1. Example of chai square test (1) false dice
When we roll dice 600 times the result is times of each pip is as follows,
Is this dice distorted by some one
1: : : : :80 6: 87
We presume the dice is accurate (Null hypothesis)
Expectation times of each pip is 100

2. 𝜒 2 = 𝑑𝑓 6−1=5

3. Chai square
probability upper limit
df df

4. Example of chai square test (2) Difference in effectiveness of medicine
There are 2 medicines. When we administered medicine A to 116 people, the medicine was effective to 66 persons. When we administered medicine B to 97 people, the medicine is effective to 56 persons.
The result is summarized as follow

5. data ratio Expectation value 122 213 91 213 116× 122 213 97× 122 213
97×
116×

6. Square of difference 𝑓 𝑖 − 𝑒 𝑖 2 𝑒 𝑖A B
observed
expected
difference
Effective 66 55
ineffective 50 41
Square of difference
𝑓 𝑖 − 𝑒 𝑖 𝑒 𝑖
Observed value of 𝜒 2 =
Degree of freedom 2−1 2−1 =1

7. Chai square
probability upper limit
df df

8. Example of chai square test (3) Social class
Social status is not richness. Even poorest people may think that they are belonging high society, because of their cultural refinement. It is relating to the culture of the area. Following of the data of interview survey asking social level of people. Which do you think you are belonging, high society, middle society or low society. The survey is implemented in A, B, C and D city. We want to know whether there are differences among cities.

9. Expectation value
Difference between observed value and expectation value

10. Square of difference
Calculation of 𝜒 2 value 𝑓 𝑖 − 𝑒 𝑖 𝑒 𝑖
Observed 𝜒 2 is
Degree of freedom of city is 4-1=3
Degree of freedom of social class is 3-1=2
So, the freedom of this dataset is (4-1)(3-1)=6

11. Chai square
probability upper limit
df df

12. ⋯ ⋯ Student’s t test Calculation of observed t t= 𝑀−𝜇 𝜎 𝑠𝑎𝑚𝑝𝑙𝑒 𝑛
when μ=0,
t= 𝑀 𝜎 𝑠𝑎𝑚𝑝𝑙𝑒 𝑛
Paired t
Comparison between two varieties planted in same pots ⋯ ⋯
Pot 1
Pot 2
Pot N

13. ⋮ 𝑑 𝑛 − = 𝑖=1 𝑛 𝑑 𝑖 Sum 𝑀= 1 𝑛 𝑖=1 𝑛 𝑑 𝑖 𝑆𝑆= 𝑖=1 𝑛 𝑑 𝑖 −𝑀 2
𝑀= 1 𝑛 𝑖=1 𝑛 𝑑 𝑖
𝑆𝑆= 𝑖=1 𝑛 𝑑 𝑖 −𝑀 2
𝜎 2 = 𝑆𝑆 𝑛−1 ⋮
S.E= 𝜎 𝑛
Pot n = −
𝑑 𝑛
t= 𝑀 𝑆.𝐸 = 𝑀 𝜎 𝑛
𝑖=1 𝑛 𝑑 𝑖
Sum

14. Threshold of t value at p≤0.01 df=4 is 4.604 t≥4.604
Example i A B
difference 1 3 2 5 4 7 6 9 n df
sum 8
average
1.6 SS
1.2
σ 2
0.3
𝑀=1.6
𝜎 𝑐 = 0.3
t= 𝑀 𝑐 𝜎 𝑐 𝑛 = =
Threshold of t value at p≤0.01 df=4 is t≥4.604
We can conclude that the difference is significant at p≤0.01 and A is larger than B

15. Unpaired t Fertilizer B 𝑛 𝐴 𝑛 𝐵 𝑑𝑓 𝐴 = 𝑛 𝐴 −1 𝑑𝑓 𝐵 = 𝑛 𝐵 −1
𝑑𝑓 𝐴 = 𝑛 𝐴 −1
𝑀 𝐴 = 1 𝑛 𝐴 𝑖=1 𝑛 𝐴 𝑑 𝐴𝑖
𝑆𝑆 𝐴 = 𝑖=1 𝑛 𝐴 𝑑 𝐴𝑖 − 𝑀 𝐴 2
𝜎 𝐴 2 = 𝑆𝑆 𝐴 𝑛 𝐴 −1
𝑛 𝐵
𝑑𝑓 𝐵 = 𝑛 𝐵 −1
𝑀 𝐵 = 1 𝑛 𝐵 𝑖=1 𝑛 𝐵 𝑑 𝐵𝑖
𝑆𝑆 𝐵 = 𝑖=1 𝑛 𝐵 𝑑 𝐵𝑖 − 𝑀 𝐵 2
𝜎 𝐵 2 = 𝑆𝑆 𝐵 𝑛 𝐵 −1

