Please only answer if you can complete each section, thank you! Aa Aa E 11. Two-

Question

Please only answer if you can complete each section, thank you!

Aa Aa E 11. Two-factor ANOVA 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 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 stress. As the physiological factor, you choose three levels of cardiovascular reactivity. The outcome is number of injuries in the previous 12 months, and the research participants are rhesus monkeys. 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. Stress has no effect on number of injuries. There is no interaction between stress and cardiovascular reactivity The effect of stress on number of injuries is no different from the effect of cardiovascular reactivity Cardiovascular reactivity has no effect on number of injuries The results of your study are summarized by the corresponding sample mean below. Each cell reports the average number of injuries for 11 rhesus monkeys.

Explanation / Answer

1. The df of Factor A = 2-1 = 1

Sum of square of Factor A is 3.4091 * 1 =3.4091

The df of Factor B = 3-1 = 2

Sum of square of Factor A is 1.3788 * 2 = 2.7576

2. Sum of Square of With treatment = 22.4394 - 6.2575 =16.1819

3.   The df of Factor A = 2-1 = 1

The df of Factor B = 3-1 = 2

The df of Interaction = 5 - 1 - 2 = 2

The df of within treatment = 65 - 5 = 60

4. The Means square of A * B is 0.0908 / 2 =0.0454

  The Means square of within treatment is 16.1819 / 60 = 0.269698

F-ration = 0.269698 / 0.0454 = 5.94049

5. F critical value = 4

Here F value > F critical value So we do not accept H0

i..e there is significance difference between the level of factor A

6.

F critical value =  3.15

Here F value > F critical value So we do not accept H0

i..e there is significance difference between the level of factor B

7.

F critical value = 3.15

Here F value > F critical value So we do not accept H0

i..e there is significance difference between interaction the levels of factor A and factor B

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