In: Statistics and Probability
Please include steps 7. In a survey of n=550 randomly selected individuals, X=260 felt that Ronald Reagan spent too much time away from Washington. Find a 92% confidence interval for the proportion
1)Level of Significance, α = 0.08
2)
Number of Items of Interest, x =
260
Sample Size, n = 550
Sample Proportion , p̂ = x/n =
0.4727
3)z -value = Zα/2 = 1.960
[excel formula =NORMSINV(α/2)]
4) Standard Error , SE = √[p̂(1-p̂)/n] =
0.021288
5) margin of error , E = Z*SE = 1.960
* 0.02129 = 0.0373
92% Confidence Interval is
6) Interval Lower Limit = p̂ - E = 0.47273
- 0.03727 = 0.435
7) Interval Upper Limit = p̂ + E = 0.47273
+ 0.03727 = 0.510
92% confidence interval is (
0.4355 < p < 0.5100
)