3. Example-3
Solve using Two-way ANOVA method
| Observation | A | B | C | | 1 | 1,4,0,7 | 13,5,7,15 | 9,16,18,13 | | 2 | 15,6,10,13 | 6,18,9,15 | 14,7,6,13 |
Solution:
Given problem
| Observation | `A` | `B` | `C` | | `1` | 1, 4, 0, 7 | 13, 5, 7, 15 | 9, 16, 18, 13 | | `2` | 15, 6, 10, 13 | 6, 18, 9, 15 | 14, 7, 6, 13 |
Row and column sums
| `A` | `B` | `C` | Row total `(x_(a))` | | `1` | 12 | 40 | 56 | `108` | | `2` | 44 | 48 | 40 | `132` | | Col total `(x_(b))` | `56` | `88` | `96` | `240` |
`sum x^2=1^2 + 4^2 + 0^2 + ... + 7^2 + 6^2 + 13^2=3010 ->(A)`
`(sum x_(b)^2)/(ra)=1/(4*2)(56^2+88^2+96^2)`
`=1/8(3136+7744+9216)`
`=1/8(20096)`
`=2512 ->(B)`
`(sum x_(a)^2)/(rb)=1/(4*3)(108^2+132^2)`
`=1/12(11664+17424)`
`=1/12(29088)`
`=2424 ->(C)`
`(sum sum x_(ab)^2)/r=1/4(12^2+40^2+56^2+44^2+48^2+40^2)`
`=1/4(144+1600+3136+1936+2304+1600)`
`=1/4(10720)`
`=2680 ->(C)`
`(sum x)^2/(rab)=(240)^2/(4*2*3)`
`=57600/24`
`=2400 ->(D)`
Sum of squares total
`SS_T=sum x^2 - (sum x)^2/(n)=(A)-(D)`
`=3010-2400`
`=610`
Sum of squares between rows
`SS_A=(sum x_(a)^2)/(rb) - (sum x)^2/(n)=(C)-(D)`
`=2424-2400`
`=24`
Sum of squares between columns
`SS_B=(sum x_(b)^2)/(ra) - (sum x)^2/(n)=(B)-(D)`
`=2512-2400`
`=112`
Sum of squares between columns
`SS_(AB)=(sum sum x_(ab)^2)/(r) - (sum x)^2/(n) - SSA - SSB = (B)-(D)- SSA - SSB`
`=2680-2400-24-112`
`=144`
Sum of squares Error (residual)
`SS_E=SS_T - SS_A - SS_B - SS_(AB)`
`=610-24-112-144`
`=330`
ANOVA table
| Source of Variation | Sums of Squares
SS | Degrees of freedom
DF | Mean Squares
MS | F | `p`-value | | A | `SS_A = 24` | `a-1 = 1` | `MS_R=24/1=24` | `24/18.3333=1.3091` | `0.2675` | | B | `SS_B = 112` | `b-1 = 2` | `MS_C=112/2=56` | `56/18.3333=3.0545` | `0.0721` | | AB | `SS_(AB) = 144` | `(a-1)(b-1) = 2` | `MS_(AB)=144/2=72` | `72/18.3333=3.9273` | `0.0384` | | Error (residual) | `SS_E = 330` | `rab-ab = 18` | `MS_E=330/18=18.3333` | | | | Total | `SS_T = 610` | `rab-1 = 23` | | | |
Conclusion:
1. F for between columns
`F(1,2)` at `0.05` level of significance
`=4.4139`
As calculated `F_R=1.3091 < 4.4139`
So, `H_0` is accepted, Hence there is no significant differentiating between rows
2. F for between columns
`F(2,2)` at `0.05` level of significance
`=3.5546`
As calculated `F_C=3.0545 < 3.5546`
So, `H_0` is accepted, Hence there is no significant differentiating between columns
This material is intended as a summary. Use your textbook for detail explanation.
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