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Lecture

RCBD II.pdf

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Department
Statistics
Course
STAT368
Professor
Douglas Wiens
Semester
Winter

Description
74 10. RCBD II The theoretical ANOVA table for a RCBD is Source SS df MS F0 Treat. 1 = 1 0 = Blocks 1 = 1 Error ( 1)· = ( 1) Total 1 The expected mean squares can be derived as in as- signment 1, and are P 2 =1 2 [ ] = + P 1 2 =1 2 [ ] = + 1 [ ] = 2 Notice a pattern? 75 For the hardness data the R output is: > g anova(g) Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) treatments 3 0.38500 0.12833 14.438 0.0008713 *** blocks 3 0.82500 0.27500 30.938 4.523e-05 *** Residuals 9 0.08000 0.00889 Thus at any level 00087, we would reject the null hypothesis of no treatment e ects ( 0: 1= · · · = = 0). It also appears that the blocks have a sig- nicant e ect. A caution here though - the random- ization alone ensures that the F-test for treatments is approximately valid even if the errors are not very nor- mal. Because of the randomization restriction, the same is not true for testing the signicance of blocks by looking at . Thus the -value of 4 523 05 for blocks should be used only as a guide, unless one is sure of the normality. 76 Suppose that we erroneously analyzed these data as a CRD? What would the ANOVA be? The must still be accounted for, and if it can’t be anywhere else it ends up in experimental error (since = for a CRD): Analysis of Variance Table Response: y Df Sum Sq Mean Sq F value Pr(>F) treatments 3 0.38500 0.12833 1.7017 0.2196 Residuals 9 0.08000 0.07542 +3 +.82500 = 12 = 0.90500 ( = 90500 12; 0 = 12833 07542 ) 77 To verify our guess about the LSEs, we show that they minimize n h io X X 2 ( ) = =1 =1 X X n o = 2 =1 =1 P P ˆ subject toˆ = = 0. Write this (s ) = ³ ´ X X ¯ ¯ + ¯ 2 ³ ´ = =1 =1 ( ˆ) ( ˆ ) ˆ 2 X 2 X ³ ´2 + ( ˆ) + ( ˆ ) + ˆ =1 =1 The constraints are satised, and are used to show that the cross-
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