Question

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 oral hygiene. As the physiological factor, you choose three levels of cortisol value. The outcome is number of cavities, 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. The effect of oral hygiene on number of cavities is no different from the effect of cortisol value. There is no interaction between oral hygiene and cortisol value. Cortisol value has no effect on number of cavities. Oral hygiene has no effect on number of cavities. The results of your study are summarized by the corresponding sample means below. Each cell reports the average number of cavities for 7 kindergartners. Low Factor B: Cortisol Value Medium High Low M-3.29 M - 2.71 T= 23 T-19 SS - 1.42865S - 1.4286 M-1.86 T = 13 ss - 0.8571 TROW - 55 Factor A: Oral Hygiene EX2 = 303 High M-1.86 T-13 SS - 0.8571 M-2.86 T-20 SS - 2.8571 M-3.00 T-21 SS - O TROW2 = 54

Answer 1

Null hypotheses are,
There is no interaction between oral hygiene and cortisol value.
Cortisol value has no effect on number of cavities.
Oral hygiene has no effect on number of cavities.

Source	SS	df	MS	F
Between Treatment	12.6905	5
Factor A	0.0238	1	0.0238	0.12
Factor B	0.9048	2	0.4524	2.19
A x B Interaction	11.7619	2	5.88095	28.50027
Within Treatments	7.4285	36	0.2063472

dfA = a - 1 = 2 - 1 = 1

dfB = b - 1 = 3 - 1 = 2

df Interaction = (a-1)(b-1) = (2-1)(3-1) = 2

df Within = df Total - (dfA + dfB + df Interaction) = 41 - (1 + 2 + 2) = 36

SS A = MS A * dfA = 0.0238 * 1 = 0.0238

SS B = MS B * dfB = 0.4524 * 2 = 0.9048

MS Interaction = SS Interaction / df Interaction = 11.7619/2 = 5.88095

SS Within = SS Total - (SS A + SS B + SS Interaction) = 20.1190 - (0.0238 + 0.9048 + 11.7619) = 7.4285

MS Within = SS Within / df Within = 7.4285 / 36 = 0.2063472

F Interaction = MS Interaction / MS Within = 5.88095 / 0.2063472 = 28.50027

Critical value of F at 0.01 significance level and df = 1, 36 is 7.40

Critical value of F at 0.01 significance level and df = 2, 36 is 5.25

Since the observed F for factor A is less than the critical value (7.40), there is no significant main effect of factor A.

Since the observed F for factor B is less than the critical value (5.25), there is no significant main effect of factor B.

Since the observed F for Interaction is greater than the critical value (5.25), there is a significant Interaction effect of factor A.

Thus,

The main effect due to factor A is not significant.
The main effect due to factor B is not significant.
The interaction between two factors is significant.