An experimenter is investigating the effects of two drugs on the activity of rat
ID: 3263389 • Letter: A
Question
An experimenter is investigating the effects of two drugs on the activity of rats. Drug A is a depressant, and Drug B is a stimulant. Half of the subjects receiving either drug are given a low dosage and half a high dosage. The experimenter also runs a control group that is given an injection of an inert substance such as a saline solution. Five different groups are represented in the experiment, each containing n=4 rats assigned randomly from the stock of laboratory rats on hand. The animals are injected and then their activity is observed for a fixed period of time. The MS within= 37.00. To conduct pair wise comparisons, how do you determine which n to use?
Drug A: Depressant Drug B: Stimulant
Control Low High Low High
A1 A2 A3 A4 A5
15 13.75 8.0 16.5 23.0
Explanation / Answer
1. Control vs. the combined experimental groups
(1)(15) + (-1/4)(13.75) + (-1/4)(8.0) + (-1/4)(16.5) + (-1/4)(23) = 15-15.3125 = -.3125
(0)(15) + (1/2)(13.75) + (1/2)(8.0) + (-1/2)(16.5) + (-1/2)(23) = 10.875 – 19.75 = -8.875
(0)(15) + (1/2)(13.75) + (-1/2)(8.0) + (1/2)(16.5) + (-1/2)(23) = 15.125 – 15.5 = -.375
Comp 1 = -.3125
Comp 2 = -8.875
Comp 3 = -.375
SS contrast = n L2/ aj2
Sum of squared weights
L1 = -.3125, L2=-8.875, L3 = -.375
SSc1 = 4(-.3125)2/1.25 = .3125
SSc2 = 4(-8.875)2/1 = 315.0625
SSc3 = 4(-.375)2/1 = .5625
F comp1 = .3125/37 <1
F comp 2 = 315.0625/37 = 8.515*
F comp3 = .5625/37 <1
Critical F need for significance with 1 and 15 df = 4.54
MSwithin = 37
Can also test the significance of a contrast using t tests instead of the sum of squares formula
So t = L/SQRT ((sum of squared coefficients * MS error)/n)
Sum of squared weights
Comp 1 = -.3125
Comp 2 = -8.875
Comp 3 = -.375
T1= -.3125/SQRT ((1.25 * 37)/4) = -.3125/3.4003 = <1
T2= -8.875/SQRT ((1 * 37)/4) = -8.875/ 3.04= 2.919*
T3= -.375/SQRT ((1 * 37)/4) = -.375/ 3.04 = <1
Critical t with 20 df 2 tailed = 2.131, so notice that the same contrasts that were significant using the SS tests are significant using the t-tests.
Using a Tukey test and a Neuman Keuls test conduct all possible pairwise comparisons and compare the results of each test.
8 13.75 15 16.5 23
8 -- 5.75 7 8.5 15
13.75 --- 1.25 2.75 9.25
15 --- 1.5 8
16.5 6.5
23 -- ----
Tukey
CD = 4.37 x SQRT (37/4) = 13.29
Conclusions 23 > 8 all other comparisons are not significant
Neuman Keuls
4.37 x SQRT (37/4) = 13.29
CD5steps = 4.37 x SQRT (37/4) = 4.37 x 3.041 = 13.29
CD4steps = 4.08 x SQRT (37/4) = 12.407
CD3steps = 3.67 x SQRT (37/4) = 11.16
CD2steps = 3.01 x SQRT (37/4) = 9.15
Conclusions 23 > 8 all other comparisons are not significant
Fisher’s Least Significant Difference
LSD = 2.131 x SQRT (2x37/4) = 9.165
Conclusions 23 > 8 and 23 > 13.75
Bonferroni T
10 total comparisons
CD = 3.29 x SQRT (2x37/4) = 14.15
Conclusions 23 > 8 all other comparisons are not significant
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