Topic 21
Analyzing Two Treatments—I
Activity 21-4: Memorization Methods and Gender
The SPSS data file MEM2WAY.SAV contains the results of a two factor factorial experiment. It compares two
memorization methods—repetition and visualization, and gender. We shall use memorization as Treatment A and
gender as Treatment B.
a)
Have SPSS draw the main effects and interaction plots. What do they show?
Estimated Marginal Means of Score
Estimated Marginal Means of Score
7
6.8
Estimated Marginal Means
Estimated Marginal Means
7
6.5
6
5.5
6.6
6.4
6.2
6
5.8
5.6
5.4
5
Repetition
Female
Male
Imagery
Method
Gender
Estimated Marginal Means of Score
Method
Repetition
Imagery
Estimated Marginal Means
7
6
5
4
Female
Male
Gender
There appears to be a difference in mean score according to both gender and method. There may or may
not be an interaction.
b) What are the value for a, b and n?
A = b = 2, n = 10
c) Compute Y , the Yi's, the Y j 's and the individual variances as was done in Activity 21-2.
Answers will vary according to which effect is the A and B effect. For these answers we will let gender be
the A effect and method be the B effect.
Y  6.20, Y1  7.20, Y2  5.20, Y1  5.45, Y2  6.95
Repitition
Imagery
Female
7.333
10.489
Male
12.544
18.722
Activity 21-5: Memorization Methods and Gender (cont’d.)
Use the results of Activity 21-4.
a)
Compute SSA and SSB.
SSA  2(10) (7.2  6.2) 2  (5.2  6.2) 2   40
SSB  2(10) (5.45  6.2) 2  (6.95  6.2) 2   22.5
b)
c)
Compute SSTotal. SSTotal = 39(13.241) = 516.399
Compute SSE.
d)
Compute SSAB.
SSE  9(7.333)  9(12.544)  9(10.489)  9(18.722)  441.792
SSAB  516.399  40  22.5  441.792  12.107
Activity 21-6: Memorization Methods and Gender (cont’d.)
Use the results of Activities 21-4 and 21-5.
a)
Compute the mean squares.
MSA 
b)
40
22.5
12.107
441.792
 40, MSB 
 22.5, MSAB 
 12.107, MSE 
 12.272
1
1
1
36
Compute the F ratios for the test of the two main effects and the test of the interaction. What are the
number of degrees of freedom associated with each one?
40
 3.259, d . f .  (1,36)
12.272
22.5
B: F 
 1.833, d . f .  (1,36)
12.272
12.107
AB: F 
 .987, d . f .  (1,36)
12.272
A: F =
c)
Compute the p-value for the three tests.
A: p -value  Pr( F(1,36)  3.259)  .05  p -value  .1
B: p-value  Pr( F(1,36)  1.833)  .1
AB: p-value  Pr( F(1,36)  .987)  .1
d)
What do you conclude?
There is some evidence of a difference due to gender, but no evidence of a difference due to method and
no evidence of an interaction.
Activity 21-7: Butterfat Content of Milk
The SPSS data file BUTTERFAT.SAV contains the butter fat content of milk from five breeds of dairy cattle and
whether or not the individual cow was mature or two years old or less. For each breed there were data for twenty
cattle—ten mature cattle and ten younger cattle. Let the breed of cattle be Treatment A and age be Treatment B.
The data are summarized below.
Breed
Ayrshire
Canadian
Guernsey
Holstein
Jersey
a)
Mature
S
0.315
0.348
0.466
0.167
0.674
Y
4.06
4.44
4.95
3.67
5.29
Young
s
0.193
0.386
0.503
0.329
0.547
Age
Young
Mature
Y
4.53
4.44
Have SPSS draw the main effects and interaction plots. What do they indicate?
Estimated Marginal Means of Butterfat
Estimated Marginal Means of Butterfat
4.55
5.50
Estimated Marginal Means
Estimated Marginal Means
4.525
5.00
4.50
4.00
4.50
4.475
4.45
4.425
4.40
3.50
Mature (At least 5 yrs. old)
Ayrshire
Canadian
Guernsey
Holstein
Jersey
Young (2 yrs. old)
Age
Breed
Estimated Marginal Means of Butterfat
Age
5.50
Estimated Marginal Means
Mature (At
least 5 yrs.
old)
Young (2 yrs.
old)
5.00
4.50
4.00
3.50
Ayrshire
Canadian Guernsey
Holstein
Jersey
Breed
b)
c)
There appears to be a difference due to breed, a difference due to age, but no interaction.
What are the values for a, b and n?
If we let breed be the A effect, and age be the B effect, then a = 5, b = 2, and n = 10
Compute SSA and SSB.
