A TA in statistics thinks that the format of the conferences and the timing of the assignments
may affect students’ assignment marks (out of 10). He decided to design three conferences
with a different format: (1) with lecturing, (2) with demonstrating problems, and (3) with
involving students in active problem-solving. Twelve students attended one of the conferences
(four in each format) and asked to submit their assignments at week 3. Another twelve
students attended one of the conferences but asked to submit their assignments at week 5.
The data are tabled below:
Format
Assignment
Lecture
Demo problems
Active problems
Week 3
6,7,5,7
7,8,8,9
9,10,10,10
Week 5
5,5,4,5
6,7,7,6
8,9,9,10
Note:
Perform the ANOVA test and table the results, stating all critical values. State and explain what
should be the follow-up analysis after the overall F-test if any (NO calculation but just explain).
Equations
Equations you may need are given as follows:
For the one-way independent groups (and simple main effects):
2
1 ⎛ K 2 ⎞ T..
SSG = ⎜ ∑ T.j ⎟ −
= UG − U
n ⎝ j=1 ⎠ N
K n
1 ⎛ K 2⎞
2
SSS(G) =∑ ∑ X − ⎜ ∑ T.j ⎟ = ∑ ∑ Xij
− UG
n ⎝ j=1 ⎠
j=1 i=1
j=1 i=1
K
n
2
ij
For the two-way independent-groups ANOVA with equal n’s:
2
1 ⎛ A 2 ⎞ T…
SSA =
∑ T − = UA − U
Bn ⎜⎝ a=1 .a. ⎟⎠ N
SSB =
2
1 ⎛ B 2 ⎞ T…
T
−
= UB − U
∑
An ⎜⎝ b=1 ..b ⎟⎠ N
2
1⎛ A B 2 ⎞ 1 ⎛ A 2 ⎞
1 ⎛ B 2 ⎞ T…
SSAB = ⎜ ∑ ∑ T.ab ⎟ −
∑ T.a. −
∑ T..b + = U AB − U A − U B + U
n ⎝ a=1 b=1
⎠ Bn ⎜⎝ a=1 ⎟⎠ An ⎜⎝ b=1 ⎟⎠ N
A B n
A B n
1⎛ A B 2 ⎞
2
2
SSS(AB) =∑ ∑ ∑ Xiab
− ⎜ ∑ ∑ T.ab
=
Xiab
− U AB
∑
∑
∑
⎟
n ⎝ a=1 b=1
⎠ a=1 b=1 i=1
a=1 b=1 i=1
For planned and post-hoc tests:
t=
Q=
F=
X·1 − X·2
MSerror MSerror
+
n1
n2
=
X·1 − X·2
= 2t
MSerror
# obs
# obs X·1 − X·2
(
)=t
2 MSerror
SScomparison
(∑ w T )
=
2
j ·j
#obs ∑ w 2j
2
X·1 − X·2
2 MSerror
# obs
A professor wishes to compare the number of hours spent studying for the midterm by the
Faculty of Arts/Social Science and the Faculty of Science students in three randomly
selected courses in chemistry, statistics, and psychology. Complete the ANOVA table,
showing the critical value and df for each test. The n’s are equal. Perform any
appropriate Scheffé post-hoc comparisons if needed (means appear below).
Source
SS
A = course
50
df
B = faculty
AB
S(AB)
MS
F
9
2
.35
12
Arts/Social Science students, chemistry: mean = 5
Arts/Social Science students, statistics: mean = 11
Arts/Social Science students, psychology: mean = 9
Science students, chemistry: mean = 6
Science students, statistics: mean = 10
Science students, psychology: mean = 7
Fcrit
As part of a summer school survey, twenty-four students were asked to rate their approval
(0=low, 10=high) of only one of four possible green efforts as a part of their statistics course.
The raw data and a part of the ANOVA table (i.e., Sums of Squares) are provided below.
No air conditioning
Electronic submission Dim lights
of assignments
No coursepack
printing
8
9
10
4
5
6
9
6
7
7
5
4
8
8
10
6
6
7
8
5
5
8
8
5
Source
SS
E = effort
37
S(E)
36.33
a) Complete the rest of the ANOVA table.
b) The researcher hypothesizes that the classroom efforts (no air conditioning and dim
lights) may be less popular than the more familiar paper efforts (electronic submission of
assignments and no coursepack printing). The researcher also thinks that no air
conditioning will be less popular than an electronic assignment submission. Perform the
planned comparisons. Are the comparisons orthogonal? Demonstrate.
Equations
Equations you may need are given as follows:
For the one-way independent groups (and simple main effects):
2
1 ⎛ K 2 ⎞ T..
SSG = ⎜ ∑ T.j ⎟ −
= UG − U
n ⎝ j=1 ⎠ N
K n
1 ⎛ K 2⎞
2
SSS(G) =∑ ∑ X − ⎜ ∑ T.j ⎟ = ∑ ∑ Xij
− UG
n ⎝ j=1 ⎠
j=1 i=1
j=1 i=1
K
n
2
ij
For the two-way independent-groups ANOVA with equal n’s:
2
1 ⎛ A 2 ⎞ T…
SSA =
∑ T − = UA − U
Bn ⎜⎝ a=1 .a. ⎟⎠ N
SSB =
2
1 ⎛ B 2 ⎞ T…
T
−
= UB − U
∑
An ⎜⎝ b=1 ..b ⎟⎠ N
2
1⎛ A B 2 ⎞ 1 ⎛ A 2 ⎞
1 ⎛ B 2 ⎞ T…
SSAB = ⎜ ∑ ∑ T.ab ⎟ −
∑ T.a. −
∑ T..b + = U AB − U A − U B + U
n ⎝ a=1 b=1
⎠ Bn ⎜⎝ a=1 ⎟⎠ An ⎜⎝ b=1 ⎟⎠ N
A B n
A B n
1⎛ A B 2 ⎞
2
2
SSS(AB) =∑ ∑ ∑ Xiab
− ⎜ ∑ ∑ T.ab
=
Xiab
− U AB
∑
∑
∑
⎟
n ⎝ a=1 b=1
⎠ a=1 b=1 i=1
a=1 b=1 i=1
For planned and post-hoc tests:
t=
Q=
F=
X·1 − X·2
MSerror MSerror
+
n1
n2
=
X·1 − X·2
= 2t
MSerror
# obs
# obs X·1 − X·2
(
)=t
2 MSerror
SScomparison
(∑ w T )
=
2
j ·j
#obs ∑ w 2j
2
X·1 − X·2
2 MSerror
# obs
