Answer:
A) 2.02 standard deviations from 600 - a win
B) The last interval was a close call.
Step-by-step explanation:
n = 2300
p = 0.47, q = 1-p = 0.53
μ = np = 1081
σ = √(npq) = 23.94
Z.025 = 1.96
B) Finding both confidence interval:
Confidence Interval = (μ-1.96(23.94)/√2300, μ+1.96(23.94)/√2300)
= (1081 - 0.978, 1081 + 0.978)
= (1080.022, 1081.978)
(1080-600)/23.94 = 2.02 standard deviations from 600 - a win
B)
AS above
n=450, p=.54, q=.46
μ=np=243, σ=√(npq)=10.6
confidence interval (243-1.96(10.6)/√450 , 243+1.96(10.6)/√450)
= 243-.98, 243+.98) = (242.0, 243.98)
(244-225)/10.6=1.8 standard deviations from win