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# solution2 - Solution to Homework#2-Stat c151/c225 Winter...

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1 Solution to Homework #2 ----Stat c151/c225 Winter 2011 3.7 (a) All entries can be determined as below. Source Degrees of Freedom Sum of Squares Mean Squares Block 2 520 260 Treatment 4 498 124.5 Residual 8 40 5 total 14 (b) The t statistics for Tukey method are calculated below (using (2.11) with ˆ σ = 5 , b = 3 ) Group A(45) B(58) C(46) D(45) E(56) A(45) 7.12 0.55 0 6.02 B(58) -6.57 -7.12 -1.10 C(46) -0.55 5.48 D(45) 6.02 E(56) 68 . 4 414 . 1 62 . 6 2 1 2 1 01 . 0 , 8 , 5 ), 1 )( 1 ( , = = = q q k b k α . By comparing the t values with 4.68, Tukey method declares that the following pairs are significantly different: A&B, A&E, B&C, B&D, C&E and D&E. (c) From (b), Tukey method declares 6 pairs of treatments to be significantly different at level 0.01. The null hypothesis of the F test is that all the treatments are same to each other. Based on the conclusion of (b), we know the null hypothesis of F test is not true at level 0.01. Hence the F test will reject the null hypothesis at level 0.01. 3.8 (a) This is a paired comparison design because a pair of catalysts is compared in each of the six batches. Each batch can be considered as a block of size 2. In this design, the two catalysts are compared on two portions of the same batch, this eliminates a source of variation due to the difference among different batches, which were known to be quite different from one another. (b) Use paired t-test, 1 2 3 4 5 6 A 9 19 28 22 18 8 B 10 22 30 21 23 12 d i 1 3 2 -1 5 4 33 . 2 6 14 1 1 = = = = N i i d N d , = = = 2 / 1 1 2 ) ( 1 1 N i i d d d N s 2.16.

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