Thomas Boyce, a professor and pediatrician at the University of British Columbia

Question

Thomas Boyce, a professor and pediatrician at the University of British Columbia, Vancouver, has studied interactions between individual differences in physiology ans difference 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 9 physical health outcome. As the environmental factor, you choose two level of stressful life events. As the physiological factor, you choose three levels of immune reactivity. The outcome is number of respiratory 1 to 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. Winch 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. The effect of stressful life events on number of respiratory illnesses is no different from the effect of immune reactivity. Immune reactivity has no effect on number of respiratory illnesses. 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. 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 fallowing ANOVA table. Work through the following steps to complete the preceding ANOVA table. The ma n effect for factor A evaluates the mean differences between the levels of factor A. The man 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. Select the correct value for the will thin-treatments sum of squares in the ANOVA table. Select the correct degrees of freedom for all the sums of scares in the ANOVA table. 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. Use the results from the completed ANOVA table and the F distribution table (dick on the following dropdown menu to access the table)to make the following conclusions. Table entries in lightface type are combat values for the .05 level of significance. Boldface type values are for the .01 level of significance. At the significance level alpha = .01, 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 ____.

Explanation / Answer

For null hypothesis,

A, C and D option. There is no interaction between stressful life events and immune reactivity, Immune reactivity has no effect on number of respiratory illness and Stressful life events have no effect on number of respiratory illness.

1)Factor A, dfA = 2 – 1 = 1 (A has 2 levels)

Factor B, dfB = 3 – 1 = 2 (B has 3 levels)

Given, MSA = 0.4630

MSB = 1.5556

Therefore MS = SS/df

SSA = MSA*dfA = 0.4630 * 1 = 0.4630

SSB = MSB*dfB = 1.5556 * 2 = 3.1112

2) SSTOTAL = SSBetween + SSWithin

SSBetween = 5.5000

SSTotal = 14.8333

SSWithin = 14.8333 – 5.5000 = 9.3333

3) Factor A, dfA = 2 – 1 = 1 (A has 2 levels)

Factor B, dfB = 3 – 1 = 2 (B has 3 levels)

DF of interaction effect, dfA*B = (a-1)(b-1) = dfA * dfB = 1 * 2 = 2

DFWithin = ab(n-1) = 3*2(9-1) = 6*8 = 48 (n=9 as each treatment is for 9 kindergrtners)

4) MSA*B = SSA*B / dfA*B = 1.9257 / 2 = 0.96285

SSWithin = 14.8333 – 5.5000 = 9.3333

DFWithin = 48

MSWithin = 0.1944

FA*B = MSA*B / MSW = 0.96285/0.1944 = 4.9529

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