[1] "Aufgabe 5" 1 2 3 4 1.04 0.84 1.03 0.91 Lilliefors (Kolmogorov-Smirnov) normality test data: DS$konzent D = 0.1072, p-value = 0.02379 Analysis of Variance Table Response: konzent Df Sum Sq Mean Sq F value Pr(>F) bed.f 3 2003.5 667.82 36.406 1.083e-14 *** Residuals 76 1394.1 18.34 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Call: lm(formula = konzent ~ bed.f, data = DS) Residuals: Min 1Q Median 3Q Max -8.250 -3.250 0.125 2.450 9.700 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 23.2500 0.9577 24.277 < 2e-16 *** bed.f2 -7.7000 1.3544 -5.685 2.30e-07 *** bed.f3 -9.0500 1.3544 -6.682 3.47e-09 *** bed.f4 -13.9500 1.3544 -10.300 4.50e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 4.283 on 76 degrees of freedom Multiple R-squared: 0.5897, Adjusted R-squared: 0.5735 F-statistic: 36.41 on 3 and 76 DF, p-value: 1.083e-14 Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = konzent ~ bed.f, data = DS) $bed.f diff lwr upr p adj 2-1 -7.70 -11.257678 -4.142322 0.0000014 3-1 -9.05 -12.607678 -5.492322 0.0000000 4-1 -13.95 -17.507678 -10.392322 0.0000000 3-2 -1.35 -4.907678 2.207678 0.7516904 4-2 -6.25 -9.807678 -2.692322 0.0000912 4-3 -4.90 -8.457678 -1.342322 0.0029348 Pairwise comparisons using t tests with pooled SD data: DS$konzent and DS$bed.f 1 2 3 2 2.3e-07 - - 3 3.5e-09 0.32204 - 4 4.5e-16 1.6e-05 0.00053 P value adjustment method: none Pairwise comparisons using t tests with pooled SD data: DS$konzent and DS$bed.f 1 2 3 2 9.2e-07 - - 3 1.7e-08 0.3220 - 4 2.7e-15 4.7e-05 0.0011 P value adjustment method: holm