In: Statistics and Probability
A committee initiated raising fund for clinic building has presented its proposal to city dwellers. 345 of the 365 town dwellers in attendance at the meeting were in favor of the proposal. Assume those in attendance is a random sample of the town people:
Given the confidence interval of 95% is used for the population proportion of town dwellers that are in favor of the proposal ( 0.92, 0.97). If there has been 1095 town dwellers present in the meeting and 1035 of those 1095 has been in favor of the proposal, what would happen to this interval? Show your work in the space provided.
Level of Significance, α =
0.05
Number of Items of Interest, x =
1035
Sample Size, n = 1095
Sample Proportion , p̂ = x/n =
0.945
z -value = Zα/2 = 1.960 [excel formula
=NORMSINV(α/2)]
Standard Error , SE = √[p̂(1-p̂)/n] =
0.0069
margin of error , E = Z*SE = 1.960
* 0.0069 = 0.0135
95% Confidence Interval is
Interval Lower Limit = p̂ - E = 0.945
- 0.0135 = 0.9317
Interval Upper Limit = p̂ + E = 0.945
+ 0.0135 = 0.9587
95% confidence interval is (
0.932 < p < 0.959 )
interval get narrower