Question
A Sample of 18 widgets has a mean of 47. And standard deviation of 99.550, At 99% confidence, the upper limit with 3 decimal places
A Sample of 18 widgets has a mean of 47. And standard deviation of 99.550, At 99% confidence, the upper limit with 3 decimal places
the solution to that is:
upper limit =47 +2.898*99.550/sqrt(18) =114.999
How did the 1.833 came about when I check the Zscore of a 99% confidence it would be (1-.99) = .01/2 = .005 Zscore of ..005 would be 3.355
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Financial Algebra advanced algebra with financial applications
Authors: Robert K. Gerver
1st edition
978-1285444857, 128544485X, 978-0357229101, 035722910X, 978-0538449670