16. s 2 = 1 𝑚 𝜎 𝐴−𝐵 2 + 1 𝑛 𝜎 𝐴−𝐵 2 = 1 𝑚 + 1 𝑛 𝜎 𝐴−𝐵 2
Degree of freedom
𝑑𝑓 𝐴−𝐵 = 𝑑𝑓 𝐴 + 𝑑𝑓 𝐵 = 𝑛 𝐴 + 𝑛 𝐵 −2
Quadratic moment of around 𝜇 (center of parent population)
We presume homoscendasticity
𝜎 𝐴 2 = 𝜎 𝐵 2 = 𝜎 𝐴−𝐵 2 = 𝜎 𝐴+𝐵 2
E 𝑀 𝐴 2 = 1 𝑛 𝐴 𝜎 𝐴 2 = 1 𝑛 𝐴 𝜎 𝐴−𝐵 2
E 𝑀 𝐵 2 = 1 𝑛 𝐵 𝜎 𝐵 2 = 1 𝑛 𝐵 𝜎 𝐴−𝐵 2
s 2 =E 𝑀 𝐴−𝐵 2 =E 𝑀 𝐴+𝐵 2 =E 𝑀 𝐴 2 +E 𝑀 𝐵 2
s 2 = 1 𝑚 𝜎 𝐴−𝐵 𝑛 𝜎 𝐴−𝐵 2 = 1 𝑚 + 1 𝑛 𝜎 𝐴−𝐵 2
s= 𝜎 𝐴−𝐵 1 𝑚 + 1 𝑛
E 𝑀 𝐴−𝐵 2 = E 𝑀 𝐴 𝐸 𝑀 𝐴 𝐸 𝑀 𝐵 +E 𝑀 𝐵 2
= E 𝑀 𝐴 2 +E 𝑀 𝐵 2
∵𝐸 𝑀 𝐴 =𝐸 𝑀 𝐵 =0
From null hypothesis liner moment of observed average is 0.

17. Definition of standard error: 𝜎 𝐴−𝐵 𝑛 𝐴−𝐵 =𝑠,
𝜎 𝐴−𝐵 𝑛 𝐴−𝐵 = 𝜎 𝐴−𝐵 𝑛 𝐴 𝑛 𝐵
𝑛 𝐴−𝐵 = 𝑛 𝐴 𝑛 𝐵 = 𝑛 𝐴 𝑛 𝐵 𝑛 𝐴 + 𝑛 𝐵
𝑛 𝐴−𝐵 = 𝑛 𝐴 𝑛 𝐵 𝑛 𝐴 + 𝑛 𝐵
t= 𝑀 𝐴 − 𝑀 𝐵 𝜎 𝐴−𝐵 𝑛 𝐴 𝑛 𝐵

18. Variance of A-B is similar to that of A+B, it can be obtained as weighted mean by each degree of freedom.
𝜎 𝐴−𝐵 2 = 𝜎 𝐴+𝐵 2 = 𝑛 𝐴 −1 𝜎 𝐴 𝑛 𝐴 −1 + 𝑛 𝐵 − 𝑛 𝐵 −1 𝜎 𝐵 𝑛 𝐴 −1 + 𝑛 𝐵 −1
= 𝑛 𝐴 −1 𝜎 𝐴 𝑛 𝐵 −1 𝜎 𝐵 2 𝑛 𝐴 + 𝑛 𝐵 −2
𝜎 𝐴−𝐵 2 = 𝜎 𝐴+𝐵 2 = 𝑆𝑆 𝐴 + 𝑆𝑆 𝐵 𝑛 𝐴 + 𝑛 𝐵 −2

19. Example A B 1 5 6 8 n 3 4 df 2
sum 12 20
average SS 14 26
𝜎 2 7
8.6667
A : 𝑛 𝐴 =3, df 𝐴 =2, 𝑀 A ＝4, 𝑆𝑆 𝐴 =14
B： 𝑛 𝐵 =4, df 𝐵 =3, 𝑀 𝐵 =5, 𝑆𝑆 𝐵 =26
A-B :
𝑛 𝐴−𝐵 = 𝑛 𝐴 𝑛 𝐵 𝑛 𝐴 + 𝑛 𝐵 = 3×4 3+4 =
Df 𝐴−𝐵 = df 𝐴 + df 𝐵 =2+3=5

20. 𝑀 𝐴 − 𝑀 𝐵 =5−4=1
𝑆𝑆 𝐴−𝐵 = 𝑆𝑆 𝐴 + 𝑆𝑆 𝐵 =14+26=40
σ 𝐴−𝐵 2 = 𝑆𝑆 𝐴−𝐵 df 𝐴−𝐵 = 40 5 =8
σ 𝐴−𝐵 = 8
t= 𝑀 𝐴 − 𝑀 𝐵 𝜎 𝐴−𝐵 𝑛 𝐴−𝐵 = =
Threshold of t value (𝑝≤0.05) is 2.571
t<2.571
We cannot deny the null hypothesis that 𝑀 𝐴 − 𝑀 𝐵 =0, ( 𝑀 𝐴 = 𝑀 𝐵 ).
From this, we cannot conclude that averages of A and B is different, and we cannot say that sample populations A and B are taken from the different parent population.

21. F test ：Logical structure of F test is simple.
It is only comparison between two variables.
In the case of example of unpaired t test
We need to confirm equality of variances before t test.　　　　　　　 A B 1 5 6 8 n 3 4 df 2
sum 12 20
average SS 14 26
𝜎 2 7
8.6667
𝐹= 𝜎 𝐵 𝜎 𝐴 2 = =

22. Ｆdistribution　（𝑝=0.05 one side）f1 f
Theoretical threshold at df of denominator 2 and df of numerator is 3 is
Observed F is This value is far smaller than , we can conclude that there is no reason
to deny similarity of variances between denominator and numerator.

23. Technical difficulty of F test is separation of variances of data.
The variance of data is composed from several causes.
When we consider combined effect among plural causes the structure of
variance is sometimes complicated depending on the experimental or survey
design. However, there three categories of causes of variance
1. Main effect: variances caused by experimental conditions or categories
of data to compare. Difference explained by designed factor
2. Residuals: Variance caused by unknown factors. Randomly sampled data
include influence of unknown factors.
3. Interaction: Unexpectedly obtained variances caused by combination
of designed condition.

