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QUESTION 2 A research engineer for a tire manufacturer is investigating tire life for a new her compound and has built 36 tim
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Answer #1

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.02 /2, 36- 1 ) = 2.438
62.1 ± t(0.02/2, 36 -1) * 2.5/√(36)
Lower Limit = 62.1 - t(0.02/2, 36 -1) 2.5/√(36)
Lower Limit = 61.084
Upper Limit = 62.1 + t(0.02/2, 36 -1) 2.5/√(36)
Upper Limit = 63.116
98% Confidence interval is ( 61.084 , 63.116 )

( 61.084 ; 63.116 )

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