3. Two-factor analysis of variance Emphasis on calculations w. Thomas Boyce, a p

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3. Two-factor analysis of variance Emphasis on calculations w. Thomas Boyce, a professor and pediatrician at the University of British Columbia, vancouver, has studied interactions between individual differences in physiology and differences in experience in determining health and wel-being. Dr. Boyce found that some children are more sensitive to their environments. They do exceptionally well when the environment is supportive but are much more likely to have mental and physical health problems when the environment has chalenges. You decide to do a study, conducting a factorial experiment to test the effectiveness of one environmental factor and one physiological factor on a environmental factor, you choose two levels of stressful life events. As physical health outcome. As the the physiological factor, you choose three levels of immune reactivity. The outcome is number of respiratory illnesses in the previous 12 months, and the research participants are kindergartners. A) You conduct a two-factor ANOvA on the data. The two-factor ANovA involves several hypothesis tests. Which of the following are null hypotheses that you could use this ANOVA to test? Check all that apply. O Immune reactivity has no effect on number of respiratory illnesses. The effect of stressful life events on number of respiratory inesses is no different from the effect ofimmune reactivity. There is no interaction between stressful life events and immune reactivity. Stressful life events have no effect on number of respiratory illnesses. The results of your study are summarized by the corresponding sample means below. Each cell reports the average number of respiratory illnesses for 9 kindergartners. Factor B: Immune Reactivity High Medium M 2.44 M m 2.22 M 3.22 T 20 TROw1 71 Low T 29 T m 22 SS m 1.5556 SS 2.2222 SS m 1.5556 EX 2 415 Factor A: Stressful Life Events M m 2.89 M m 2.78 M m 2.78 High T 26 T m 25 T m 25 TROw2 m 76 SS m 0.8889 SS m 1.5556 SS 1.5556 Toou 55 Too12 m 47 Toou 45 You perform an ANOVA to test that there are no main effects of factor A, no main effects of factor B, and no interaction between factors A and B. Some of the results are presented in the following ANOVA table. ANOVA Table Source

Explanation / Answer

A) Immune reactivity has no effect on number of respiratory illnesses.
There is no interaction between stressful life events and immune reactivity
Stressful life events have no effect on number of respiratory illnesses

1) Corrected Mean, CM = 400.1667, SS(factor A) = ?Ai2/rb - CM, Ai = Sum of each row,
r = number of replicates = 9, b = number of levels in factor B = 3
SS(factor A) = [(71)2 + (76)2]/(9*3) - 400.1667 = 0.4630

SS(factor B) = ?Bi2/ra - CM = [(55)2 + (47)2 + 452]/(9*2) - 400.1667 = 3.111

2) SS(within) = SS(total) - SS(treatments) = 14.8333 - 5.5000 = 9.3333

3) df(factor A) = a - 1 = 2 - 1 = 1 (no. of groups - 1)
df(factor B) = b - 1 = 3 - 1 = 2
df(A x B interaction) = (a-1)*(b-1) = 1*2 = 2
df(within) = df(total) - df(treatments) = 53 - 5 = 48

4) MS(A x B interaction) = SS(A x B interaction)/df(A x B interaction) = 1.9257/2 = 0.96285
MS(Within) = SS(within)/df(within) = 9.3333/48 = 0.1944

F(A x B interaction) = MS(A x B interaction)/MS(Within) = 0.96285/0.1944 = 4.9518

5B) p-value(Factor A) = F1,48(2.38) = 0.1295 > 0.01, not significant
p-value(factor B) = F2,48(8.00) = 0.0010 < 0.01, significant
p-value(A x B interaction) = F2,48(4.9518) = 0.0111 > 0.01, not significant

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