24. Experimental design (On way ANOVA)
𝑛 𝐴
𝑛 𝐵
𝑛 𝐶
Main effect: Differences of fertilizer ✔
Residuals: Randomness ✔
3. Interaction: not exist −

25. Data
F 1, F 2, ⋯ F 𝑔 𝑑 𝑑 ⋯ 𝑑 𝑔1
𝑑 𝑑 ⋯ 𝑑 𝑔2
⋮ ⋮ ⋱ ⋮
𝑑 1 𝑛 1
𝑑 1 𝑛 ⋮
𝑑 𝑔 𝑛 𝑔
Size 𝑛 𝑛 ⋯ 𝑛 𝑔
Sum 𝑖=1 𝑛 1 𝑑 1𝑖 , 𝑖=1 𝑛 2 𝑑 2𝑖 ⋯ 𝑖=1 𝑛 𝑔 𝑑 𝑔𝑖
Average 𝑀 1 = 1 𝑛 1 𝑖=1 𝑛 1 𝑑 1𝑖 , 𝑀 2 = 1 𝑛 2 𝑖=1 𝑛 2 𝑑 2𝑖 ⋯, 𝑀 𝑔 = 1 𝑛 𝑝 𝑖=1 𝑛 𝑔 𝑑 𝑔𝑖
SS SS 1 = 𝑖=1 𝑛 𝑑 1𝑖 − 𝑀 SS 2 = 𝑖=1 𝑛 𝑑 2𝑖 − 𝑀 ⋯ SS 𝑔 = 𝑖=1 𝑛 𝑑 𝑔𝑖 − 𝑀 𝑔 2
σ σ 1 2 = SS 1 𝑛 1 , σ 2 2 = SS 2 𝑛 2 , ⋯, σ 𝑔 2 = SS 𝑔 𝑛 𝑔

26. SS 𝑇 = 𝑗=1 𝑔 𝑖=1 𝑛 𝑗 𝑑 𝑗𝑖 − 𝑀 𝑇 2 , σ 𝑇 2 = SS 𝑇 𝑁−1
Total
𝑁= 𝑗=1 𝑔 𝑛 𝑗 , Sum 𝑇𝑜𝑡𝑙𝑎 = 𝑗=1 𝑔 𝑖=1 𝑛 𝑔 𝑑 𝑗𝑖 Average: 𝑀 𝑇 1 𝑁 𝑗=1 𝑔 𝑖=1 𝑛 𝑔 𝑑 𝑗𝑖
SS 𝑇 = 𝑗=1 𝑔 𝑖=1 𝑛 𝑗 𝑑 𝑗𝑖 − 𝑀 𝑇 , σ 𝑇 2 = SS 𝑇 𝑁−1
Separation of SS
𝑑 𝑗𝑖 − 𝑀 𝑇 = 𝑑 𝑗𝑖 − 𝑀 𝑗 + 𝑀 𝑗 − 𝑀 𝑇 = 𝑒 𝑖𝑗 + 𝐸 𝑗
𝑒 𝑖𝑗 : deviation of data from mean of subsample
𝐸 𝑗 : deviation of mean of subsample from total mean
= 𝑖=1 𝑛 𝑗 𝑑 𝑗𝑖 − 𝑀 𝑇 2 = 𝑖=1 𝑛 𝑗 𝑒 𝑖𝑗 + 𝐸 𝑗 2 = 𝑖=1 𝑛 𝑗 𝑒 𝑖𝑗 𝐸 𝑗 𝑖=1 𝑛 𝑗 𝑒 𝑖𝑗 + 𝑖=1 𝑛 𝑗 𝐸 𝑗 2
= 𝑖=1 𝑛 𝑗 𝑒 𝑖𝑗 2 + 𝑛 𝑗 𝐸 𝑗 2 = SS 𝑗 + 𝑛 𝑗 𝐸 𝑗 2
SS 𝑇 = 𝑗=1 𝑔 𝑖=1 𝑛 𝑗 𝑑 𝑗𝑖 − 𝑀 𝑇 2 = 𝑗=1 𝑔 SS 𝑗 + 𝑛 𝑗 𝐸 𝑗 = 𝑗=1 𝑔 SS 𝑗 + 𝑗=1 𝑔 𝑛 𝑗 𝐸 𝑗 2

27. SS 𝑇 = 𝑗=1 𝑔 SS 𝑗 + 𝑗=1 𝑔 𝑛 𝑗 𝐸 𝑗 2
𝑗=1 𝑔 𝑛 𝑗 𝐸 𝑗 2 = SS 𝑇 − 𝑗=1 𝑔 SS 𝑗
SS 𝑗 is residual removeing maineffect, randomness
SS 𝑇 is comosed from SSresidual and SS main effect
𝑗=1 𝑔 𝑛 𝑗 𝐸 𝑗 2 is maine SS of main effect
SS 𝑇 = 𝑆𝑆 𝑓𝑎𝑐𝑡𝑜𝑟 + 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
𝑆𝑆 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑗=1 𝑔 𝑛 𝑗 𝐸 𝑗 2
𝑑𝑓 𝑡𝑜𝑡𝑎𝑙 =𝑁−1= 𝑗=1 𝑔 𝑛 𝑗 −1
𝑑𝑓 𝑓𝑎𝑐𝑡𝑜𝑟 =𝑔−1
𝑑𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 = 𝑁−1 − 𝑔−1 = 𝑗=1 𝑔 𝑛 𝑗 −1 =𝑁−𝑔
𝜎 𝑓𝑎𝑐𝑡𝑜𝑟 2 = 𝑆𝑆 𝑓𝑎𝑐𝑡𝑜𝑟 𝑑𝑓 𝑓𝑎𝑐𝑡𝑜𝑟 , 𝜎 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 2 = 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 𝑑𝑓 𝑟𝑒𝑠𝑖𝑒𝑢𝑎𝑙
𝐹= 𝜎 𝑓𝑎𝑐𝑡𝑜𝑟 𝜎 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 2

