SOLUTION: construct a 99% confidence interval estimate for a population of 50, a standard deviation of .02 and a sample mean of .995 Then a 95% interval and how it changes from the 99% in

Question 1178895: construct a 99% confidence interval estimate for a population of 50, a standard deviation of .02 and a sample mean of .995
Then a 95% interval and how it changes from the 99% interval
Found 2 solutions by Boreal, ewatrrr:
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
99% half-interval is z(0.995)*sigma/sqrt(n)=2.576*0.02/sqrt(50)=0.0073
mean +/- half-interval is (0.9877, 1.0223)
95% interval is going to be narrower but with less confidence.
half-interval is 1.96*0.02/sqrt(50)=0.0055
so interval is (0.9895, 1.0005)
Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

Hi ME =, = .0028 n > 40 and population σ = .02 µ =.995 99% confidence interval estimate ME = = .0072 .995- .007 < < .995 + .007 CI: (.988, 1.002) 95% interval: ME = = .0055 .995 - .0055< < .995 + .0055 CI: (.9895, 1.0055) CI is wider at 99% Wish You the Best in your Studies.

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