Homework Answers
Given problem
| Observation | A | B | C |
| 1 | 60, 51, 57 | 66, 69, 63 | 66, 54, 60 |
| 2 | 48, 51, 54 | 60, 54, 57 | 72, 69, 69 |
Row and column sums
| A | B | C | Row total (xa) | |
| 1 | 168 | 198 | 180 | 546 |
| 2 | 153 | 171 | 210 | 534 |
| Col total (xb) | 321 | 369 | 390 | 1080 |
∑x2=602+512+572+...+722+692+692=65700→(A)
∑x2bra=13⋅2(3212+3692+3902)
=16(103041+136161+152100)
=16(391302)
=65217→(B)
∑x2arb=13⋅3(5462+5342)
=19(298116+285156)
=19(583272)
=64808→(C)
∑∑x2abr=13(1682+1982+1802+1532+1712+2102)
=13(28224+39204+32400+23409+29241+44100)
=13(196578)
=65526→(C)
(∑x)2rab=(1080)23⋅2⋅3
=116640018
=64800→(D)
Sum of squares total
SST=∑x2-(∑x)2n=(A)-(D)
=65700-64800
=900
Sum of squares between rows
SSA=∑x2arb-(∑x)2n=(C)-(D)
=64808-64800
=8
Sum of squares between columns
SSB=∑x2bra-(∑x)2n=(B)-(D)
=65217-64800
=417
Sum of squares between columns
SSAB=∑∑x2abr-(∑x)2n-SSA-SSB=(B)-(D)-SSA-SSB
=65526-64800-8-417
=301
Sum of squares Error (residual)
SSE=SST-SSA-SSB-SSAB
=900-8-417-301
=174
ANOVA table
| Source of Variation | Sums of Squares SS |
Degrees of freedom DF |
Mean Squares MS |
F | p-value |
| A | SSA=8 | a-1=1 | MSR=81=8 | 814.5=0.5517 | 0.4719 |
| B | SSB=417 | b-1=2 | MSC=4172=208.5 | 208.514.5=14.3793 | 0.0007 |
| AB | SSAB=301 | (a-1)(b-1)=2 | MSAB=3012=150.5 | 150.514.5=10.3793 | 0.0024 |
| Error (residual) | SSE=174 | rab-ab=12 | MSE=17412=14.5 | ||
| Total | SST=900 | rab-1=17 |
Conclusion:
1. F for between columns
F(1,2) at 0.05 level of significance
=4.7472
As calculated FR=0.5517<4.7472
So, H0 is accepted, Hence there is no significant differentiating
between rows
2. F for between columns
F(2,2) at 0.05 level of significance
=3.8853
As calculated FC=14.3793>3.8853
So, H0 is rejected, Hence there is significant differentiating
between columns
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