In: Statistics and Probability
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.
Solution :
Given that,
n = 300
x = 81
Point estimate = sample proportion = = x / n = 81/300=0.27
1 - = 1-0.27 =0.73
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 *((0.27*0.73) / 300)
E = 0.042
A 90% confidence interval for proportion p is ,
- E < p < + E
0.27-0.042 < p < 0.27+0.042
0.228< p < 0.312