MidtermI_2008_PracticeExamIsol-1

MidtermI_2008_PracticeExamIsol-1 - MidtermI_PracticeExamI...

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MidtermI_PracticeExamI Solutions 1. The appropriate hypotheses are REG IMP a REG IMP H H : : 0 (1st – Assuming equal variances): The test statistic will come from a t-distribution with 123 degrees of freedom. Approximating this distribution using a 120 t distribution, our decision rule is Reject 0 H if 98 . 1 120 , 025 . 0 t t , and do not reject otherwise The pooled estimate of standard deviation for the two populations is 1536 . 8 4815 . 66 2 75 50 ) 232 . 9 )( 74 ( ) 178 . 6 )( 49 ( 2 2 p s The test statistic is then 3714 . 2 4886 . 1 53 . 3 75 1 50 1 ) 1536 . 8 ( 99 . 31 46 . 28 t This value exceeds the critical value from the decision rule. Hence, we reject 0 H and conclude that the “improved” design does appear to change the level of hydrocarbon vapors emitted. (2nd – Assuming unequal variances): The test statistic assuming unequal variances is then 5611 . 2 3783 . 1 53 . 3 75 ) 232 . 9 ( 50 ) 178 . 6 ( 99 . 31 46 . 28 2 2 t We will use a t-test with DF=100. the t -table does not contain an entry for DF = 100, so we can approximate using either 60 t or 120 t (in this case, we play it conservatively and use 60 t ). Our decision rule is thus Reject 0 H if 2 60 , 025 . 0 t t , and do not reject otherwise. Our conclusion is the same as with the equal variance assumption: we reject 0 H ; hence, there is significant evidence to suggest that the “improved” design changes the level of hydrocarbon vapors emitted.
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2 (a) Summary of Fit (for the Regular device) RSquare _0.776_ Root Mean Square Error 4.40 Mean of Response 31.99 Observations 75 Analysis of Variance
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MidtermI_2008_PracticeExamIsol-1 - MidtermI_PracticeExamI...

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