11 two-factor anova - emphasis on calculations courses.aplia.com 11. Two-factor
ID: 3357565 • Letter: 1
Question
11 two-factor anova - emphasis on calculations courses.aplia.com 11. Two-factor ANOVA-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 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. Stressful life events have no effect on number of respiratory illnesses. Immune reactivity has 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 6 kindergartners. Factor B: Immune Reactivity Medium Low High M-3.17 T= 19 M=4.17 T-25 | | Mit 3.00 T = 18 TROwl = 62 Low SS- SS 0.8333 SS 0.8333 EX2 452 Factor A: Stressful Life Events M 3.83 M 3.50 T=21 M 3.33 T = 23 TROW2 = 64Explanation / Answer
Answer:
Source
SS
DF
MS
F
A
0.1111
1
0.1111
0.62
B
4.6668
2
2.3334
13.12
AB
0.8890
2
0.4445
2.50
Error
5.3331
30
0.1778
Total
11.0000
35
F distribution with df denominator 30.
Main effect Factor A is not significant
Main effect Factor B is significant
Interaction effect is not significant.
Source
SS
DF
MS
F
A
0.1111
1
0.1111
0.62
B
4.6668
2
2.3334
13.12
AB
0.8890
2
0.4445
2.50
Error
5.3331
30
0.1778
Total
11.0000
35
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