9 01 in each tail 2 005 3 df22 t 2 1717

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Unformatted text preview: B ) − (µ A − µ B ) T= A 1 1 Sp + n A nB Sp = 2 2 ( nA − 1) S A + ( nB − 1) SB (12 − 1) 33792 + (12 − 1) 59802 = nA + nB − 2 12 + 12 − 2 2) α = 1 − 0.9 = 0.1 , in each tail α 2 = 0.05 3) d.f.=22, tα 2 = 1.717 (by using tinv() in Excel or Table A.4) = 4857 1 1 + nA nB µ A − µ B = ( X A − X A ) ± tα 2 S p = ( 35200 − 35442 ) ± 1.717 × 4857 × 11 + 12 12 = −242 ± 3404 The 90% confidence interval for the difference between the two population means is −242 ± 3405 or ( ­3647, 3163). b) Find a 90% confidence interval for the difference between the two population means, assuming the population variances are not equal. The population variance is unknown, but unequal between two populations, T distribution is applied and d.f. =nA+nB ­2=12+12 ­2=22. 2/19 CEE 202 – Engineering Risk & Uncertainty Spring 2011 (Bond) Homework 11 Diff. between Means; Least ­Squares 1) T distribution (X − X B ) − (µ A − µ B ) T= A 2 2 S A SB + n A nB 4) α = 1 − 0.9 = 0.1 , in each tail α 2 = 0.05 5) d.f.=22, tα 2 = 1.717 (by using tinv() in Excel or Table A.4) µ A − µB = ( X A − X B ) ± tα / 2 2 2 S A SB + n A nB 2 = (35200 − 35442) ± 1.717 × 3379 5980 + 12 12 2 = −242 ± 3405 The 90% confidence interval for the difference between the two population means is −242 ± 3405 or ( ­3647, 3163). Comments: 1) This problem was given to show that the two assumptions (equal variance and unequal variance) come out the same or very close. 2) 3 points are off if the degree of freedom is not right. 2) Test data on the deformation and hardness of a certain type of steel have are given in the table below: Deformation H (hardness, (mm) kg mm2) 6 68 11 65 13 53 22 10 28 37 35 32 a) Graph the data on a two ­dimensional plot. Comment on the results....
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