Y  4.4821, Y1  4.06, Y2  4.4385, Y3  4.95, Y4  3.6695
Y5  5.2925, Y1  4.4366, Y2  4.5276
SSA  2(10)[ 4.06  4.4821  (4.4385  4.4821) 2  (4.95  4.4821) 2
2
(3.6695  4.4821) 2  (5.2925  4.4821) 2 ]  34.321
SSB  5(10)[(4.4366  4.4821) 2  (4.5276  4.4821) 2 ]  .207
d)
Compute SSE.
SSE  9(.037)  9(.149)  9(.253)  9(.108)  9(.299)
9(.099)  9(.121)  9(.217)  9(.027)  9(.455)
 15.885
e)
The value for SSTotal is 50.68. Use this to compute SSAB.
SSAB  50.68  34.321  .207 15.885  .267
Activity 21-8: Butterfat Content of Milk (cont’d.)
Use the information and results from Activity 21-7.
a)
Compute the mean squares.
34.321
 8.58
4
.207
MSB 
 .207
1
.267
MSAB 
 .067
4
15.885
MSE 
 .1765
90
MSA 
b)
Compute the F ratios for the test of the two main effects and the test of the interaction. What are the
number of degrees of freedom associated with each one?
8.58
 48.61, d . f .  (4,90)
.1765
.207
B: F 
 1.173, d . f .  (1,90)
.1765
.067
AB: F 
 1.512, d . f .  (4,90)
.1765
A: F =
c)
Compute the p-value for the three tests.
A: p-value  Pr( F(4,90)  48.61)  .001
B: p-value  Pr( F(1,90)  1.173)  .1
AB: p-value  Pr( F(4,90)  1.512)  .1
d)
What conclusions do you reach?
There is very strong evidence of a difference in mean butterfat content due to breed, no evidence of a
difference in mean butterfat content due to age, and no evidence of an interaction between breed and age.
Activity 21-9: Insurance Rates
The SPSS data file INSURANCE.SAV the rates paid for term life insurance. The main effects are the gender of the
insure and whether or not he/she smokes. The goal is to see how the amount paid varies according to these two
main effects and to see if there is in interaction.
a)
Have SPSS draw the main effects and interaction plots. Describe what they show.
Estimated Marginal Means of Premium
Estimated Marginal Means of Premium
280
260
Estimated Marginal Means
Estimated Marginal Means
250
240
230
220
260
240
220
200
210
180
200
Smoker
Nonsmoker
Smoke
Male
Female
Gender
Estimated Marginal Means of Premium
Smoke
325
Smoker
Nonsmoker
Estimated Marginal Means
300
275
250
225
200
175
Male
Female
Gender
There appears to be a difference due to gender, a difference due to whether or not a person smokes, but no
interaction.
b) What are the values of a, b, n, and N?
If we let the A effect be gender, and the B effect be smoker, then a = b = 2, and n = 23
c) Compute SSA and SSB.
Y  228.48, Y1  253.26, Y2  203.70, Y1  271.74, Y2  185.22
SSA  2(23)[ 253.26  228.48   (203.7  228.48) 2 ]  56942.4528
2
SSB  2(23)[(271.74  228.48) 2  (185.22  228.48) 2 ]  172171.3392
d) Compute SSE.
SSE  22(31911.265)  22(14551.976)  22(9872.53)  22(6843.182)
 1389936.966
e)
Compute SSTotal.
f)
Compute SSAB.
SSTotal  92(18001.505)  1656138.46
SSAB  1656138.46  56942.4528 172171.3392 1389936.966
 37087.702
Activity 21-10: Insurance Rates (cont’d.)
Use the information and results from Activity 21-9.
a)
Compute the mean squares.
56942.4528
 56942.4528
1
172171.3392
MSB 
 172171.3392
1
37087.702
MSAB 
 37087.702
1
1389936.966
MSE 
 15794.73825
88
MSA 
b) Compute the F ratios for testing the two main effects and the interaction. What are the degrees of freedom
for each of your F ratios?
56942.4528
 3.605, d . f .  (1,88)
15794.73825
172171.3392
B: F 
 10.900, d . f .  (1,88)
15794.73825
37087.702
AB: F 
 2.348, d . f .  (1,88)
15794.73825
A: F =
c)
Compute the p-value for each of your F ratios.
A: p-value  Pr( F(1,88)  3.605)  .05  p-value  .1
B: p-value  Pr( F(1,88)  10.9)  .001
AB: p-value  Pr( F(1,88)  2.348)  .1
d) What conclusions do your reach?
There is some evidence of a difference due to gender, very strong evidence of a difference between
smokers and non-smokers, but no evidence of an interaction.