CHAPTER 10
ANOVA
- One way ANOVa
One-way ANOVA
• - test the hypothesis that the means of
three or more populations are the same.
• The alternative hypothesis is that not all
population means are the same.
• Example: Students taught by 3 different methods
(Methods I, Methods II and Methods III) aimed at
improving their Mathematics scores
• The null hypothesis is: All three population
means are the equal
Ho: µ1 = µ2 = µ3
• The alternative hypothesis: Not all three
population means are equal
H1: Not all three means are equal
Assumptions of One-way ANOVA
• 1) the populations from which the samples
are drawn are (approximately) normally
distributed.
• 2) The populations from which the
samples are drawn have the same
variance (or standard deviation)
• 3) The samples drawn from different
populations are random and independent
• The ANOVA test is done by calculating 2 estimates of the
variance, α2, of the population distributions:
1) the variance between samples also called Mean
Square between samples or MSB
2) the variance within samples also called Mean
Square within samples, or MSW
The value of the F statistic for ANOVA is given by:
VarianceBe tweenSamples MSB
F

VarianceWi thinSamples
MSW
Symbols used
• x = score of students
• k = the number of different samples (or
treatments)
• ni = the size of sample i
• Ti = the sum of the values in sample i
• n = the number of values in all samples = n1 + n2
+ n3 + n4 + ….
• Σx = the sum of the values in all samples = T1 +
T2 + T3 + T4 + ….
• Σx2 = the sum of the squares of the values in all
samples
• To calculate MSB and MSW, first calculate
between samples sum of squares (SSB) and
within-samples sum of squares (SSW)
 T1 T
  x 
T
SSB   

 ...... 
n

n
n
n
2
3
 1

2
2
2
2
3
2
2
2
2


T
T
T
2
3
1
2
SSW   X   

 ......
 n1 n1 n3

SSB
MSB 
k 1
SSW
MSW 
nk
Example of One-way ANOVA Calculation
•
Fifteen students were taught using 3 different methods I, II and III and the
Mathematics scores obtained are as follows:
Method 1
48
73
51
65
87
Method II
55
85
70
69
90
Method III
84
68
95
74
67
T1 = 324
T2 = 369
T3 = 388
n1 = 5
n2 = 5
n3 = 5
Σx = T1 + T2 + T3 = 324 + 369 + 388 = 1081
n = n1 + n2 + n3 = 5 + 5 + 5 = 15
Σx2 = square of all the scores in the 3 samples and add them
= (48)2 + (73)2 + (51)2 + (65)2 + …..
= 80,709
 3242 369 2 3882  10812

SSB  


 432.133

5
5 
15
 5
 3242 369 2 3882 
  2372.8000
SSW  80,709  


5
5 
 5
SST  SSB  SSW  432.1333  2372.8000  2804.9333
MSB 
MSW 
SSB 432.1333

 216.0667
k 1
3 1
SSW 2372.8000

 197.7333
nk
15  3
MSB 216.0667
F

 1.09
MSW 197.7333
Degrees of freedom for the numerator of F = k – 1 = 3-1 = 2
Degrees of freedom for the denominator of F = n – k = 15 – 3 = 12
Look in the F distribution table:
For the 2 df for numerator and 12 df for the denominator, the critical value F is 6.93
Do not reject Ho
Reject Ho
α = .01
6.93
Critical value of F
DECISION: The F value calculated (1.09) is less than the critical value 6.93. It falls
In the non-rejection region. So, do not reject Ho. There are no significant
differences (p < .01) between the three groups. Which means that the treatment
(teaching of mathematics using the 3 different methods) do not affect the mathematics
Scores of students.
How to produce the One-way ANOVA table for your
research report (see my SPSS book p119)
EDUC
Sum of
Squares
Between Groups
df
Mean Square
432.133
2
216.067
Within Groups
2372.800
12
197.733
Total
2804.933
14
F
Sig.
1.090 NS
How to report the results of your One-way ANOVA
table
(See my SPSS book p 122)
Exercise 1
A Science teacher wanted to test three different methods of teaching
a difficult topic. She randomly selected 21 students and randomly divided
them into three groups and taught the three groups in 3 different ways. The
following are the marks obtained by the students:
Method I
Method II
Method III
15
11
9
8
16
17
17
9
11
7
16
8
26
24
15
12
20
6
8
19
14
a) State the null and the alternative hypothesis
b) Show the rejection and nonrejection regions on the F distribution curve for
α = .025
c) Calculate SSB, SSW and SST
d) What are the degrees of freedom for the numerator and the denominator
e) Calculate the between-samples and within-samples variances
f) What is the critical value of F for α = .025?
g) What is the calculated value of the test statistic F?
h) Write the ANOVA table for this exercise
i) Will you reject the null hypothesis stated at α = .025?