, 81 said that they were strongly opposed to having a state lottery. Assuming that the 300 people in the poll are a proper simple random sample, find the 90% confidence interval for the proportion of the state’s population strongly opposed to the lottery.

Answer 1

Solution :

Given that,

n = 900

x = 81

Point estimate = sample proportion = = x / n = 81/900=0.09

1 -   = 1- 0.09 =0.91

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z  0.05 = 1.645 ( Using z table )

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

= 1.645 *((0.09*0.91) /900 )

E = 0.016

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.09-0.016 < p < 0.09+0.016

0.074< p < 0.106