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A sample of new Widener Engineering graduates reveals an average starting salary of 74,000 with a...

A sample of new Widener Engineering graduates reveals an average starting salary of 74,000 with a standard deviation of 9,000, an n=25, compute an interval estimate for their salary at a .04 level of confidence?

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Answer #1

96% Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.04 /2, 25- 1 ) = 2.172
74000 ± 2.172 * 12/√(25)
Lower Limit = 74000 - 2.172 * 12/√(25)
Lower Limit = 73995
Upper Limit = 74000 + 2.172 * 12/√(25)
Upper Limit = 74005
96% Confidence interval is ( 73995 , 74005)

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