Question

3. Two-factor analysis of variance Emphasis on calculations Aa Aa 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 well-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 challenges You decide to do a similar study, conducting a factorial experiment to test the effectiveness of one environmental factor and one physiological factor on a physical health outcome. As the environmental factor, you choose two levels of stressful life events. As 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. 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 Stressful life events have no effect on number of respiratory illnesses. □ Immune reactivity has no effect on number of respiratory illnesses. The effect of stressful life events on number of respiratory illnesses is no different from the effect of immune reactivity There is no interaction between stressful life events and immune reactivity 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 Medium Low High M 3.22 T 29 SS= 1.5556 M= 2.44 T 22 SS= 2.2222 M 2.22 T=20 SS= 1.5556 Low TROwi-71 Factor A: X2 415 Stressful Life Events M 2.89 T=26 SS = 0.8889 M= 2.78 T 25 SS = 1.5556 M = 2.78 T=25 SS = 1.5556 High TRow2 = 76 TCOLI = 55 COL2 47 TcoL3 45

Answer 1

According to the question, there are two factors, such as immune reactivity and stressful life events. Immune reactivity has three levels and stressful life events has two level. The summarized data is provided in the question. Two-way ANOVA can be used to analyze the data. The null hypotheses that can be analyzed using two-way ANOVA are defined as follows: 1. Immune reactivity has no effect on the respiratory illnesses. 2. Stressful life events has no effect on the respiratory illnesses. 3. There is no interaction effect between immune reactivity and stressful life events. (1) The sum of squares for the effect A is calculated as follows: SS. = MS, degrees of freedom for A = 0.4630x(2-1) = 0.4630 The sum of squares for the effect B is calculated as follows: SS, = MS, degrees of freedom for B =1.5556x(3-1) = 3.1112 Hence, the sum of squares for factor A, and B are 0.4630, and 3.1112, respectively. (2) The value of within treatment sum of squares is calculated as follows: SS within treatments = SS pat-SS between treatments = 14.8333-5.5000 = 9.3333 Hence, the within treatment sum of squares is 9.3333. (3) The degrees of freedom (df) for all sum of squares are calculated as follows:

df. = Number of levels in factor A-1 = 2-1 = 1 dfg = Number of levels in factor B-1 = 3-1 = 2 (Number of levels in factor A -1) $ (Number of levels in factor B-1) =1x2 = 2 dfwithin treatments = dfront - df between treatments = 53-5 = 48 Hence, the degrees of freedom for factor A, factor B, interaction between A and B, and error are 1, 2, 2, and 48, respectively. The mean square of interaction between A and B is calculated as follows: SSAB MS 18df AB 1.9257 2 = 0.9629 Therefore, the mean square of interaction is 0.9629. The value of within treatment mean sum of squares is calculated as follows: 2 MS within treats ents SS within reatments of within treatments 9.3333 48 = 0.1944 Hence, the within treatment mean sum of squares is 0.1944. The F-ratio of the interaction is calculated as follows:

MS AB FB MS-invest 0.9629 0.1944 = 4.95 Therefore, F-ratio of the interaction is 4.95. The complete ANOVA table is shown below. Source Between treatments Factor A Factor B Ax B interaction Within treatments Total Sum of squares df Mean squares F-ratio 5.5000 5 0.4630 1 0.4630 238 3.1112 1.5556 8.00 1.9257 0.9629 4.95 9.3333 48 0.1944 14.8333 | 53 | Excel function is used to obtain the critical value for factor A corresponding to 1 numerator degrees of freedom and 48 denominator degrees of freedom. The screenshot is shown below. fx =F.INV.RT(0.05,1,48) 4.043 As the obtained critical value (4.043) is greater than the F-ratio for factor A (2.38), it can be concluded that the stressful life events does not affect the respiratory illness. Excel function is used to obtain the critical value for factor B corresponding to 2 numerator degrees of freedom and 48 denominator degrees of freedom. The screenshot is shown below. fr =F.INV.RT(0.05,2,48) DE F 3.191 As the obtained critical value (3.191) is less than the F-ratio for factor B (8.00), it can be concluded that the immune reactivity affects the respiratory illness.

Excel function is used to obtain the critical value for interaction corresponding to 2 numerator degrees of freedom and 48 denominator degrees of freedom. The screenshot is shown below. =F.INV.RT(0.05,2,48) D 3.191 As the obtained critical value (3.191) is less than the F-ratio for interaction factor (4.95), it can be concluded that the interaction factor affects the respiratory illness. Therefore, the main effect due to factor A is insignificant, the main effect due to factor B is significant, and the interaction effect between the two factors is significant