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. There is no interaction between stressful life events and immune reactivity. 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. 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 7 kindergartners. Factor A: Stressful Life Events Source Between treatments Factor A Factor B AX B interaction Within treatments Total ANOVA Table SS 12.6905 11.7619 Low 20.119 High df 5 Factor B: Immune Reactivity Low M = 3.29 T = 23 SS= 1.4286 Medium M= 2.71 T = 19 SS= 1.4286 41 M = 1.86 T = 13 SS=0.8571 TCOLI = 36 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. MS M= 2.86 T = 20 SS= 2.8571 TCOL2 = 39 0.0238 0.4524 F High M = 1.86 T = 13 SS=0.8571 0.12 2.19 M = 3.00 T = 21 SS=0 TCOL3 = 34 TROWI= 55 TROW2 = 54 EX² = 303
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. There is no interaction between stressful life events and immune reactivity. 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. 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 7 kindergartners. Factor A: Stressful Life Events Source Between treatments Factor A Factor B AX B interaction Within treatments Total ANOVA Table SS 12.6905 11.7619 Low 20.119 High df 5 Factor B: Immune Reactivity Low M = 3.29 T = 23 SS= 1.4286 Medium M= 2.71 T = 19 SS= 1.4286 41 M = 1.86 T = 13 SS=0.8571 TCOLI = 36 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. MS M= 2.86 T = 20 SS= 2.8571 TCOL2 = 39 0.0238 0.4524 F High M = 1.86 T = 13 SS=0.8571 0.12 2.19 M = 3.00 T = 21 SS=0 TCOL3 = 34 TROWI= 55 TROW2 = 54 EX² = 303
Holt Mcdougal Larson Pre-algebra: Student Edition 2012
1st Edition
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Chapter11: Data Analysis And Probability
Section: Chapter Questions
Problem 8CR
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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.
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