in part a appears to be more powerful, as it leads to a higher z score.
2. z = (.55 .50) / (.5)(.5) / 320 = .05/.028 = 1.79, one-tailed p = .0367;
Reject H0 at .05 level; two-tailed p = .0367*2 = .073
b)
Estimated Marginal Means of MEASURE_1
12
Estimated Marginal Means
10
8
6
drug group
4
placebo
caffeine
2
1
2
3
4
DISTRACT
You can see some interaction in the graph, but it is not dramatic, so it is
1. a)
Source
Subject
Treatment
Interaction
Total
SS
230.75
25
30
285.75
df
7
1
7
15
MS
F
p
25
4.286
5.83
< .05
The matched t was 2.41 for these data; 2.412 = 5.81. The slight discrepancy with the F fo
Because 7.0 > 4.08, the contrast is significant at the .05 level.
b) The planned contrast was significant, even though the omnibus ANOVA was not. This contrast
takes advantage of the fact that the mea
b)
Xi 96.08 ; X = +1.5 (2.7) + 96.08 = 4.05 + 96.08 = 100.13;
i
2.7
1.5
Xi 96.08 ; X = -.8 (2.7) + 96.08 = -2.16 + 96.08 = 93.92
i
2.7
.8
2. a) area above z = -.4 = .1544 + .5 = .6544, so 65.44%;
area