SOLUTION: Find the margin of error and confidence interval. 95% confidence, n=100, xbar=9500, standard deviation = 12,345

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Question 1001734: Find the margin of error and confidence interval.
95% confidence, n=100, xbar=9500, standard deviation = 12,345
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
The SE is s/sqrt(n)=1234.5
n=100 is large enough to use the z-distribution, and z(0.95)-1.96
1.96*SE=2419.62, That is the half-width of the CI and the margin of error when said +/-.
(9500-2419.6, 9500+2419.6)
(7080.4, 11919.6)