Solve using Two-way ANOVA method

Observation	A	B	C	D	E	F
1	1200	1000	980	900	750	800
2	1000	1100	700	800	500	700
3	890	650	1100	900	400	350

Solution:
Given problem

Observation	`A`	`B`	`C`	`D`	`E`	`F`
`1`	1200	1000	980	900	750	800
`2`	1000	1100	700	800	500	700
`3`	890	650	1100	900	400	350

	`A`	`B`	`C`	`D`	`E`	`F`	Row total `(x_(r))`
`1`	1200	1000	980	900	750	800	`5630`
`2`	1000	1100	700	800	500	700	`4800`
`3`	890	650	1100	900	400	350	`4290`
Col total `(x_(c))`	`3090`	`2750`	`2780`	`2600`	`1650`	`1850`	`14720`

`sum x^2=13010000 ->(A)`

`(sum x_(c)^2)/(r)=1/3(3090^2+2750^2+2780^2+2600^2+1650^2+1850^2)`

`=1/3(9548100+7562500+7728400+6760000+2722500+3422500)`

`=1/3(37744000)`

`=12581333.3333 ->(B)`

`(sum x_(r)^2)/(c)=1/6(5630^2+4800^2+4290^2)`

`=1/6(31696900+23040000+18404100)`

`=1/6(73141000)`

`=12190166.6667 ->(C)`

`(sum x)^2/(n)=(14720)^2/18`

`=216678400/18`

`=12037688.8889 ->(D)`

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

`=13010000-12037688.8889`

`=972311.1111`

Sum of squares between rows
`SS_R=(sum x_(r)^2)/(c) - (sum x)^2/(n)=(C)-(D)`

`=12190166.6667-12037688.8889`

`=152477.7778`

Sum of squares between columns
`SS_C=(sum x_(c)^2)/(r) - (sum x)^2/(n)=(B)-(D)`

`=12581333.3333-12037688.8889`

`=543644.4444`

Sum of squares Error (residual)
`SS_E=SS_T - SS_R - SS_C`

`=972311.1111-152477.7778-543644.4444`

`=276188.8889`

Note : You can find p-value using F Distribution calculator

ANOVA table

Source of Variation	Sum of Squares SS	df	Mean Squares MS	F	`p`-value
Between rows	`SS_R=152477.7778`	`r-1=2`	`152477.7778/2=76238.8889`	`76238.8889/27618.8889=2.7604`	FDist(2.7604,2,10) `=0.111`
Between columns	`SS_C=543644.4444`	`c-1=5`	`543644.4444/5=108728.8889`	`108728.8889/27618.8889=3.9368`	FDist(3.9368,5,10) `=0.0311`
Error (residual)	`SS_E=276188.8889`	`(r-1)(c-1)=10`	`276188.8889/10=27618.8889`
Total	`SS_T=972311.1111`	`rc-1=17`

Conclusion:
1. F for between rows
`F(2,10)` at `0.05` level of significance

`=4.1028`

As calculated `F_R=2.7604 < 4.1028`

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

2. F for between columns
`F(5,10)` at `0.05` level of significance

`=3.3258`

As calculated `F_C=3.9368 > 3.3258`

So, `H_0` is rejected, Hence there is significant differentiating between columns

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