28. Facilitation of calculation of sum of square
𝑆𝑆= 𝑖=1 𝑛 𝑥 𝑖 − 𝑥 2 = 𝑖=1 𝑛 𝑥 𝑖 2 −2 𝑖=1 𝑛 𝑥 𝑖 𝑥 + 𝑖=1 𝑛 𝑥 2 = 𝑖=1 𝑛 𝑥 𝑖 2 −2 𝑥 𝑖=1 𝑛 𝑥 𝑖 +𝑛 𝑥 2
= 𝑖=1 𝑛 𝑥 𝑖 2 − 𝑖=1 𝑛 𝑥 𝑖 𝑛 =𝑆− 𝑇 2 𝑛
∵ 𝑥 = 𝑖=1 𝑛 𝑥 𝑖 𝑛
𝑆 𝑖 = 𝑗=1 𝑛 𝑖 𝑥 𝑖𝑗 2 , 𝑇 𝑖 = 𝑗=1 𝑛 𝑖 𝑥 𝑖𝑗
𝑆𝑆 𝑖 = 𝑆 𝑖 − 𝑇 𝑖 2 𝑛 𝑖
𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 =𝑆 𝑆 1 +𝑆 𝑆 2 +…+𝑆 𝑆 1
= 𝑆 1 − 𝑇 𝑛 1 + 𝑆 2 − 𝑇 𝑛 2 +⋯+ 𝑆 𝑖 − 𝑇 𝑖 2 𝑛 𝑖 +⋯+ 𝑆 𝑛 − 𝑇 𝑛 2 𝑛 𝑛
𝑆− 𝑇 𝑛 𝑇 𝑛 2 +⋯+ 𝑇 𝑖 2 𝑛 𝑖 +⋯+ 𝑇 𝑛 2 𝑛 𝑛
𝑆= 𝑆 1 + 𝑆 2 +⋯+ 𝑆 𝑖 + ⋯+𝑆 𝑛
SS 𝑡𝑜𝑡𝑎𝑙 =S− 𝑇 2 𝑛 𝑡𝑜𝑡𝑎𝑙

29. SS 𝑡𝑜𝑡𝑎𝑙 = SS 𝑓𝑎𝑐𝑡𝑜𝑟 + SS 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
= 𝑇 𝑛 𝑇 𝑛 2 +⋯+ 𝑇 𝑖 2 𝑛 𝑖 +⋯+ 𝑇 𝑛 2 𝑛 𝑛 − 𝑇 2 𝑛
= 𝑖=1 𝑁 𝑇 𝑖 2 𝑛 𝑖 − 𝑇 2 𝑛 𝑡𝑜𝑡𝑎𝑙
SS 𝑓𝑎𝑐𝑡𝑜𝑟 = 𝑖=1 𝑁 𝑇 𝑖 2 𝑛 𝑖 − 𝑇 2 𝑛 𝑡𝑜𝑡𝑎𝑙

30. 𝐸𝑥𝑎𝑚𝑝𝑙𝑒
Sub population 1 2 3 4
sum 10 8 9 5 7 15 12 13 14
𝑛 𝑖 6 24 N
𝑇 𝑖 31 67 27 61
186 T
𝑥 𝑖
9.25
5.4
𝑆 𝑖
199
791
163
699
1852 S
𝑇 𝑖 2 𝑛 𝑖
145.8

31. 　 𝑛 𝑡𝑜𝑡𝑎𝑙 =24
T=186
S=1852
𝑖= 𝑇 𝑖 2 𝑛 𝑖 =
𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 =1852− =1852−1441.5=410.5
𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 =1852− =
𝑆𝑆 𝑓𝑎𝑐𝑡𝑜𝑟 = − = −1441.5=
df 𝑡𝑜𝑡𝑎𝑙 =24−1=23
　 df 𝑓𝑎𝑐𝑡𝑜𝑟 =4−1=3
df 𝑓𝑎𝑐𝑡𝑜𝑟 =23−3=20
𝜎 𝑓𝑎𝑐𝑡𝑜𝑟 2 = =41.973
　 𝜎 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 2 = =
F ratio= 𝜎 𝑓𝑎𝑐𝑡𝑜𝑟 𝜎 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 2 = ≈2.9498
Threshold of F in F distribution table (p=0.05) at df 𝑛𝑢𝑚𝑒𝑟𝑎𝑡𝑜𝑟 =3, df 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 =20 is
　Observed F=2.9498<3.0983
We cannot conclude that there is different among sub populations.

32. Sum of square degree of freedom mean square F 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 p F 𝑡ℎ𝑟𝑒𝑠ℎ𝑜𝑙𝑑
An example of expression of the result of analysis variance table
　　　　 Sum of square　degree of freedom　mean square F 𝑜𝑏𝑠𝑒𝑟𝑣𝑒𝑑 p F 𝑡ℎ𝑟𝑒𝑠ℎ𝑜𝑙𝑑
Source　　　　　 (SS) (df) (MS= σ 2 )
among population 　　　 　3　　　　 　　　
residual　　　　　 　　　

