In: Statistics and Probability
QUESTION 3
Factor A: Caffeinated |
||||
No Caffeine |
Caffeine |
|||
Factor B: Extroversion |
Extrovert |
n = 10 M = 13 T = 130 SS = 18 |
n = 10 M = 11 T = 110 SS = 36 |
TExtrovert =240 MExtrovert = 12 |
Introvert |
n = 10 M = 14 T = 140 SS =18 |
n = 10 M = 9 T = 90 SS = 28 |
TIntrovert =230 MExtrovert=11.5 |
|
T No Caffeine =270 |
T Caffeine =200 |
∑X2 = 5770, G = 470, N = 40
Sol:
SS | df | MS | F | P value | Fcrit | |
A | 2.500 | 1 | 2.500 | 0.900 | 0.3490 | 4.113 |
B | 122.500 | 1 | 122.500 | 44.096 | 0.0000 | 4.113 |
AB | 22.500 | 1 | 22.500 | 8.099 | 0.0070 | 4.113 |
Error | 100.000 | 36 | 2.778 | |||
Total | 247.500 | 39 |
(d) The Hypothesis:
FACTOR A
H0: There is no difference in the means of Factor A
Ha: At least one of the means differs from the others.
FACTOR B
H0: There is no difference in the means of Factor B.
Ha: At least one of the means differs from the others.
INTERACTION
H0: There is no interaction between the two factors. They are independent.
Ha: There is an interaction between the two factors. They are not independent.
The Decision Rule:
The Critical Value Method: If F test is > F critical, Then Reject H0
P Value Method: If p value is < , Then Reject H0.
The Decision:
For Factor A
F test (0.009) is < F Critical (4.113), We Fail to Reject H0
P value (0.349) is > (0.01). We Fail to Reject H0
The Conclusion: There isn't sufficient evidence to conclude that at least one mean differs from the others.
For Factor B
F test (44.096) is > F Critical (4.113), We Reject H0
P value (0.000) is < (0.01). We Reject H0
The Conclusion: There is sufficient evidence to conclude that at least one mean differs from the others.
For Interaction
F test (8.099) is > F Critical (4.113), We Reject H0
P value (0.007) is < (0.01). We Reject H0
The Conclusion: There is sufficient evidence to conclude there is an interaction between the 2 factors.
The Effect Sizes:
= SS effect / SS Total - SS Other 2 effects
= SS A / SS Total - SS B - SS AB = 2.5 / (972.55 - 122.5- 22.5) = 0.003 (Small effect)
= SS B / SS Total - SS A - SS AB = 122.5 - (972.55 - 2.5 - 22.5) = 0.13 (Small Effect)
= SS AB / SS Total - SS A - SS B = 84.05 / (972.55 - 2.5 - 832.05 ) = 0.61 (Medium effect)
Conclusion:
Extroversion and caffeine might contribute to ability to memorize, but that effect may vary across different levels of extroversion or amount of caffeine. A two way ANOVA tested reaction time for 40 people with No Caffeine and Some caffeine and with Introverts and Extroverts..
Different levels of caffeine did not show a difference in memory power F(1,39) = 0.9, p = 0.349, = 0.003 which a small effect, Different levels of extroversion showed a significant difference in memory power F(1,36) = 44.096, p = 0.0000, = 0.13 which had a small effect. The interaction between Extroversion and Caffeine was also significant F(1,36) = 8.099, p = 0.007, = 0.61, the effect being medium at the most.