ANOVA

1
ANOVA

• Analysis of Variance
• Method to compare more than 2 means
• As with comparing more than 2 ps, we cant use
  subtraction
• Will use sum of squares

2
ANOVA

• Think of an extension to 2 sample t problem
• Need an estimate of SD
• In 2 sample t, we always assumed the two distns
  had the same SD
• Had to combine our 2 estimates (one from each
  sample) to get a single est

3
ANOVA

• Review 2 sample t test
• Given N, Avg, SD for 2 groups
• SPpool(s1,n1,s2,n2)
• PVtprob(0,SP?(1/n11/n2),av1-av2,999)
• (double for two sided)

4
ANOVA

• Pool(s1,n1,s2,n2)
• ? (n1-1)s12 (n2-1) s22 / (n1-1 n2-1)
• Concentrate on the numerator and ignore sqrt
• ? (n-1)s2
• SSE, Sum of Squares Error
• (Error not in sense of mistake)

5
ANOVA

• If we ignored the fact that our data came from
  (potentially) different groups, we might have
  started with
• ? (x-avg)2
• SSTotal
• Note that ?SSTotal/(N-1) would be our estimator
  of SD using all the data (and ignoring grouping)

6
ANOVA

• The Sum of Squares Identity
• SSTotal SSE SSTreatment
• Where
• SSTr ? nj (avgj-AVG)2
• Except for nj this is like the sum of squares of
  the averages

7
ANOVA

• See Ex 62, p. 488
• 3 levels of blockage
• Measure flow rate where collapse occurs
• In book as fig11.4
• Not a good form for Excel
• See ANOVA1.xls
• First col tells which blockage level
• Second col is the flowrate

8
ANOVA

• To extract the values that correspond to flow
  rate1, use
• Dt(,1)1
• means all rows
• Equality is
• Single is assignment
• Use above expression to select rows
• Dt( Dt(,1)1, )

9
ANOVA

• To count how many, use length()
• Length(Dt( Dt(,1)1, ))
• Average is mean()
• SD is std()
• Most convenient to store these in lists

10
ANOVA

• For i13,
• Ns(i) Length(Dt( Dt(,1)i, ))
• Avs(i)
• Sds(i)
• End

11
ANOVA

• These are rows
• We can display conveniently by making them cols
  of a matrix
• ns avs sds
• 11.0000 11.2091 1.8987
• 14.0000 15.0857 2.1504
• 10.0000 17.3300 2.1680

12
ANOVA

• Alternate plan is to compute Ns, Avs, SDs as a
  matrix in the first place
• Write a function SUMSTATS(X, grp) which returns
  such a matrix
• The number of rows of the answer will be the
  number of groups
• Assume grp contains 1,groups

13
ANOVA

• Notice that SDs are quite similar in our example
• Good if we are going to assume the underlying SDs
  are all equal
• Easy to get SSE from here
• gtgt ssesum((ns-1).sds.2)
• 138.4672
• (Close to p. 497 in book)

14
ANOVA

• Exercises
• 1. Fig 11.5, p. 489, has data on roadway
  aggregate. Do sumstats.
• 2. Fig 11.6, p. 491, has data on carpet fibers.
  Do sumstats.

15
ANOVA

• To get SSTr, we need overall average
• Generally NOT the avg of the avgs
• Have to weight by the Ns
• gtgt grandsum(ns.avs)/sum(ns)
• 14.5086
• gtgt sstrsum(ns.(avs-grand).2)
• 204.0202

16
ANOVA

• Dont actually need SST, but it would be SSTrSSE
  342.4874
• gtgt (length(d114)-1)std(dt(,2)).2
• 342.4874

17
ANOVA

• H0 all means equal
• Ha not all means equal
• (Not the same as all means different)
• It can be shown that when H0 is true, then each
  of the SSs is a (multiple of) ?2

18
ANOVA

• Degrees of Freedom
• Usual est of SD has dfN-1
• So, if H0 true, Total df N-1
• Each term of SSE has df ni-1
• Fact Sums of (ind) Chi2 has df given by sum of
  df
• So df for SSE S (ni-1)
• Df for SSTr is difference
• groups-1 G-1

19
ANOVA
20
ANOVA

• Recall that the mean of a ?2 df
• For comparison, we should divide SS/df
• Called Mean Square
• MSTr, MSE (dont usually do MST)

21
ANOVA
22
ANOVA
23
ANOVA

• If H0 is true, then MSTr MSE
• Consider their ratio (to get rid of the common
  multiple)
• F MSTr / MSE
• Near 1 -gt H0
• F large -gt avgs are more spread out than expected
• -gt Ha

24
ANOVA

• F distn
• Depends on df for numerator and df for
  denominator
• FPROB(df1, df2, lo, hi)
• Can calculate p-value

25
ANOVA

• For our problem, F102/4.33 23.57
• gtgt fprob(2,32,23.57,999)
• 5.1058e-007
• So if H0 were true, it would be nearly impossible
  to get F this large.
• Can be quite sure that H0 is not true.
• Means are not all exactly the same.

26
ANOVA

• Write a function that takes SUMSTATS and returns
  ANOVA
• Since ANOVA table is not a full matrix, suggest
  we return a matrix for SS, df, MS and then return
  F and PV as scalars
• F,pv,SS ANOVA(sumstats)

27
ANOVA

• Exercises
• 1. Fig 11.5, p. 489, has data on roadway
  aggregate. Do ANOVA.
• 2. Fig 11.6, p. 491, has data on carpet fibers.
  Do ANOVA.