33. ⋮ ⋱ ⋯ Main effect: Differences of fertilizer ✔
Two way ANOVA without replication
Phosphate level ⋯ ⋮ ⋱
𝑁=1,
𝑃=1
𝑁=1,
𝑃=2
𝑁=1,
𝑃= 𝜑 𝑝
𝑁=1,
𝑃= 𝜑 𝑝
Combined impacts of
Phosphate level and nitrogen level
in fertilize on the growth of a plant
Nitrogen level
𝑁= 𝜑 𝑛 ,
𝑃=2
𝑁= 𝜑 𝑛 ,
𝑃= 𝜑 𝑝
𝑁= 𝜑 𝑛 ,
𝑃=1
Main effect: Differences of fertilizer ✔
Interaction + Residuals: Randomness+ ✔

34. Factor A size sum mean
Factor B ⋯ 𝑛 𝑎
𝑑 𝑑 ⋯ 𝑑 1 𝑛 𝑎 𝑛 𝑎 𝑇 𝑏1 = 𝑗=1 𝑛 𝑎 𝑑 1𝑗 𝑀 𝑏1 = 𝑇 𝑏1 𝑛 𝑎
𝑑 𝑑 ⋯ 𝑑 2 𝑛 𝑎 𝑛 𝑎 𝑇 𝑏2 = 𝑗=1 𝑛 𝑎 𝑑 2𝑗 𝑀 𝑏2 = 𝑇 𝑏2 𝑛 𝑎
⋮ ⋮ ⋮ ⋱ ⋮ ⋮ ⋮ ⋮
𝑛 𝑏 𝑑 𝑛 𝑏 𝑑 𝑛 𝑏 2 ⋯ 𝑑 𝑛 𝑏 𝑛 𝑎 𝑛 𝑎 𝑇 𝑏 𝑛 𝑏 = 𝑗=1 𝑛 𝑎 𝑑 𝑛 𝑏𝑗 𝑀 𝑏 𝑛 𝑏 = 𝑇 𝑏 𝑛 𝑏 𝑛 𝑎
Size 𝑛 𝑏 𝑛 𝑏 ⋯ 𝑛 𝑏 − 𝑁= 𝑛 𝑎 𝑛 𝑏 −
Sum 𝑇 𝑎 𝑇 𝑎 𝑇 𝑎 𝑛 𝑎 𝑁= 𝑛 𝑎 𝑛 𝑏 − 𝑇 𝑛 𝑎
Average 𝑀 𝑎1 𝑀 𝑎2 ⋯ 𝑀 𝑎 𝑛 𝑎 − 𝑇 𝑛 𝑏 𝑀= 𝑇 𝑛 𝑎
𝑇 𝑎𝑗 = 𝑖=1 𝑛 𝑏 𝑑 𝑖𝑗 , 𝑀 𝑎𝑗 = 𝑇 𝑎𝑗 𝑛 𝑏 , 𝑇= 𝑖=1 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝑑 𝑖𝑗
𝑀 𝑎 = 1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑀 𝑎𝑗 , 𝑀 𝑏 = 1 𝑛 𝑏 𝑗=1 𝑛 𝑏 𝑀 𝑏𝑗

35. 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑑 𝑖𝑗 −𝑀 2
𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑑 𝑖𝑗 −𝑀 2
= 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑑 𝑖𝑗 − 𝑀 𝑎𝑗 −𝑀 + 𝑀 𝑏𝑖 −𝑀 +𝑀 + 𝑀 𝑎𝑗 −𝑀 + 𝑀 𝑏 −𝑀 2
𝑒 𝑖𝑗 =𝑑 𝑗𝑘 − 𝑀 𝑎𝑗 −𝑀 + 𝑀 𝑏𝑖 −𝑀 +𝑀
𝐸 𝑎𝑗 = 𝑀 𝑎𝑗 −𝑀, 𝐸 𝑏𝑖 = 𝑀 𝑎𝑖 −𝑀
𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑒 𝑖𝑗 + 𝐸 𝑎𝑗 + 𝐸 𝑏𝑖 2
= 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑒 𝑖𝑗 𝐸 𝑎𝑗 𝐸 𝑏𝑖 𝑒 𝑖𝑗 𝐸 𝑎𝑗 + 𝐸 𝑎𝑗 𝐸 𝑏𝑖 + 𝐸 𝑏𝑖 𝑒 𝑖𝑗
= 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑒 𝑖𝑗 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝐸 𝑎𝑗 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝐸 𝑏𝑖 2
= 𝑆𝑆 𝑟𝑒𝑠𝑖𝑢𝑑𝑢𝑎𝑙 + 𝑛 𝑏 𝑆𝑆 𝑎 + 𝑛 𝑎 𝑆𝑆 𝑏
𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑒 𝑖𝑗 𝐸 𝑎𝑗 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝐸 𝑎𝑗 𝐸 𝑏𝑖 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝐸 𝑏𝑖 𝑒 𝑖𝑗 =0
𝜎 𝑎 2 = 𝑆𝑆 𝑎 𝑑𝑓 𝑎 , 𝜎 𝑏 2 = 𝑆𝑆 𝑏 𝑑𝑓 𝑏 𝜎 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 2 = 𝑆𝑆 𝑟𝑒𝑖𝑑𝑢𝑎𝑙 𝑑𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
𝐹 𝑎 = 𝜎 𝑎 𝜎 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 2 , 𝐹 𝑏 = 𝜎 𝑏 𝜎 𝑟𝑒𝑠𝑖𝑐𝑢𝑎𝑙 2

