ADDITIONAL PP 15.1
Let’s first do a bare bones problem involving the one-way, independent groups ANOVA.
Twelve individuals are randomly sampled from the population and randomly assigned 4 each to
one of three groups. The conditions for each group are made as similar as possible, except for
the independent variable. The level of the independent variable differs for each group, with
Group 1 receiving the lowest level, Group 2 the next lowest, and Group 3 the highest. The
dependent variable is measured for each subject in each group. The table below shows the
results.
Group 1
5
2
1
4
a.
b.
c.
Group 2
6
4
2
3
Group 3
7
6
6
8
What is the alternative hypothesis?
What is the null hypothesis?
What is the conclusion? Use  = 0.05
SOLUTION is on next page.
SOLUTION
Group 1
X1
5
2
1
4
12
Group 2
X1
25
4
1
16
46
2
X2
6
4
2
3
15
n1 = 4
12
 3.00
X1 
4
all
scores
N = 12
a.
b.
c.
Group 3
X2
36
16
4
9
65
2
X3
7
6
6
8
27
n2 = 4
15
 3.75
X2 
4
 X  54
X32
49
36
36
64
185
n3 = 4
27
 6.75
X3 
4
all
scores
all
scores
 X 2  296
XG 
 X  54  4.50
N
12
Alternative hypothesis: The alternative hypothesis states that the independent variable has
an effect on the dependent variable. Therefore, at least one of the means (1, 2, or 3)
differs from at least one of the others.
Null hypothesis: The null hypothesis states that the independent variable does not have an
effect on the dependent variable. Therefore, the three sets of sample scores are random
samples from populations where 1 = 2 = 3.
Conclusion, using  = 0.05
A. Calculate Fobt.
STEP
1: Calculate the between-groups sum of squares, SSB.
 all

 scores 


  X 12  X 2 2  X 3 2    X 


SS B  

N
n2
n3 
 n1
 122 152 27 2  54



 31.5

4
4  12
 4
STEP
2: Calculate the within-groups sum of squares, SSW.
  X 12  X 2 2  X 3 2 



SS W   X  
n2
n3 
 n1
all
scores
2
 122 152 27 2 
 296  


  21.5
4
4
4


2
STEP
3: Calculate the total sum of squares, SST.
2
 all

 scores 
all
  X
2
scores

54 


2
 296 
 53.0
SS T   X 
N
12
Note, this step is a check on the calculations for SSB and SSW.
SST = SSB + SSW
53.0 = 31.5 + 21.5
53.0 = 53.0
The calculations are correct.
STEP
4: Calculate the degrees of freedom for each estimate.
dfB = k – 1 = 3 – 1 = 2
dfW = N – k = 12 – 3 = 9
dfT = N – 1 = 12 – 1 = 11
STEP
5: Calculate the between-groups variance estimate, sB2.
2
sB 
STEP
STEP
SS B 31.5

 15.750
2
2
6: Calculate the within-groups variance estimate, sW2.
2
sW 
SS W 21.5

 2.389
9
df W
F obt 
2
sB  15.750  6.59
2
2.389
sW
7: Calculate Fobt.
B. Evaluate Fobt. From Table F, with  = 0.05, dfnumerator = 2, and dfdenominator = 9,
Fcrit = 4.26
Since Fobt > Fcrit, we reject H0. The independent variable has had an effect on the
dependent variable. Increasing levels of the independent variable appear to increase the
value of the dependent variable. A summary of the solution is shown in the following
table.
Source
SS
df
Between groups
31.5
2
Within groups
21.5
9
Total
53.0
11
*With  = 0.05, Fcrit = 4.26. Therefore, H0 is rejected.
s2
15.750
2.389
Fobt
6.59*