chapter 1

Chapter 10: The Analysis of Variance2 22229. E( SSTr ) = E( ΣJ)iXi• − nX••= ΣJiE( Xi•) − nE( X••)i2= ΣJ[ ( ) ( ( ) ] [ ( ) ( ( ) ]2iVar Xi• + E Xi•− n Var X••+ E X••( ΣJµ )222⎡σ⎡⎤2⎤ σi i= ΣJi⎢ + µ i ⎥ − n⎢+ ⎥⎣ Ji ⎦ ⎢⎣n n ⎥⎦2= I − σ + ΣJ 2iµ + α − ΣJµ + α( 1) ( ) [ ( )] 2ii i222( 1) 2[ ] 222= I − σ + ΣJiµ+ µ ΣJiαi+ ΣJiαi− µ ΣJi( I −1) σ + ΣJ iαiwhich E(MSTr) is obtained through division by ( I −1).= , from30.a. = α 0α , α = 1 , α 11 2=3−4=and from figure (10.5), power ≈ . 90 .b. Φ 2 = .5J, so = .707 J9 looks to be sufficient., soΦ and = 4( J 1)2 2 2 2( 0 + 0 + ( −1)+ 1 )= 4,2 2Φ =Φ = 2,1ν . By inspection of figure (10.5), J =2−c. µ1= µ2= µ3= µ4, µ5= µ1+ 1, so µ = µ + 11 5, 1α1= α2= α3= α4= − ,5α ( ), 220225== 44 5Φ1= 1 .60 Φ =1. 26 , ν1= 4 , ν2= 45≈ . .of figure (10.6), power 55. By inspection31. With σ = 1 (any other σ would yield the same Φ ), α = 1, α =α 0 , 12 2 2 2( 5( − 1) + 5( 0) + 5( 0) + 5( 1))= 2. 51−2 3=α ,2 .25Φ =, Φ =1. 58 , ν1= 3 , ν2= 141power ≈ . 62 .4=, andy = x . This givesy1 •=15.43 , y2 •=17. 15 , y3 •=19. 12 , y4 •= 20. 01, y••= 71. 71,32. With Poisson data, the ANOVA should be done usingij ijΣΣy 2 ij= 263.79 , CF = 257.12, SST = 6.67, SSTr = 2.52, SSE = 4.15, MSTr = .84, MSE =.26, f = 3.23. Since F 5. 29 , H o cannot be rejected. The expected number of flaws. 01,3, 16=per reel does not seem to depend upon the brand of tape.306