1. 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)`
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`
ANOVA table| Source of Variation | Sums of Squares
SS | Degrees of freedom
DF | Mean Squares
MS | F | `p`-value |
| Between samples | SSB = 50 | `k-1` = 3 | MSB = 16.6667 | 1.2821 | 0.3144 |
| Within samples | SSW = 208 | `n-k` = 16 | MSW = 13 | | |
| Total | SST = 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
