chapter 1

Chapter 10: The Analysis of Varianceb. With xi•= 8.27, 7.50, 6.35, and 5.82 for I = 1, 2, 3, 4, we calculate the residualsxx for all observations. A normal probability plot appears below, and indicatesij− i•that the distribution of residuals could be normal, so the normality assumption isplausible.Normal Probability Plot for ANOVA Residuals21resids0-1-2-2-10prob12c. Q 3. 96 and. 05,4,20=Wij2.09 ⎛ ⎞⎜1 1= 3 .96⋅⎟+ , so the Modified Tukey2⎝JiJj ⎠intervals are:Pair Interval Pair Interval1,2 77 2. 37. ± 2,3 1 .15±2. 451,3 .92 2. 251 ± 2,4 1 .68±2. 451,4 2 .45 ± 2. 25 * 3,4 . 53 ± 2. 344 3 2 1Only Brands 1 and 4 are different from each other.28. SSTr Σ Σ( X − X )2222{i•••} = Σ Ji( Xi•− X••) = ΣJiXi•− 2X••Σ JiXi•+ X•ΣJi=•i jiii2222 22 2iXi• − 2 X••X••+ nX••= ΣJiXi•− 2nX••+ nX••= ΣJiXi•− n••.ii= Σ JXii305