2. Example-2
Solve using One-way ANOVA method
| Observation | A | B | C | | 1 | 8 | 7 | 6 | | 2 | 10 | 7 | 8 | | 3 | 6 | 8 | 10 | | 4 | 7 | 9 | 6 | | 5 | 9 | 8 | 4 | | 6 | 0 | 5 | 5 | | 7 | 0 | 0 | 7 |
Solution:
| `A` | `B` | `C` | | 8 | 7 | 6 | | 10 | 7 | 8 | | 6 | 8 | 10 | | 7 | 9 | 6 | | 9 | 8 | 4 | | 0 | 5 | 5 | | 0 | 0 | 7 | | `sum A=40` | `sum B=44` | `sum C=46` |
| `A^2` | `B^2` | `C^2` | | 64 | 49 | 36 | | 100 | 49 | 64 | | 36 | 64 | 100 | | 49 | 81 | 36 | | 81 | 64 | 16 | | 0 | 25 | 25 | | 0 | 0 | 49 | | `sum A^2=330` | `sum B^2=332` | `sum C^2=326` |
Data table
| Group | `A` | `B` | `C` | Total | | N | `n_1=5` | `n_2=6` | `n_3=7` | `n=18` | | `sum x_i` | `T_1=sum x_1=40` | `T_2=sum x_2=44` | `T_3=sum x_3=46` | `sum x=130` | | `sum x_(i)^2` | `sum x_1^2=330` | `sum x_2^2=332` | `sum x_3^2=326` | `sum x^2=988` | | Mean `bar x_i` | `bar x_1=8` | `bar x_2=7.3333` | `bar x_3=6.5714` | Overall `bar x=7.2222` | | Std Dev `S_i` | `S_1=1.5811` | `S_2=1.3663` | `S_3=1.9881` | |
Let k = the number of different samples = 3
`n=n_1+n_2+n_3=5+6+7=18`
Overall `bar x=130/18=7.2222`
`sum x=T_1+T_2+T_3=40+44+46=130 ->(1)`
`(sum x)^2/n=130^2/18=938.8889 ->(2)`
`sum T_i^2/n_i=(40^2/5+44^2/6+46^2/7)=944.9524 ->(3)`
`sum x^2=sum x_(1)^2+sum x_(2)^2+sum x_(3)^2=330+332+326=988 ->(4)`
ANOVA:
Step-1 : sum of squares between samples
`"SSB"= (sum T_i^2/n_i) - (sum x)^2/n = (3)-(2)`
`=944.9524-938.8889`
`=6.0635`
Or
`"SSB"=sum n_j * (bar x_j - bar x)^2`
`=5xx(8-7.2222)^2+6xx(7.3333-7.2222)^2+7xx(6.5714-7.2222)^2`
`=6.0635`
Step-2 : sum of squares within samples
`"SSW"= sum x^2 - (sum T_i^2/n_i) = (4)-(3)`
`=988-944.9524`
`=43.0476`
Step-3 : Total sum of squares
`"SST"="SSB"+"SSW"`
`=6.0635+43.0476`
`=49.1111`
Step-4 : variance between samples
`"MSB"=("SSB")/(k-1)`
`=6.0635/(2)`
`=3.0317`
Step-5 : variance within samples
`"MSW"=("SSW")/(n-k)`
`=43.0476/(18-3)`
`=43.0476/(15)`
`=2.8698`
Step-6 : test statistic F for one way ANOVA test
`F=("MSB")/("MSW")`
`=3.0317/(2.8698)`
`=1.0564`
the degree of freedom between samples
`k-1=2`
Now, degree of freedom within samples
`n-k=18-3=15`
ANOVA table
| Source of Variation | Sums of Squares
SS | Degrees of freedom
DF | Mean Squares
MS | F | `p`-value | | Between samples | SSB = 6.0635 | `k-1` = 2 | MSB = 3.0317 | 1.0564 | 0.3722 | | Within samples | SSW = 43.0476 | `n-k` = 15 | MSW = 2.8698 | | | | Total | SST = 49.1111 | `n-1` = 17 | | | |
`H_0` : There is no significant differentiating between samples
`H_1` : There is significant differentiating between samples
`F(2,15)` at `0.05` level of significance
`=3.6823`
As calculated `F=1.0564 < 3.6823`
So, `H_0` is accepted, Hence there is no significant differentiating between samples
This material is intended as a summary. Use your textbook for detail explanation.
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