Question
Using the built in data in R “ToothGrowth”. Why factor(does) resulted in a different p-value using anova function.
headerH TRUE) 33 Corn <- factor(datal, 1]) 34 Yield <- datal, 21 35 Corn 36 table (Corn) 37 Yield 38 tapply(Yield, list (Corn) mean) group means 39 boxplot(Yield~datal,1]) 40 41 InsectSprays 42 table(InsectSprays) 43 Jungkook InsectSprays, 1] 44 Jungkook 45-Jin= InsectSprays [, 2] 46 Jin 47 boxplot (Jungkook-Jin) 48 49 pairwise.t.test(vield, Corn, pool.sd FALSE, p. adjust.method none 50 Insectsprays 51 does 52 ToothGrowth 53 factor (supp) 54 50:13 Clop Level Console Terminal x C/Users/Edward/Desktop/statistics lab/ > anova(1mlen factor (supp) + factor (dose), ToothGrowth)) Analysis of Variance Table Response: len Df Sum Sq Mean Sq F value Pr OF) factor(supp) 1 205.35 205.35 14.017 0.0004293 factor (dose) 2 2426.43 1213.22 82.811 < 2.2e-16 Residuals 56 820.43 14.65 signif. codes: 00.0010.010.050.1 > anova (1mClen factor (supp) + dose, ToothGrowth)) Analysis of Variance Table 1 ToothGrowth) Analysis of var iance Table? Response: len Df Sum Sq Mean q F value Pr(>F) 123.989 6. 314e-16 О.01.*. О.05 factor (supp) 1 205.35 205.35 11.447 0.001301 dose Residuals 57 1022.56 17.94 1 2224.30 2224.30 0.1/11 signif. codes : o does 0 . , 0.001 [11 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 0.5 1.0 1.0 1.0 1.0 1.0 1.0 1.0 1. [19] 1.0 1.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 2.0 0.5 0.5 0.5 0.5 0.5 0. [13 9.23
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