36. Facilitation of calculation
𝑛 𝑏 𝑆𝑆 𝑎 = 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 − 𝑆𝑆 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 + 𝑛 𝑎 𝑆𝑆 𝑏
𝑛 𝑎 𝑆𝑆 𝑏 = 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 − 𝑆𝑆 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 + 𝑛 𝑏 𝑆𝑆 𝑎
We calculate SS in a row
𝑆𝑆 𝑟𝑖 = 𝑗=1 𝑛 𝑎 𝑑 𝑖𝑗 − 1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑑 𝑖𝑗 2
𝑗=1 𝑛 𝑎 𝑑 𝑖𝑗 = 𝑗=1 𝑛 𝑎 𝑒 𝑖𝑗 + 𝐷 𝑏𝑖 + 𝐷 𝑎𝑗 +𝑀 = 𝑗=1 𝑛 𝑎 𝑒 𝑖𝑗 + 𝑛 𝑎 𝐷 𝑏𝑖 + 𝑛 𝑎 𝑀+ 𝑗=1 𝑛 𝑎 𝐷 𝑎𝑗 = 𝑗=1 𝑛 𝑎 𝑒 𝑖𝑗 + 𝑛 𝑎 𝐷 𝑏𝑖 + 𝑛 𝑎 𝑀
𝑗=1 𝑛 𝑎 𝐷 𝑎𝑗 =0
1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑑 𝑖𝑗 = 1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑒 𝑖𝑗 + 𝐷 𝑏𝑖 +𝑀
𝑑 𝑖𝑗 − 1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑑 𝑖𝑗 = 𝑒 𝑖𝑗 + 𝐷 𝑏𝑖 + 𝐷 𝑎𝑗 +𝑀− 1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑒 𝑖𝑗 + 𝐷 𝑏𝑖 +𝑀 = 𝑒 𝑖𝑗 − 1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑒 𝑖𝑗 + 𝐷 𝑎𝑗 = 𝜀 𝑖𝑗 + 𝐷 𝑎𝑗
∵𝜀 𝑖𝑗 = 𝑒 𝑖𝑗 − 1 𝑛 𝑎 𝑗=1 𝑛 𝑎 𝑒 𝑖𝑗

37. 𝑆𝑆 𝑟𝑖 = 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 + 𝐷 𝑎𝑗 2 = 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 2 +2 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 𝐷 𝑎𝑗 + 𝑗=1 𝑛 𝑎 𝐷 𝑎𝑗 2
𝑖=1 𝑛 𝑏 𝑆𝑆 𝑟𝑖 = 𝑖=1 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 𝑖=1 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 𝐷 𝑎𝑗 + 𝑖=1 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝐷 𝑎𝑗 2 = 𝑖=1 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝐷 𝑎𝑗 2
∵ 𝑖=1 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 𝐷 𝑎𝑗 =0
𝑆𝑆 𝑟 = 𝑖=1 𝑛 𝑏 𝑆𝑆 𝑟𝑖 , 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 = 𝑖=1 𝑛 𝑏 𝑗=1 𝑛 𝑎 𝜀 𝑖𝑗 2
𝑆𝑆 𝑟 = 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 + 𝑛 𝑏 𝑆𝑆 𝐴
Similarly
𝑆𝑆 𝑐 = 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 + 𝑛 𝑎 𝑆𝑆 𝐵
𝑆𝑆 𝑐 ;𝑆𝑆 𝑜𝑓 𝑐𝑜𝑙𝑢𝑚𝑛
𝑛 𝑏 𝑆𝑆 𝐴 = 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 − 𝑆𝑆 𝑐
𝑛 𝑎 𝑆𝑆 𝐵 = 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 − 𝑆𝑆 𝑟

38. Example of dataset
A: contents of phosphate
B: contents of nitrogen A1 A2 A3 B1 11 8 B2 10 13 19 B3 9 18 B4 14

39. Calculation process A1 A2 A3 𝑛 𝑇 𝑖 𝑆 𝑖 𝑖=1 𝑛 𝑇 𝑖 2 𝑛
𝑖=1 𝑛 𝑇 𝑖 2 𝑛
𝑆 𝑖 − 𝑖=1 𝑛 𝑇 𝑖 2 𝑛 B1 11 8 3 30
306
300 6 B2 10 13 19 42
630
588 B3 9 18 45
729
675 54 B4 14 51
881
867 4 12
168
2546
2352
116
𝑇 𝑗 44 60 64
𝑆 𝑗
498
938
1110
𝑗=1 𝑛 𝑇 𝑗 2 𝑛
484
900
1024
𝑆 𝑗 − 𝑗=1 𝑛 𝑇 𝑗 2 𝑛 38 86
138
194
𝑆𝑆 𝑟 = 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 + 𝑛 𝑏 𝑆𝑆 𝐴
𝑆𝑆 𝑐 = 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 + 𝑛 𝑎 𝑆𝑆 𝐵
𝑆𝑆 𝑡𝑜𝑡𝑎𝑙
𝑆𝑆 𝑐 = 𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 + 𝑛 𝑎 𝑆𝑆 𝐵

40. 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 =194
4 𝑆𝑆 𝐴 =194−138=56
3 𝑆𝑆 𝐵 =194−116=78
𝑆𝑆 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =194−56−78=60
𝑆𝑆 𝐴 = 56 4 =14
𝑆𝑆 𝐵 = 78 3 =26

41. Analysis of variance table
Source　　　 SS df MS Observed F p threshold F
A　　　　 　　　　　　2 　　　8　　　　
B 　
Residual
Sum　　
Source　　　 SS df MS Observed F p
A　　　　 　　　　　　2 　　　8　　　　 nnnn
B 　 nnnn
Residual
Sum　　

