chapter 1

Chapter 10: The Analysis of Variance7.Source Df SS MS FTreatments 3 75,081.72 25,027.24 1.70Error 16 235,419.04 14,713.69Total 19 310,500.76H : µ = µ = µ = µ0 vs. H : aat least two ' is.70 < F 2.46 , so p-value > .10, and we fail to reject H o.The hypotheses are1 2 3 41. 10,3, 16=µ are unequal.x1 •= 2332. , x2 •= 2576. 4 ,3 •= 2625. 9x4 •= 2851.5 , x5 •= 3060. 2 ,••=13,446. 543,992.55MSTr = = 10,998.4,31,475.0310,998.14MSE = = 1049.17 and f = = 10. 48301049.17.48 ≥ F 4. ,8. The summary quantities are 5x ,x , so CF = 5,165,953.21, SST = 75,467.58,SSTr = 43,992.55, SSE = 31,475.03, 1410. 01,4, 30=displayed in an ANOVA table as requested.) Since 02: µ = µ = µ = µ = µ0 1 2 3 4 5. (These values should beH is rejected. There are differences in the true average axialstiffness for the different plate lengths.9. The summary quantities are 3x1 •= 34. , x2 •= 39. 6 , x3 •= 33. 0 ,4 •= 41. 9( 148.8)x••=148.8 , ΣΣx2 ij= 946. 68 , so CF =24= 922.56 ,SST = 946 .68 −922.56= 24.12 ,2( 34.3) + ... + ( 41.9)22x ,SSTr =− 922.56 = 8.98 , SSE = 24 .12 −8.98= 15. 14 .6Source Df SS MS FTreatments 3 8.98 2.99 3.95Error 20 15.14 .757Total 23 24.123.10= F.< 3.95 < 4. 94 = F , . 01 < − value < . 05Since05,3,20.01,3, 20rejected at level .05.p and H o is297