Analysis of Variance (ANOVA). W&W, Chapter 10. The Results. Many other factors may determine the salary level, such as GPA. The dean decides to collect new data selecting one student randomly from each major with the following average grades. New data.
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W&W, Chapter 10
Many other factors may determine the salary level, such as GPA. The dean decides to collect new data selecting one student randomly from each major with the following average grades.
A+ 41 45 51 M(b1)=45.67
A 36 38 45 M(b2)=39.67
B+ 27 33 31 M(b3)=30.83
B 32 29 35 M(b4)=32
C+ 26 31 32 M(b5)=29.67
C 23 25 27 M(b6)=25
M(t)1=30.83 M(t)2=33.5 M(t)3=36.83
Now the data in the 3 samples are not independent, they are matched by GPA levels. Just like before, matched samples are superior to unmatched samples because they provide more information. In this case, we have added a factor that may account for some of the SSE.
Now SS(total) = SST + SSB + SSE
Where SSB = the variability among blocks, where a block is a matched group of observations from each of the populations
We can calculate a two-way ANOVA to test our null hypothesis.
Ho: 1 =2 = 3
HA: 1 2 3
We are testing the same hypothesis as in the completely randomized design.
SST = b(M(t)j - )2
where b = the number of blocks
M(t)j = the mean for each sample
= grand mean
SST = (6)(30.83-33.72)2 + (6)(33.5-33.72)2 + (6)(36.83-33.72)2 = 108.4
This captures the variation across our samples (majors).
SSB = k(M(b)i -)2
where k = the number of samples
M(b)i = the mean for each block
= grand mean
SSB = (3)(45.67-33.72)2 + (3)(39.67-33.72)2 + …(3)(25-33.72)2 = 854.9
This captures the variation across our blocks (GPA levels).
SS = (Xij - )2
SS = (41-33.72)2 + (36-33.72)2 + … + (27-33.72)2 = 1015.61
We know that
SS = SST + SSB + SSE
So SSE = SS – SST – SSB
SSE = 1015.61 – 108.4 – 854.9 = 52.2
Now we can compare our results across the two designs we have discussed:
Sum of Completely Randomized Randomized Block
squaresDesign (One way ANOVA)Design (Two way)
SST 193 108.4
SSB ---- 854.9
SSE 819.5 52.2
SS 1012.5 1015.61
We can see that we have dramatically decreased the error (SSE) by accounting for GPA. In other words we have decreased the variability caused by the difference among the blocks.
Source of df Sum of Mean
Treatment k-1 SST MST=SST/(k-1)
Block b-1 SSB MSB=SSB/(b-1)
Error n-k-b+1 SSE MSE=SSE/(n-k-b+1)
Total n-1 SS=SST+SSB+SSE
We can calculate a F-statistic to test differences among samples or blocks.
F = MST = SST/(k-1)
F = 108.4/(3-1)
F = 54.2/5.2 = 10.4
Critical F, k-1, n-k-b+1 = F.05, 2, 10 = 4.1
Because our calculated F (10.4) exceeds our critical F (4.1), we reject the null hypothesis that the means across the samples are equal.
We conclude that there is a difference in the mean salary levels across the 3 business majors.
We could also test whether the blocks are different from each other, or whether students with higher GPA’s earn more money.
F = MSB = SSB/(b-1)
F = 854.9/(6-1)
F = 170.982/5.2 = 32.76
Critical F, b-1, n-k-b+1 = F.05, 5, 10 = 3.33
We can also reject the null hypothesis of no difference among blocks because our calculated F (32.76) exceeds our critical F (3.33).
It is interesting to note that MSE is similar to the pooled variance sp2 which we calculated earlier for a matched samples confidence interval.
MSE = (n1 – 1)s12 + (n2 – 1)s22 +..+(nk – 1)sk2
(n – k)
Thus MSE is an unbiased estimate of 2.
W&W show that you can substitute MSE for s in the calculation of a confidence interval (1 - 2).