42. ⋯ ⋯ ⋮ ⋮ ⋮ ⋱ ⋯ N=1, P=1 N=1, P=2 N=1, P= 𝑛 𝑝 N=2, P=1 N=2, P=2
Two way ANOVA with replications
N=1, P=1 ⋯
N=1, P=2
N=1, P= 𝑛 𝑝 ⋯
N=2, P=1
N=2, P=2
N=2, P= 𝑛 𝑝 ⋮ ⋮ ⋮ ⋱ ⋯
N= 𝑛 𝑛 , P= 𝑛 𝑝
N= 𝑛 𝑛 , P=1
N= 𝑛 𝑛 , P=2

43. Factor A
Factor B ⋯ 𝑛 𝑎 sum mean
𝑑 𝑑 ⋯ 𝑑 1 𝑛 𝑎 1
𝑑 𝑑 ⋯ 𝑑 1 𝑛 𝑎 2
⋮ ⋮ ⋱ ⋮
𝑑 11 𝑛 𝑔 𝑑 12 𝑛 𝑔 ⋯ 𝑑 1 𝑛 𝑎 𝑛 𝑔
sum 𝑇 𝑇 ⋯ 𝑇 1 𝑛 𝑎 𝑇 𝐵1
mean 𝑇 11 𝑛 𝑔 𝑇 12 𝑛 𝑔 ⋯ 𝑇 12 𝑛 𝑔 𝑇 𝐵1 𝑛 𝑔 𝑛 𝑎
⋮ ⋮
𝑛 𝑏 𝑑 𝑛 𝑏 𝑑 𝑛 𝑏 ⋯ 𝑑 𝑛 𝑏 𝑛 𝑎 1
𝑑 𝑛 𝑏 𝑑 𝑛 𝑏 ⋯ 𝑑 𝑛 𝑏 𝑛 𝑎 2
⋮ ⋮ ⋱ ⋮
𝑑 𝑛 𝑏 1 𝑛 𝑔 𝑑 𝑛 𝑏 2 𝑛 𝑔 ⋯ 𝑑 𝑛 𝑏 𝑛 𝑎 𝑛 𝑔
sum 𝑇 𝑛 𝑏 𝑇 𝑛 𝑏 ⋯ 𝑇 𝑛 𝑏 𝑛 𝑎 𝑇 𝐵 𝑛 𝑏
mean 𝑇 𝑛 𝑏 1 𝑛 𝑔 𝑇 𝑛 𝑏 2 𝑛 𝑔 ⋯ 𝑇 𝐵 𝑛 𝑏 𝑛 𝑔 𝑛 𝑎
Sum 𝑇 𝐴 𝑇 𝐴 ⋯ 𝑇 𝐴 𝑛 𝑎 T
Mean 𝑇 𝐴1 𝑛 𝑔 𝑛 𝑎 𝑇 𝐴2 𝑛 𝑔 𝑛 𝑎 ⋯ 𝑇 𝐴 𝑛 𝑎 𝑛 𝑔 𝑛 𝑎 𝑇 𝑛 𝑔 𝑛 𝑎 𝑛 𝑏

44. 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 1 𝑖𝑗𝑘 −𝑀 2
𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 𝑖𝑗𝑘 −𝑀 2
𝑑 𝑖𝑗𝑘 −𝑀= 𝑑 𝑖𝑗𝑘 − 𝜀 𝑖𝑗 − 𝜀 𝑖 − 𝑀 𝑎𝑗 −𝑀 + 𝑀 𝑏𝑖 −𝑀 +𝑀 + 𝜀 𝑖𝑗 − 𝜀 𝑖 + 𝑀 𝑎𝑗 −𝑀 + 𝑀 𝑏 −𝑀
𝛿 𝑖𝑗𝑘 =𝑑 𝑖𝑗𝑘 − 𝜀 𝑖𝑗 − 𝜀 𝑖 + 𝑀 𝑎𝑗 −𝑀 + 𝑀 𝑏𝑖 −𝑀 +𝑀
𝜖 𝑖𝑗 = 𝜀 𝑖𝑗 − 𝜀 𝑖 , 𝐸 𝑎𝑗 = 𝑀 𝑎𝑗 −𝑀, 𝐸 𝑏𝑖 = 𝑀 𝑎𝑖 −𝑀,
𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 𝑒 𝑖𝑗𝑘 + 𝜖 𝑖𝑗 + 𝐸 𝑎𝑗 + 𝐸 𝑏𝑖 2
𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 𝑒 𝑖𝑗𝑘 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 𝜖 𝑖𝑗 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 𝐸 𝑎𝑗 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 𝐸 𝑏𝑖 2
𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑘=1 𝑛 𝑔 𝑒 𝑖𝑗𝑘 𝑛 𝑔 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝜖 𝑖𝑗 𝑛 𝑏 𝑛 𝑔 𝑗=1 𝑛 𝑎 𝐸 𝑎𝑗 𝑛 𝑏 𝑛 𝑔 𝑖=1 𝑛 𝑏 𝐸 𝑏𝑗 2
= 𝑆𝑆 𝑟𝑒𝑠𝑖𝑢𝑑𝑢𝑎𝑙 + 𝑛 𝑔 𝑆𝑆 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛(𝑎×𝑏) + 𝑛 𝑏 𝑛 𝑔 𝑆𝑆 𝑎 + 𝑛 𝑎 𝑛 𝑔 𝑆𝑆 𝑏
𝜎 𝑎 2 = 𝑆𝑆 𝑎 𝑑𝑓 𝑎 , 𝜎 𝑏 2 = 𝑆𝑆 𝑏 𝑑𝑓 𝑏 𝜎 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 2 = 𝑆𝑆 𝑟𝑒𝑖𝑑𝑢𝑎𝑙 𝑑𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 , 𝜎 (𝑎×𝑏) 2 = 𝑆𝑆 (𝑎×𝑏) 𝑑𝑓 (𝑎×𝑏) ,
𝐹 𝑎×𝑏 /𝑟 = 𝜎 (𝑎×𝑏) 𝜎 𝑟 2 , 𝐹 𝑎/𝑟 = 𝜎 𝑎 𝜎 𝑟 2 , 𝐹 𝑏/𝑟 = 𝜎 𝑏 𝜎 𝑟 𝐹 𝑎/ 𝑎×𝑏 = 𝜎 𝑎 𝜎 𝑎×𝑏 2 , 𝐹 𝑏/ 𝑎×𝑏 = 𝜎 𝑏 𝜎 𝑎×𝑏 2

