Solve using One-way ANOVA method
| Observation | A | B | C |
| 1 | 25 | 31 | 24 |
| 2 | 30 | 39 | 30 |
| 3 | 36 | 38 | 28 |
| 4 | 38 | 42 | 25 |
| 5 | 31 | 35 | 28 |
Solution:
| `A` | `B` | `C` |
| 25 | 31 | 24 |
| 30 | 39 | 30 |
| 36 | 38 | 28 |
| 38 | 42 | 25 |
| 31 | 35 | 28 |
| `sum A=160` | `sum B=185` | `sum C=135` |
| `A^2` | `B^2` | `C^2` |
| 625 | 961 | 576 |
| 900 | 1521 | 900 |
| 1296 | 1444 | 784 |
| 1444 | 1764 | 625 |
| 961 | 1225 | 784 |
| `sum A^2=5226` | `sum B^2=6915` | `sum C^2=3669` |
Data table
| Group | `A` | `B` | `C` | Total |
| N | `n_1=5` | `n_2=5` | `n_3=5` | `n=15` |
| `sum x_i` | `T_1=sum x_1=160` | `T_2=sum x_2=185` | `T_3=sum x_3=135` | `sum x=480` |
| `sum x_(i)^2` | `sum x_1^2=5226` | `sum x_2^2=6915` | `sum x_3^2=3669` | `sum x^2=15810` |
| Mean `bar x_i` | `bar x_1=32` | `bar x_2=37` | `bar x_3=27` | Overall `bar x=32` |
| Std Dev `S_i` | `S_1=5.1478` | `S_2=4.1833` | `S_3=2.4495` | |
Let k = the number of different samples = 3
`n=n_1+n_2+n_3=5+5+5=15`
Overall `bar x=480/15=32`
`sum x=T_1+T_2+T_3=160+185+135=480 ->(1)`
`(sum x)^2/n=480^2/15=15360 ->(2)`
`sum T_i^2/n_i=(160^2/5+185^2/5+135^2/5)=15610 ->(3)`
`sum x^2=sum x_(1)^2+sum x_(2)^2+sum x_(3)^2=5226+6915+3669=15810 ->(4)`
ANOVA:
Step-1 : sum of squares between samples
`"SSB"= (sum T_i^2/n_i) - (sum x)^2/n = (3)-(2)`
`=15610-15360`
`=250`
Or
`"SSB"=sum n_j * (bar x_j - bar x)^2`
`=5xx(32-32)^2+5xx(37-32)^2+5xx(27-32)^2`
`=250`
Step-2 : sum of squares within samples
`"SSW"= sum x^2 - (sum T_i^2/n_i) = (4)-(3)`
`=15810-15610`
`=200`
Step-3 : Total sum of squares
`"SST"="SSB"+"SSW"`
`=250+200`
`=450`
Step-4 : variance between samples
`"MSB"=("SSB")/(k-1)`
`=250/(2)`
`=125`
Step-5 : variance within samples
`"MSW"=("SSW")/(n-k)`
`=200/(15-3)`
`=200/(12)`
`=16.6667`
Step-6 : test statistic F for one way ANOVA test
`F=("MSB")/("MSW")`
`=125/(16.6667)`
`=7.5`
the degree of freedom between samples
`k-1=2`
Now, degree of freedom within samples
`n-k=15-3=12`
ANOVA table
| Source of Variation | Sums of Squares
SS | Degrees of freedom
DF | Mean Squares
MS | F | `p`-value |
| Between samples | SSB = 250 | `k-1` = 2 | MSB = 125 | 7.5 | 0.0077 |
| Within samples | SSW = 200 | `n-k` = 12 | MSW = 16.6667 | | |
| Total | SST = 450 | `n-1` = 14 | | | |
`H_0` : There is no significant differentiating between samples
`H_1` : There is significant differentiating between samples
`F(2,12)` at `0.05` level of significance
`=3.8853`
As calculated `F=7.5 > 3.8853`
So, `H_0` is rejected, Hence there is significant differentiating between samples
This material is intended as a summary. Use your textbook for detail explanation.
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