chapter 1

Chapter 10: The Analysis of Variance10.a. E( X )••=ΣE( Xi•)I=Σµi = µI.222 σ 2b. E ( X ) Var ( X ) + [ E( X )] = + µi• =i•i•i.222 σ 2c. ( X ) Var( X ) + [ E( X )] = + µE .E••SSTr=••••d. ( ) [ ]⎟ i • •• ⎜ 2 ⎟⎝ + ⎠⎜ 2J µi ⎝ IJ + µ ⎠JIJ2 2⎛= E JΣX− IJX = J∑⎜σ2⎞ ⎛⎟ − IJ⎜( I −1 ) σ2 + JΣ( µ − µ ) 22 2 2 2= I σ + JΣµi− σ − IJµ=iE( MSTr )Eσ2, so( SSTr ) 2 2 2[ ]( µiµ )= E JΣXi•− IJX•= + J∑=•I −1e. When H o is true, µ ... = µ = µ1=i, so Σ ( − µ ) 2 = 02>2When H o is false, Σ ( µ i− µ ) 0 , so ( MSTr ) > σoverestimates2σ ).σ .⎞−I − 12µ iand ( MSTr ) = σ2E .E (on average, MSTrSection 10.2272.8Q. 05,5, 15= , = 4 .37 = 36. 09411. 4. 37w .3 1 4 2 5437.5 462.0 469.3 512.8 532.1The brands seem to divide into two groups: 1, 3, and 4; and 2 and 5; with no significantdifferences within each group but all between group differences are significant.298