45. Total degree of freedom: df 𝑡𝑜𝑡𝑎𝑙 = 𝑛 𝑎 𝑛 𝑏 𝑛 𝑔 −1
Degree of freedom among A: df 𝐴 = 𝑛 𝑎 −1
Degree of freedom among B: df 𝐵 = 𝑛 𝑏 −1
Degree of freedom of interaction: df 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =( 𝑛 𝑎 −1)( 𝑛 𝑏 −1)
Degree of freedom of residual: df 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 = 𝑛 𝑏 𝑛 𝑏 ( 𝑛 𝑔 −1)
𝑛 𝑎 𝑛 𝑏 𝑛 𝑔 −1= 𝑛 𝑎 −1 + 𝑛 𝑏 −1 + 𝑛 𝑎 −1 𝑛 𝑏 −1 + 𝑛 𝑏 𝑛 𝑏 ( 𝑛 𝑔 −1)

46. Sum of SS in the Row: SS 𝑡𝑜𝑡𝑎𝑙−𝐵 𝑛 𝑎 𝑛 𝑔 SS 𝐵 = SS 𝑡𝑜𝑡𝑎𝑙 − SS 𝑡𝑜𝑡𝑎𝑙−𝐵
Facilitation of calculation
𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑆 𝑡𝑜𝑡𝑎𝑙 − 𝑇 2 𝑛 𝑡𝑜𝑡𝑎𝑙
SS 𝑡𝑜𝑡𝑎𝑙 = 𝑛 𝑏 𝑛 𝑔 SS 𝐴 + 𝑛 𝑎 𝑛 𝑔 SS 𝐵 + 𝑛 𝑔 SS 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 + SS 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
=sum of sum of squares of replications in each cell (combnation of condtions)
= 𝑗=1 𝑛 𝑎 𝑖=1 𝑛 𝑏 𝑠𝑠 𝑖𝑗
Sum of SS in the Row: SS 𝑡𝑜𝑡𝑎𝑙−𝐵
𝑛 𝑎 𝑛 𝑔 SS 𝐵 = SS 𝑡𝑜𝑡𝑎𝑙 − SS 𝑡𝑜𝑡𝑎𝑙−𝐵
Sum of SS in the columns: SS 𝑡𝑜𝑡𝑎𝑙−𝐴
𝑛 𝑎 𝑛 𝑔 SS 𝐴 = SS 𝑡𝑜𝑡𝑎𝑙 − SS 𝑡𝑜𝑡𝑎𝑙−𝐴
𝑛 𝑔 SS 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = SS 𝑡𝑜𝑡𝑎𝑙 − 𝑛 𝑏 𝑛 𝑔 SS 𝐴 + 𝑛 𝑎 𝑛 𝑔 SS 𝐵 − SS 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙  

48. 𝑆𝑆 𝑡𝑜𝑡𝑎𝑙 = 𝑆 𝑡𝑜𝑡𝑎𝑙 − 𝑇 2 𝑛 𝑡𝑜𝑡𝑎𝑙 =7728− 504 2 36 =7728−7056=672
SS 𝑡𝑜𝑡𝑎𝑙 =4×3× SS 𝐴 +3×3× SS 𝐵 +3× SS 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 +SS 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
𝑆𝑆 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙
=sum of sum of squares of replications in each cell (combnation of cndtions)
= 𝑖=1 4 𝑗=1 3 𝑠𝑠 𝑖𝑗 = =90
Sum of SS in the Row: SS 𝑡𝑜𝑡𝑎𝑙−𝐵 = =438
3×3×SS 𝐵 = SS 𝑡𝑜𝑡𝑎𝑙 − SS 𝑡𝑜𝑡𝑎𝑙−𝐵 =672−438=234
Sum of SS in the columns: SS 𝑡𝑜𝑡𝑎𝑙−𝐴 = =504
4 ×3×SS 𝐴 = SS 𝑡𝑜𝑡𝑎𝑙 − SS 𝑡𝑜𝑡𝑎𝑙−𝐴 =672−504=168
3× SS 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = SS 𝑡𝑜𝑡𝑎𝑙 −4×3× SS 𝐴 − 3×3×SS 𝐵 − SS 𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 −168−234−90=180
SS 𝐵 = =26
SS 𝐴 = =14
SS 𝑖𝑛𝑡𝑒𝑟𝑎𝑐𝑡𝑖𝑜𝑛 = =60

49. Source　 SS df MS 𝐹 0𝑏𝑠𝑒𝑟𝑣𝑒𝑑 1 𝐹 𝑡ℎ𝑟𝑒𝑠ℎ𝑜𝑙𝑑 1 𝐹 0𝑏𝑠𝑒𝑟𝑣𝑒𝑑 2 𝐹 𝑡ℎ𝑟𝑒𝑠ℎ𝑜𝑙𝑑 2
A　　 　　　2 　　　 7 　　　　
B 　　　
Interaction
residual　　
Sum

50. Importance of confirmation of equality of variance (Particularly ratio data)