Question

aov (hw labels) >summary (fit) > fit Df Sum Sq Mean Sq F value Pr(>F) 2 493.3 labels 246.7 5.763 0.0176 Residuals 12 513.6 42.8 Signif. codes: 0 0.001 0.01 0.05 0.1 1 12. For this problem, you will use R to conduct an ANOVA F-test. The R code from #11 should be useful for this problem. In #11, the dean only collected samples of size five from each engineering majors. Suppose he collects larger samples (n, 20, i 1,2,3) and the data is given in the table below. EE 33 22 22 19 17 24 22 16 20 20 16 28 19 24 18 29 18 16 29 24 ME 26 25 23 10 22 15 14 18 18 17 24 19 18 14 22 11 26 21 10 22 CE 10 11 14 14 11 20 17 17 10 15 15 11 4 20 17 15 20 13 10 20 (a) Generate an ANOVA table in R and attach this to your homework. (b) Complete a hypothesis test at a = 0.01 to test the claim that all three engineering majors have the same mean homework load. (c) The dean is interested in which majors have a difference in mean homework load. Com- plete determine which, if any, of the pairwise differences are statistically significant. TukeyHSD multiple comparisons adjustment in R (attach to your homework) and a plot(TukeyHSD(fit)) plot(TukeyHSD(fit) once you have completed the Hint: In R, ANOVA calculations. use or
please provide the output in R

Answer 1

R RGui (32-bit) Help Misc Packages Windows File Edit View Ep STOF R Console > hwload-data.frame (marks-c (33,22,22,19,17,24, 22,16,20, 20,16,28,19,24,18,29,18,16, 29, 24, 26, 25, 23, 10,22,15, 14,18,18,17,24,19,18,14,22,11,26, 21,10,22, + 10,11, 14, 14,11,20,17,17,10,15,15,11,4,20,17,15,20,13, 10,20), tsubject factor (c (rep ("EE", 20), rep ("ME", 20),rep ("CE", 20))) ) >ajuste 1m (hwload$marks hwload$ subject) > anova (ajuste) Analysis of Variance Table Response: hwload$marks Pr (>F) Df Sum Sq Mean Sq 585.1 292.550 value hwload$subject 12.728 2.692e-05 57 1310.2 22.985 Residuals 0.05 . 0.1 0.001 0.01 Signif. codes:

Connder yi denting work Caad th udent tram en nmaigr suscet tue 'k ce vespeftely let ialr 23 Lorespond to EE HE tee the radel s 1al /43 2124-- 20 Mit eli gur We to tert ane N S tAll i's aLe Rot eqnal at tewel of gfaee 2 001 We do ANOVA fimd lunee hat ence n nbet's stndent han rking fut vaing-statitle 2 692y105 p-malue (0.077 ull ne Aamu at y. el ft stgnaftenee mewn bad Now lN Vjectng to apphy Tkeu t ten t chiek whick ubiects haemuakedly difuent orlkins Load m the ploting iuleaerlto ne ane Andhydes the LD a Mull an ME-EE ne can instgniticant ME-CE das net cumtain hense gainC of bbut mut o far fnom 'o. So, it's

1gnificanee s modenntz. Pnt when Come to the et o ECE, 1ddoes ni m onen Contam mueh monefar homunsk LoadEE ditenence is CE engineering students se EE - CE the difeaene Hneer stali eally stgnicant

>al <aov (hwload$marks posthoc <- TukeyHSD (x-al, hwload$subject) 'hwload$subject', conf.level-0.95) > posthoc Tukey multiple comparisons of means 95% family-wise confidence level hwload$marks hwload$subject) Fit: aov (formula = hwload$subjec diff lwr p adj Idn 7.60 EE-CE 3.9516695 11.2483305 0.0000164 ME-CE 4.55 8.1983305 0.0109551 0.9016695 ME-EE -3.05 -6.6983305 0.5983305 0.1185817 > plot (TukeyHSD (a1, 'hwload$subject'))

95% family-wise confidence level 5 0 5 10 Differences in mean levels of hwload$subject ME-EE ME-CE EE-CE -LO