Comparison Among Groups more than 2 groups, 1 factor

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Comparison Among Groups(more than 2 groups, 1
factor)
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Comparison Among k Groups (k gt2) With One Factor

• The test to be considered determines if all k
  groups have the same central value (median or
  mean, depending on the test), or at least one of
  the groups differs from the others.

Parametric Approach

• When data within each of the groups are normally
  distributed and possess identical variances -
  Analysis of Variance (ANOVA).
• ANOVA tests whether each groups mean is
  identical.
• If k 2, ANOVA t-test.

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Nonparametric Approach

• If the assumptions of normality and equal
  variance are not met - KRUSKAL-WALLIS test.
• K-W test is much like the rank-sum test extended
  to more than two groups.
• The K-W test compares the medians of groups
  differentiated by one explanatory variable or one
  factor e.g. months, seasons, locations.

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Null Hypothesis Tests

• All the one factor tests have as their null
  hypothesis that each groups median (or mean) is
  identical, with the alternative hypothesis that
  at least one is different. This is the same as
  saying all groups have the same distribution vs.
  at least one distribution differs.
• However, when the null hypothesis is rejected,
  these tests do not tell which group or groups are
  different!
• To tell which groups are different - Multiple
  comparison test.
• Multiple comparison tests are performed only
  after the ANOVA or K-W null hypothesis has been
  rejected, for determining which groups differ
  from other.

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Graphical Displays

• As usual before any formal comparison is carried
  out, do side-by-side boxplots. These will
  indicated at a glance whether
• 1. data in each group are normally distributed.
• 2. variances are approximately equal.
• 3. to use parametric or nonparametric tests.

The Kruskal-Wallis Test

• Like other nonparametric tests, the K-W test may
  be computed by an exact method used for small
  sample sizes, by a large sample approximation
  (computer packages), or by ranking the data and
  performing a parametric test on the ranks.
• Luckily, the exact method is rarely required.
  Large sample approximations give p-values very
  close to their exact values. Exact values are
  needed only when k3 with sample sizes of 5 or
  less per group, or k ? 4 of size 4 or less per
  group.

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Large Sample Approximation for the K-W Test

• Situation Several groups of data are to be
  compared, to determine if their medians are
  significantly different. For a total sample
  size of N, the overall average rank will equal
  (N1)/2. If the average rank within a group
  (average group rank) differs considerably from
  this overall average, not all groups can be
  considered similar.
• Computation All N observations are jointly ranked
  from 1 to N, smallest to largest. These ranks
  Rij are then used for computation of the test
  statistic. Within each group, the average
  group rank Rj is computed

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• Tied data When observations are tied, assign the
  average of their ranks to each.
• Test Statistic The average group rank Rj is
  compared to the overall average rank R
  (N1)/2, squaring and weighing by sample size,
  to form the test statistic K
• Decision Rule To reject Ho all groups have
  identical distributions, vs.
• H1 at least one distribution
  differs
• Reject Ho if K ? x21-?,(k-1) the 1-? quantile
  of the chi- square distribution with (k-1)
  degrees of freedom, otherwise do not reject
  Ho.

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Example

• Fecal coliforms, in organisms per 100ml, were
  measured in the Waterton River. Do all four
  seasons exhibit similar values, or do one or more
  seasons differ?

Selected fecal coliform data (from Lin and Evans,
1980). counts in organisms per
100ml Summer Fall Winter
Spring 100 65 28 22 220 120 58 53
300 210 120 110 430 280
230 140 640 500 310 320 1600 1100
500 1300
PPCC 0.05 0.06 0.50 0.005
p-value
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• FECAL COLIFORM COUNTS (ORGANISMS/100ml)

2000 1500 1000 500 0



SUMMER FALL WINTER SPRING
Boxplots of Fecal Coliform data. Data from the
Illinois River
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Answer

• Should a parametric or nonparametric test be
  performed on these data? If even one of the four
  groups exhibits non-normality, the assumptions or
  parametric ANOVA are violated.
• The consequences of this violation is an
  inability to detect differences which are truly
  present - lack of power.
• Judging from the boxplots, a nonparametric test
  should be used on these data.
• Computation of the K-W test is shown below, the
  computed K value (H in Minitab) is compared to
  the chi-square distribution.

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Selected fecal coliform data (from Lin and Evans,
1980). counts in organisms per
100ml Summer Fall Winter
Spring 6 5 2 1 12 8.5 4 3 15 11 8.5
7 18 14 13 10 21 19.5 16 17 24 22
19.5 23 16 13.3 10.5 10.2
Ranks Rij
Rij
R 12.5
K 2.69 ?20.95,(3)7.815 p0.44 so, do
not reject equality of distributions.
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The Rank Transform Approximation to the K-W Test

• Computed by performing a one-factor ANOVA on the
  ranks Rij. This approximation compares the mean
  rank within each group to the overall mean rank,
  using an F-distribution for the approximation of
  the distribution of K.
• The F and chi-square approximations will result
  in very similar p-values.
• The rank transform method should properly be
  called an ANOVA on the ranks.

