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c Dng.cengage.com/static/nblul/index.html?nbld-497996&nbNodeld-176563379;&eISBN; 9781305647305 MINDTAP Two-Factor Analysis of variance (Independent Measures)-Aplia Due Tomorrow at SAM CEST 3. Two-factor analysis of variance 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 ch dren are more sensitive to their environments. They do exceptionally wel when the environment is supportive but much more likely to have mental a physcal health problems when the environment has challenges. You decide to do a simiar 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 cardiovasolar 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. O Cardiovascular reactivity has no effect on number of injuries. O The effect of stress on number of injuries is no different from the effect of cardiovascular reactivity stress has no effect on number of injuries. There is no interaction between stress and cardiovascular reactivity. The results of your study are summarized by the corresponding sample means below. Each cel reports the average number of injuries for 6 rhesus monkeys.Explanation / Answer
Answer
Which of the following are the null hypotheses that you could use this ANOVA to test:
Correct options –
1. Cardiovascular reactivity has no effect on number of injuries.
3. Stress has no effect on number of injuries.
4. There is no interaction between stress and cardiovascular reactivity.
Work through the following steps to complete the ANOVA table.
1.
Correct values for the sums of squares for factors A and B in ANOVA table are calculated below.
SS for factor A = MS Factor A * DF Factor A
= 0.1111 * 1
SS for factor A = 0.1111
SS for factor B = MS Factor B * DF Factor B
= 2.3334 * 2
SS for factor B = 4.6668
2.
Correct value for within treatment sum of squares in ANOVA table is given below.
SS Within Treatments = Total SS – SS between Treatments
SS Within Treatments = 11.0000 – 5.6667
SS Within Treatments = 5.3333
3.
Correct degree of freedom for all the sum of squares in ANOVA table is given below.
DF for factor A = 2 – 1 = 1
DF for factor B = 3 – 1 = 2
DF for AxB = 2 * 1 = 2
DF for within treatments = 35 – 5 = 30
4.
Correct values for the mean square due to AxB interaction and mean square and F-ratio are calculated below.
MS for AxB interaction = SS for AxB interaction / DF for AxB interaction
MS for AxB interaction = 0.8890 / 2
MS for AxB interaction = 0.4445
MS for within treatments = SS for within treatments / DF for within treatments
MS for within treatments = 5.3333 / 30
MS for within treatments = 0.1778
F-ratio for AxB interaction = MS for AxB interaction / MS for within treatments
F-ratio for AxB interaction = 0.4445 / 0.1778
F-ratio for AxB interaction = 2.50
5.
Completed ANOVA table is given below.
Source
SS
df
MS
F
Between Treatments
5.6667
5
Factor A
0.1111
1
0.1111
0.62
Factor B
4.6668
2
2.3334
13.12
AxB interaction
0.8890
2
0.4445
2.50
Within Treatments
5.3333
30
0.1778
Total
11.0000
35
F distribution table value for comparison of F ratio for factor A is 4.171. Thus F value for factor A (0.62) is less than critical F value (4.171). Therefore, the main effect due to factor A is not significant.
F distribution table value for comparison of F ratio for factor B is 3.316. Thus F value for factor B (13.12) is greater than critical F value (3.316). Therefore, the main effect due to factor B is significant.
F distribution table value for comparison of F ratio for AxB interaction is 3.316. Thus F value for AxB interaction (2.50) is less than critical F value (3.316). Therefore, the interaction effect is not significant.
Conclusion: At the significance level = .05, the main effect due to factor A is not significant, the main effect due to factor B is significant, and the interaction effect between the two factors is not significant.
Source
SS
df
MS
F
Between Treatments
5.6667
5
Factor A
0.1111
1
0.1111
0.62
Factor B
4.6668
2
2.3334
13.12
AxB interaction
0.8890
2
0.4445
2.50
Within Treatments
5.3333
30
0.1778
Total
11.0000
35
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