1. CHAPTER 10
ANOVA
- One way ANOVa

2. 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.

3. 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

4. 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

5. 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:

6. 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

7. To calculate MSB and MSW, first calculate between samples sum of squares (SSB) and within-samples sum of squares (SSW)

8. 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 Method II Method III
T1 = 324
T2 = 369
T3 = 388
n1 = 5
n2 = 5
n3 = 5

9. Σx = T1 + T2 + T3 = = 1081
n = n1 + n2 + n3 = = 15
Σx2 = square of all the scores in the 3 samples and add them
= (48)2 + (73)2 + (51)2 + (65)2 + …..
= 80,709

10. 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 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.

11. How to produce the One-way ANOVA table for your research report (see my SPSS book p119)
How to report the results of your One-way ANOVA table (See my SPSS book p 122)

12. 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
State the null and the alternative hypothesis
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?