28
ANOVA

• If we only have 2 groups, should we do ANOVA or 2
  sample t test?
• Consider the following
• N18, Av112.3, SD15.3
• N212, Av216.4, SD26.6

29
ANOVA

• SP6.1273
• For the two sided Ha
• gtgtpvtprob(0,spsqrt(1/N11/N2),N1-1N2-1,Av2-Av1,
  999)2
• 0.1599

30
ANOVA

• For ANOVA
• sm
• 8.0000 12.3000 5.3000
• 12.0000 16.4000 6.6000
• gtgt f,pv,ssaanova1(sm)
• f
• 2.1492
• pv
• 0.1599
• So the (two-sided) pvalue is exactly the same by
  either method

31
ANOVA

• But theres more!!
• ssa
• 80.6880 1.0000 80.6880
• 675.7900 18.0000 37.5439
• 756.4780 19.0000 39.8146
• and SP2 37.5439
• So vMSE RMSE is the analog of SP

32
ANOVA

• If we reject H0, then some means are different.
  Which ones?
• Problem of multiple comparisons
• If H0 were true (all means equal) and we did 20
  tests with alpha0.05, then we should expect that
  1 of the comparisons would appear to be different
• If we are going to do a number of 2 sample t
  tests, we will have to use a lower ? than usual

33
ANOVA

• The book uses the Studentized range method
• There are a number of different methods for
  addressing this problem
• We will use the Bonferroni method

34
ANOVA

• The probability of a union of sets could be as
  high as the sum of the probabilities (if the sets
  are mutually disjoint)
• In our cases, the sets are making a Type I
  error for each test we are doing
• If we are doing N tests and each test has a
  Prob(Type I) 0.05/N, then the overall Prob(Type
  I) 0.05

35
ANOVA

• Therefore, if we want to end up at an overall
  level of 5, say, then we need to conduct each
  test at a level of 0.05/ tests
• For G groups, there are G(G-1)/2 pairs
• For flowrate example, G3 and there are 32/2 3
  pairs to compare
• Do 2 sample t tests with ?/3 0.05/3

36
ANOVA

• If we use p-values, then we should multiply the
  (apparent) p-value by the of comparisons
• If we do confidence intervals, we should get
  wider intervals
• For 95 intervals, we usually solve for 0.025
• Now we solve for 0.025 / ( groups)

37
ANOVA

• Recall that in the 2 sample t test, we use
  SD?(1/n11/n2) for the diff of avgs
• What do we use for the pooled SD?
• This was the basis for SSE (or MSE)
• For SD, use ?MSE
• For df, use df for MSE

38
ANOVA

• The p-value for comparing 1-2 is 0.0082
• For 1-3, PV0.0003
• For 2-3, PV0.0872
• Using Bonferoni, we cannot conclude that 2 and 3
  are different, but 1 IS diff from both 2 and 3.
• Note that on p. 504, he gets a slightly diff
  result using conf intervals and Studentized range
• The Studentized range is a bit more efficient
  than Bonferroni, but note that his last CI comes
  quite close to 0

39
Rank procedures

• The underlying assumption of ANOVA is that the
  data has a normal distn
• When we are not comfortable with that assumption
  there are other methods available
• Nonparametric methods

40
Rank procedures

• A large class of NP methods are based on ranks
• Replace the data with ranks and then do
  calculations
• We should feel comfortable in knowing which
  values are larger than which other values, even
  if we dont know the probability of such a
  difference

41
Rank procedures

• Function yranks(x)
• Midranks to handle ties in the xs
• For comparing multiple groups, we use
  Kruskal-Wallis
• Based on the sum of ranks within each group

42
Rank procedures

• Text has a formula on p. 713, but not very
  meaningful
• Better formula is based on S(obs-exp)2/exp
• If we have N total obs and n1 obs in group 1,
  then the avg rank of all the data is (N1)/2
• We would expect the rank sum for group 1 to be
  n1(N1)/2, etc
• Can then compute the diff between obs and exp
• TRICK There is a fudge factor. Have to multiply
  by 6/N to get ?2

43
Rank procedures

• For artery flow
• rranks(dt(,2))
• rs(1)sum(r(dt(,1)1)) etc
• ns(1)sum(dt(,1)1) etc
• exptns(n1)/2
• gtgt c26/nsum((rs-expt).2./expt)
• 20.4380
• gtgt pvchiprob(2,c2,99)
• 3.6471e-005
• NOTE df groups-1 df for Treatment

44
Rank procedures

• Can use pooling as we did for Chi2 to determine
  which differences are significant
• But the pooling can be done on the rank sums (and
  expecteds)
• As before, df do not change

45
Rank procedures

• Carpet fiber example
• Assume weve run sumstats
• gtgt nssmst(,1)'
• ns
• 16 16 13 16 14 15
• gtgt rranksums(x(,2),x(,1))
• r
• 495 1177.5 353
  854.5 288 927
• gtgt nsum(ns)
• n
• 90

46
Rank procedures

• gtgt exptns(n1)/2
• expt
• 728 728 591.5
  728 637 682.5
• gtgt c26/nsum((r-expt).2 ./ max(1,expt))
• c2
• 49.937
• Chi2 w/ df5, so extremely large

47
Rank procedures

• gtgt (r-expt)./expt
• ans
• -0.32005 0.61745 -0.40321
  0.17376 -0.54788 0.35824
• Note that 1 and 3 look similar
• gtgt rcpool(r,1,3),exptcpool(expt,1,3),
• r
• 848 1177.5 0
  854.5 288 927
• expt
• 1319.5 728 0
  728 637 682.5
• gtgt c26/nsum((r-expt).2 ./ max(1,expt))
• c2
• 49.787
• Still large (unchanged), so diff is not due to a
  diff between 1 3