Solve using Two-way ANOVA method

Observation	A	B	C
1	1,4,0,7	13,5,7,15	9,16,18,13
2	15,6,10,13	6,18,9,15	14,7,6,13

Solution:
Given problem

Observation	`A`	`B`	`C`
`1`	1, 4, 0, 7	13, 5, 7, 15	9, 16, 18, 13
`2`	15, 6, 10, 13	6, 18, 9, 15	14, 7, 6, 13

Row and column sums

	`A`	`B`	`C`	Row total `(x_(a))`
`1`	12	40	56	`108`
`2`	44	48	40	`132`
Col total `(x_(b))`	`56`	`88`	`96`	`240`

`sum x^2=1^2 + 4^2 + 0^2 + ... + 7^2 + 6^2 + 13^2=3010 ->(A)`

`(sum x_(b)^2)/(ra)=1/(4*2)(56^2+88^2+96^2)`

`=1/8(3136+7744+9216)`

`=1/8(20096)`

`=2512 ->(B)`

`(sum x_(a)^2)/(rb)=1/(4*3)(108^2+132^2)`

`=1/12(11664+17424)`

`=1/12(29088)`

`=2424 ->(C)`

`(sum sum x_(ab)^2)/r=1/4(12^2+40^2+56^2+44^2+48^2+40^2)`

`=1/4(144+1600+3136+1936+2304+1600)`

`=1/4(10720)`

`=2680 ->(C)`

`(sum x)^2/(rab)=(240)^2/(4*2*3)`

`=57600/24`

`=2400 ->(D)`

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Sum of squares total
`SS_T=sum x^2 - (sum x)^2/(n)=(A)-(D)`

`=3010-2400`

`=610`

Sum of squares between rows
`SS_A=(sum x_(a)^2)/(rb) - (sum x)^2/(n)=(C)-(D)`

`=2424-2400`

`=24`

Sum of squares between columns
`SS_B=(sum x_(b)^2)/(ra) - (sum x)^2/(n)=(B)-(D)`

`=2512-2400`

`=112`

Sum of squares between columns
`SS_(AB)=(sum sum x_(ab)^2)/(r) - (sum x)^2/(n) - SSA - SSB = (B)-(D)- SSA - SSB`

`=2680-2400-24-112`

`=144`

Sum of squares Error (residual)
`SS_E=SS_T - SS_A - SS_B - SS_(AB)`

`=610-24-112-144`

`=330`

ANOVA table

Source of Variation	Sums of Squares SS	Degrees of freedom DF	Mean Squares MS	F	`p`-value
A	`SS_A = 24`	`a-1 = 1`	`MS_R=24/1=24`	`24/18.3333=1.3091`	`0.2675`
B	`SS_B = 112`	`b-1 = 2`	`MS_C=112/2=56`	`56/18.3333=3.0545`	`0.0721`
AB	`SS_(AB) = 144`	`(a-1)(b-1) = 2`	`MS_(AB)=144/2=72`	`72/18.3333=3.9273`	`0.0384`
Error (residual)	`SS_E = 330`	`rab-ab = 18`	`MS_E=330/18=18.3333`
Total	`SS_T = 610`	`rab-1 = 23`

Conclusion:
1. F for between columns
`F(1,2)` at `0.05` level of significance

`=4.4139`

As calculated `F_R=1.3091 < 4.4139`

So, `H_0` is accepted, Hence there is no significant differentiating between rows

2. F for between columns
`F(2,2)` at `0.05` level of significance

`=3.5546`

As calculated `F_C=3.0545 < 3.5546`

So, `H_0` is accepted, Hence there is no significant differentiating between columns

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