Question

Use the accompanying partially completed two-way ANOVA summary table to complete parts a through e below. Click the icon to view the table a) Complete the two-way ANOVA table below. Sum of Degrees of Mean Sum of Source Squares Freedom Squares Factor A 140 Factor B Interaction 20 Error 360 Total 600 44 (Type integers or decimals.)

Answer 1

Sum of Sanare l Degre of freedom Mean Sama square Source 70 luo rache A factor B и0 20 Interaction 360 Eros Total I 6oo I 44 MB - SSE SSA= TSS - SS SSAB 600 = 600 - 140 - 20 - 360 = 80 AL db - art = 2 12 = 3) - b-1 -2 16 = 3] OAB - (0-1) (6-1) = (3-1) (3-1) = 4. If an is the no. of explicates to 4 We know dhe = abl 8-1) = 36 3x3 (-) = 36 S 8.4 = 4 543= MSA = 140=70, MS2 = 80 = 40 MSAB= 20 = 5, MSE - 360 = 10 MSA 70-7, torsport hos S O -Y FAR MSAB = 3 = 3 36 MSE E MSAB M SE

b) eas ( Solved in part (0) B BD D) : Hi Facton factor A and B A ad B do interact, do interact P-valu = P(f> FAB) A FAB~ F4,36 P (F35) 0.7358 ~ 0.73 = &= 0.01 s P-value = 0.8358 So A) & Do not reject the nell hypothesls. There is insufficient evidence to conclude that factor A and B interact. d) According to ANOVA table so A part is correct but Factor A hos ? levels it should be like tres As Hot MA, MA2 MA3 the Not all factor A means are equal We don't have 4 levels so it may be other MA is ad typo

P.value Ploof) P(F= 7) = FAM F2,36 0.0027 ~ 0.003 a = bol > P-value = 0.003 A.) Reject the null hypothesis. There is sufficient evidence to conclude that not all factor A means are equal. 2 a Hii means Not all Factor B au equal. Prvalue Ple>FB) & FAN F2,36 P(6747= 0.027 d=0.01 < P-value - 0.027 c) Do not regut the null heppothesis. There is insuffin evidan to conclude that not all factod B means are equal. ddddd