Question

Given the following outputs of anova analysis: summaryCaovCanxiety$'anxiety' anxiety$'stress' Df Sum Sq Mean Sq F value Pr(F) anxietySstress 3 182.1 60.70 11.94 5.56e-05 Residuals 24 122.0 5.08 Signif. codes: 0 0.001*0.01 *'0.05 '.' 01''1 > TukeyHSD aov(anxiety$'anxiety'~ anxiety$'stress' Tukey multiple comparisons of means 95% family-wise confidence level Fit: aov(formula = anxiety$anxiety ~ anxiety$stress) S' anxiety$stress . diff lwr upr P adj 2-1 4.4285714 1.1040404 7.753102 0.0061238 3-1 3.7142857 0.3897547 7.038817 0.0245552 4-1 7.1428571 3.8183261 10.467388 0.0000230 3-2-0.7142857-4.0388167 2.610245 0.9333083 4-2 2.7142857-0.6102453 6.038817 0.1379750 4-3 3.42857140.1040404 6.753102 0.0415245 6y 11 69 in

Q: indicate probability that all four groups of experiments have the same mean and probability that group 2 and 3 have the same mean.

Answer 1

The probability that all four groups of experiments have the same mean is 5.56e-05 or 0.0000556 and the probability that group 2 and 3 have the same mean is 0.9333083.