Solve using One-way ANOVA method

Observation	A	B	C	D
1	8	12	18	13
2	10	11	12	9
3	12	9	16	12
4	8	14	6	16
5	7	4	8	15

Solution:

`A`	`B`	`C`	`D`
8	12	18	13
10	11	12	9
12	9	16	12
8	14	6	16
7	4	8	15
`sum A=45`	`sum B=50`	`sum C=60`	`sum D=65`

`A^2`	`B^2`	`C^2`	`D^2`
64	144	324	169
100	121	144	81
144	81	256	144
64	196	36	256
49	16	64	225
`sum A^2=421`	`sum B^2=558`	`sum C^2=824`	`sum D^2=875`

Data table

Group	`A`	`B`	`C`	`D`	Total
N	`n_1=5`	`n_2=5`	`n_3=5`	`n_4=5`	`n=20`
`sum x_i`	`T_1=sum x_1=45`	`T_2=sum x_2=50`	`T_3=sum x_3=60`	`T_4=sum x_4=65`	`sum x=220`
`sum x_(i)^2`	`sum x_1^2=421`	`sum x_2^2=558`	`sum x_3^2=824`	`sum x_4^2=875`	`sum x^2=2678`
Mean `bar x_i`	`bar x_1=9`	`bar x_2=10`	`bar x_3=12`	`bar x_4=13`	Overall `bar x=11`
Std Dev `S_i`	`S_1=2`	`S_2=3.8079`	`S_3=5.099`	`S_4=2.7386`

Let k = the number of different samples = 4
`n=n_1+n_2+n_3+n_4=5+5+5+5=20`

Overall `bar x=220/20=11`

`sum x=T_1+T_2+T_3+T_4=45+50+60+65=220 ->(1)`

`(sum x)^2/n=220^2/20=2420 ->(2)`

`sum T_i^2/n_i=(45^2/5+50^2/5+60^2/5+65^2/5)=2470 ->(3)`

`sum x^2=sum x_(1)^2+sum x_(2)^2+sum x_(3)^2+sum x_(4)^2=421+558+824+875=2678 ->(4)`

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ANOVA:
Step-1 : sum of squares between samples
`"SSB"= (sum T_i^2/n_i) - (sum x)^2/n = (3)-(2)`

`=2470-2420`

`=50`

Or
`"SSB"=sum n_j * (bar x_j - bar x)^2`

`=5xx(9-11)^2+5xx(10-11)^2+5xx(12-11)^2+5xx(13-11)^2`

`=50`

Step-2 : sum of squares within samples
`"SSW"= sum x^2 - (sum T_i^2/n_i) = (4)-(3)`

`=2678-2470`

`=208`

Step-3 : Total sum of squares
`"SST"="SSB"+"SSW"`

`=50+208`

`=258`

Step-4 : variance between samples
`"MSB"=("SSB")/(k-1)`

`=50/(3)`

`=16.6667`

Step-5 : variance within samples
`"MSW"=("SSW")/(n-k)`

`=208/(20-4)`

`=208/(16)`

`=13`

Step-6 : test statistic F for one way ANOVA test
`F=("MSB")/("MSW")`

`=16.6667/(13)`

`=1.2821`

the degree of freedom between samples
`k-1=3`

Now, degree of freedom within samples
`n-k=20-4=16`

p-value :
`p=FDist(F,df1,df2)=FDist(1.2821,3,16)=0.3144` (Using F Distribution calculator)

ANOVA table

Source of Variation	Sum of Squares SS	df	Mean Squares MS	F	`p`-value
Between samples	`SS_B=50`	`k-1=3`	`50/3=16.6667`	1.2821	0.3144
Within samples	`SS_W=208`	`n-k=16`	`208/16=13`
Total	`SS_T=258`	`n-1=19`

`H_0` : There is no significant differentiating between samples

`H_1` : There is significant differentiating between samples

`F(3,16)` at `0.05` level of significance

`=3.2389`

As calculated `F=1.2821 < 3.2389`

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

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