ANOVA (One way table)

1. ANOVA
Presented By : Sujay G Gowda
Guided by : Dr R K Prema

2.  ANOVA, or Analysis of Variance, is a statistical method used to compare
the means of three or more groups. It is a powerful tool for testing
hypotheses about whether there is a statistically significant difference
between the means of different groups, and it is widely used in a
variety of fields, including psychology, education, medicine, and
business.
ANOVA Meaning
 ANOVA works by partitioning the total variance in a dataset into two
components:
variance within groups
British statistician
Ronald Fisher
Limitation
 ANOVA can only tell you whether there is a statistically significant
difference between groups; it cannot tell you which groups are
different.
 If ANOVA rejects the null hypothesis, then you will need to conduct
further tests to determine which groups are different.

3. Example
 A researcher wants to test whether there is a difference in the effectiveness of three different teaching methods for
teaching students math. The researcher randomly assigns students to one of the three teaching methods and then
measures their math achievement at the end of the semester.The researcher can use ANOVA to compare the math
achievement scores of the students in the three different teaching methods. If ANOVA rejects the null hypothesis,
then the researcher can conclude that there is a statistically significant difference in the effectiveness of at least two
of the teaching methods.

4. Step 1: Calculate the mean for each group.
Step 2: Calculate the total mean.
Step 3: Calculate the SSB (Sum of Squares Between)
Step 4: Calculate the between groups degrees of freedom.
Step 5: Calculate the SSE. (Sum of squares error)
Step 6: Calculate the degrees of freedom of errors.
Step 7: Determine the MSB. (Mean square between)
Step 8: Determine the the MSE. (Mean square error)
Step 9 : ANOVA test statistic
STEPS IN ANOVA

5. Fertilizer 1 Fertilizer 2 Fertilizer 3
6 8 13
8 12 9
4 9 11
5 11 8
3 6 7
4 8 12
Three types of fertilizers are used on three groups of plants for 5 weeks. We want to check if there is a
difference in the mean growth of each group. Using the data given below apply a one way ANOVA test
at 0.05 significant level.

6. Fertilizer 1 Fertilizer 2 Fertilizer 3
6 8 13
8 12 9
4 9 11
5 11 8
3 6 7
4 8 12
𝑿𝒋 = 5 𝑿𝒋 = 9 𝑿j = 10
H0: μ1= μ2= μ3
H1: The means are not equal
Total mean, 𝑿 = 8
Here,
𝑿j = mean of the jth group,
𝑿 = overall mean
nj = sample size of the
jth group.
Step 1. Calculate the mean for each group.
Srep 2: Calculate the total mean.
Step 3: Calculate the SSB.
+ 6 (5 - 8)2
+ 6 (9 - 8)2
+ 6 (10 - 8)2
df1 = k -
1
SSB = ∑nj (𝑿𝒋 −
𝑿).
SSB = 84
Step 4: Calculate the between groups degrees of freedom.
df1 = 3 - 1 =
2

7. Fertilizer 1 (X - 5)2 Fertilizer
2
(X - 9)2 Fertilizer
3
(X - 10)2
6 1 8 1 13 9
8 9 12 9 9 1
4 1 9 0 11 1
5 0 11 4 8 4
3 4 6 9 7 9
4 1 8 1 12 4
𝑿 = 5 Total = 16 𝑿 = 9 Total = 24 𝑿 = 10 Total = 28
SSE = ∑(X−
𝑿𝒋).
Step 5: Calculate the SSE.
X = each data point in the jth group
𝑋j = mean of the jth group
SSE = 16 + 24 + 28 =
68
Step 6: Calculate the degrees of freedom of errors.
df2 = N -
k
df2 = 18 - 3 = 15

8. Using the f table at α= 0.05
the critical value is given as F(0.05, 2, 15) =
3.68
Step 7: Determine the MSB
MSB = SSB /
df1
= 84 / 2 =
42
Step 8: Determine the MSE.
MSE = SSE /
df2
= 68 / 15 = 4.53
Step 9: ANOVA test statistic.
f = MSB / MSE = 42 / 4.53 =
9.33
As f > F, thus, the null hypothesis is rejected and
it can be concluded that there is a difference in
the mean growth of the plants.

9. THANK
YOU.….