W. Thomas Boyce, a professor and pediatrician at the University of British Columbia, 1v'ancouver, has studied interactions between individual differences in physiology and differences in experience in determining health and wellbeing. 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 healm 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. I: Stressful life events have no effect on number of respiratory illnesses. I: There is no interacb'on between stressful life events and immune reactivity. I: Immune reactivity has no effect on number of respiratory illnesses. I: The effect of stressful life events on number of respiratory illnesses is no different from the effect of 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 5 kindergartners. The results of your study are summarized by the corresponding sample means below. Each cell reports the average number of respiratory illnesses for6 kindergartners. Factor B: Immune Reactivity Low Medium High M: 3.00 i\\'i=3.1'.'r |\\"l=-4-.1?r Low T = 18 T = 19 T = 25 TROWI = 62 55 = 0 55 = 0.8333 55 = 0.8333 Factor A: 2X2 = 452 stressful Life Events M = 3.33 M = 3.50 M = 3.83 High T = 20 T = 21 T = 23 Taowz = 64 55 = 1.3333 55 = 1.5000 55 = 0.8333 Toou = 38 Tom: = 40 Tom: = 48 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 Aand B. Some of the results are presented in the following ANOVA table. ANOVA Tabla Source 55 df us F Between treatments 5.666? 5 Factor A 0.1111 0.62 Factor B 2.3334 13.12 A x B interaction 0.8890 Within treatments Total 11.0000 35 Work through the following steps to complete the preceding ANOVA table. Work through the following steps to complete the preceding ANOVA table. 1. The main effect for factor A evaluates the mean differences between the levels of factor A. The main effect for factor B evaluates the mean differences between the levels of factor B. Select the correct values for the sums of squares for factors A and B in the ANOVA table. 2. Select the correct value for the within-treatments sum of squares in the ANOVA table. 3. Select the correct degrees of freedom for all the sums of squares in the ANOVA table. 4. Select the correct values for the mean square due to A X B interaction, the within treatments mean square, and the F-ratio for the A X B interaction. 5. Use the results from the completed ANOVA table and the F distribution table (click on the following dropdown menu to access the table) to make the following conclusions. The F Distribution: Table entries in lightface type are critical values for the .05 level of significance. Boldface type values are for the .01 level of significance. At the significance level o = .05, the main effect due to factor A is , the main effect due to factor B is , and the interaction effect between the two factors is