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• This approach becomes useful when we want to
  perform multiple comparison tests using Tukeys
  method which is a parametric approach. The
  reason for this is that there is no good
  nonparametric multiple comparison tests available
  at present.
• Using the previous data, the p-value 0.47 when
  the rank transform approximation is used. This
  p-value is essentially identical to that for the
  large sample approximation.
• The details of the computations will follow after
  ANOVA.

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ANOVA (One Factor)

• Parametric equivalent to the K-W test. It
  compares the mean values of each group with the
  overall mean for the entire data set.
• If the group means are dissimilar, some of them
  will differ from the overall mean. See following
  figure.
• If the group means are dissimilar, they will also
  be similar to the overall mean.
• Why should a test of differences between means be
  named an analysis of variance?

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• In order to determine if the differences between
  group means (the signal) can be seen above the
  variation with groups (the noise), the total
  noise in the data as measured by the total sum of
  squares is split into two parts.
• Total sum of squares treatment of sum of
  squares error of sum of squares
• (overall variation) (group means - overall
  mean) (variation within groups)

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Computation

• If the total sum of the square is divided by N-1,
  where N is the total number of observations, it
  equals the variance of the yijs. Thus ANOVA
  partitions the variance of the data into two
  parts, one measuring the signal and the other the
  noise. These are then compared to determine if
  the means are significantly different.

One Factor Analysis of Variance
Situation Several groups of data are to be
compared, to determine if their means are
significantly different. Each group is assumed
to have a normal distribution around its mean.
All groups have the same variance.
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• Computation The treatment mean square and error
  mean square are computed as their sum of square
  is divided by their degrees of freedom (df).
  When the treatment mean square is larger than
  the error mean square as measured by an F-test,
  the group means are significantly different.
• where k-1 treatment degrees of
  freedom
• where N-k error degrees of freedom
• Tied data No alterations necessary.

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• Test Statistic The test statistic F
• F MST/MSE
• Decision Rule To reject Ho the mean of every
  group is identical, vs.
• H1 at least one mean
  differs.
• Reject Ho if F ? F1-?, k-1, N-k the 1-?
  quantile of an F distribution with k-1 and N-k
  degrees of freedom otherwise, do not reject
  Ho.

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• The computations and results of an ANOVA are
  usually organized into an ANOVA. For a one-way
  ANOVA, the table looks like
• Source df SS MS F p-value
• Treatment k-1 SST MST MST/MSE p
• (between group)
• Error N-k SSE MSE
• (within-group)
• Total N-1 Total SS
• where
• and

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• Using the data of the example, the ANOVA table is
  given below
• Source df SS MS F p-value
• Season 3 361367 120466 0.67 0.58
• Error 20 3593088 179654
• Total 12 3954485
• In general, large F value gtgt 1 small
  p-values reject Ho.
• We should not be using ANOVA here because of
  non-normality.

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Assumptions of the ANOVA

• Since the t-test is a special case of the ANOVA,
  all the assumptions pertaining to the t-test
  apply to the ANOVA.
• 1. All samples are random samples from their
  respective populations.
• 2. All samples are independent of one another.
• 3. Departures from group mean are normally
  distributed for all groups.
• 4. All groups have equal variance.
• Violation of either the normality or constant
  variance assumption results in a loss of ability
  to see differences between means (a loss of
  power).

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• The ANOVA suffers from the same five problems as
  did the t-test
• 1. lack of power when applied to non-normal data
• 2. dependence on an additive model
• 3. lack of applicability for censored data
• 4. assumption that the mean is a good measure of
  central tendency for skewed data
• 5. difficulty in assessing whether the normality
  and equality of variance assumptions are valid
  for small sample sizes.

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Multiple Comparison Tests (MCTs)

• MCTs compare all possible pairs of treatment
  group means or medians, and are performed only
  after the null hypothesis of all medians or
  means identical has been rejected.
• Many MCTs are available in the literature e.g.
  Sheffe, Bonferroni, Fisher, Tukey, Duncans
  multiple range test, Regwq, Regwf, etc.
• Of all the methods available, Tukeys method is
  the most generally applicable and powerful MCT
  for a variety of situations. (This test is
  available on Minitab).

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• Tukeys method is a parametric procedure. For
  non-normal data, use the ANOVA on the rank
  transform, then use Tukeys method to test for
  differences in the means of the ranks.
• For more details see
• Steel and Torrie (1980) Principle and Procedures
  of Statistics - A Biometrical Approach, McGraw
  Hill.
• We only need to learn how to interpret the
  results given by Minitab.
• Use family error rate of 0.05
• If the interval does not include zero, then
  difference is statistically significant.

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Presentation of Multiple Comparison Tests

• The results are often presented in one of the two
  following formats
• 1. Letters
• y1 gt y2 gt y3 gt y4
• y1 y2 y3 y4
• A AB BC C

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• Treatment group means are ordered, and those
  having the same letter underneath them are not
  significantly different. The convenience of this
  presentation format is that letters can easily be
  positioned within side-by-side boxplots.

A
MCT results Boxes with same letter are not
significantly different
AB
BC
C
Boxplots with letters showing the result of a MCT.
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• 2. Lines
• y1 y2 y3 y4
• In this presentation format, group means
  connected by a single unbroken line are not
  significantly different. This format is suited
  for inclusion in a table listing group means or
  medians.