1) The document discusses two examples of one-way and two-way ANOVA. In the first example, ANOVA is used to analyze test score data from four schools to determine if there are significant differences in performance between the schools. The F-value calculated is less than the critical value, so there are no significant differences. 2) The second example uses two-way ANOVA to analyze sales data from four salespeople across three seasons. ANOVA indicates there are no significant differences in performance between salespeople or across seasons, as the calculated F-values are less than the critical values. 3) In both examples, ANOVA is used to test the hypothesis that samples come from the same population by comparing the mean squares between

1. ANOVA
(Short-Cut
method)
Dr. R. MUTHUKRISHNAVENI
SAIVA BHANU KSHATRIYA
COLLEGE, ARUPPUKOTTAI

2. ANOVA – One way - Illustration
• To assess the significance of possible variation in performance in a certain test
between the convent schools of a city, a common test was given to a number of
students taken at random from the senior fifth class of each of the four schools
concerned, the results are given below. Make an analysis of variance of data
Schools
A B C D
8 12 18 13
10 11 12 9
12 9 16 12
8 14 6 16
7 4 8 15

3. Solution
• Let us take the hypothesis that there is no significance difference in performance in a
certain common test among convent schools of a certain city H0: µA= µB= µC = µD
Schools
A A2 B B2 C C2 D D2
8 64 12 144 18 324 13 169
10 100 11 121 12 144 9 81
12 144 9 81 16 256 12 144
8 64 14 196 6 36 16 256
7 49 4 16 8 64 15 225
45 421 50 558 60 824 65 875
𝑜𝑟 𝑇𝑜𝑡𝑎𝑙

4. • The sum of all items of Various samples = 𝐴 + 𝐵 + 𝐶 + 𝐷 =
45+50+60+65 =220
• Correction factor =
𝑇2
𝑁
=
220
20
=
18400
20
= 2420
• Sum of Square between the samples (SSC) =
( 𝐴)
2
𝑁
+
( 𝐵)
2
𝑁
+
( 𝐶)
2
𝑁
+
( 𝐷)
2
𝑁
-
𝑇2
𝑁
=
(45)2
20
+
(50)2
20
+
(60)2
20
+
(65)2
20
- 2420 = 2470 -2420 =50
• Total sum of square(SST)= 𝐴2 + 𝐵2 + 𝐶2 + 𝐷2 -
𝑇2
𝑁
=
421+558+824+875-2420 = 258
• Sum of square within samples = SST – SSC = 258 -50 = 208

5. • Inference
• The table value for F for V1 = 3 and V2 = 16 at 5% level of
significance = 3.24 . The CV of F is less than the TV, Hence
there is no significance difference in performance in a certain
common test among convent schools of a certain city (the
samples could have come from the same universe)
Source of
Variation
SS(Sum of
Squares)
V(Degrees of
Freedom
MS(Mean
Square)
Variance Ratio of
F
Between Samples 50 3 16.7
16.7
13.0
= 1.285
Within Samples 208 16 13.0
Total 258 19

7. Two Way ANOVA - Illustration
• A tea company appoints four salesmen - A,B,C and D and observes their sales in three seasons
– summer, winter and monsoon. The figures(in lakhs) are given in the following table
• (i) Do the salesman significantly differ in performance?
• (ii) Is there significant difference between the seasons?
Seasons Salesmen Season Total
A B C D
Summer 36 36 21 35 128
Winter 28 29 31 32 120
Monsoon 26 28 29 29 112
Salesmen total 90 93 81 96 360

8. Solution
• H0 1: There is no significant difference in performance of salesmen
• H0 2: There is no significant difference in seasons sales
• Grand Average =
360
12
=30
Seasons - 30 Salesmen - 30 Season Total
A B C D
Summer +6 +6 -9 +5 +8
Winter -2 -1 +1 +2 0
Monsoon -4 -2 -1 -1 -8
Salesmen total 0 +3 -9 +6 0

9. • Correction factor = =
𝑇2
𝑁
=
02
12
= 0
• Sum of Square between salesmen(
(column total)2
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑖𝑛 𝑡𝑜𝑡𝑎𝑙
-
correction factor)=
02+32+−92+62
3
− 0 =
0+9+81+36
3
=42
• Sum of Square between seasons(
(𝑅𝑜𝑤 𝑡𝑜𝑡𝑎𝑙)2
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑖𝑡𝑒𝑚𝑠 𝑖𝑛 𝑡𝑜𝑡𝑎𝑙
-
correction factor)=
82+02+−82
4
− 0 =
64+0+64
4
= 32
• Total sum of square = individual sample2 – Correction factor
= 62
+ −22
+ −42
+ 62
+ −12
+ 62
+ −22
+ −92
+ 12
+ 52
+ 22
+
−12 − 0 = 210

10. • Inference
• H0 - 1 There is no significant difference in performance of salesmen
• Table value for V1 = 3 and V3 = 6 at 5% level of significance is 4.76. CV < TV the
hypothesis is accepted
• H0 – 2 There is no significant difference in seasons sales
• Table value for V2 = 2 and V3 = 6 at 5% level of significance is 5.14. CV < TV the
hypothesis is accepted
Sequence of
Variance
Sum of
Squares
Degrees of
freedom
Mean of
Square
Ratio of F
Between columns 42 3 14 F =
𝑀𝑆𝐶
𝑀𝑆𝐸
=
14
22.67
= 1.619
Between Rows 32 2 16 F =
𝑀𝑆𝑅
𝑀𝑆𝐸
=
16
22.67
= 1.417
Residual or Error 136* 6* 22.67
Total 210 11
210 – 42--32
11 – 2 -3