Transcript STT 430/530, Nonparametric Statistics

K-Sample Methods
• Assume X1j from treatment 1 (sample of size n1) and
and so forth… Xkj from treatment k (sample of size nk)
for a total of n1+n2+ … +nk = N observations. Assume
the populations have cdfs Fi , i=1,…,k. The null
hypothesis is that all the F’s are the same
distribution. The alternative is that at least one pair of
cdfs is different for at least one value of the
distribution. But see the Shift Alternative p.79-80. …
See Table 3.1.1 on page 80 for a complete
description of these notations…
• Recall that the usual parametric statistic for testing
the above null hypothesis is F=MST/MSE, where
MST = treatment mean squares, MSE=error mean
squares (formulas on page 80-81). Assuming normal
populations with equal variances, this statistic has an
F distribution with k-1 df in the numerator and N-k df
in the denominator. Use tables to get p-values or use
the permutation F-test as described on pages 81-85.
See especially the permutation F-test based on SSX,
defined on page 84. See the next slide for a review
of the logic of the F-test.
• Go over the “Steps in Carrying Out the Permutation
F-Test” on p.81. Implement this procedure in R – use
the lm and anova function to pull out the specific
values of the F statistic.
1. MSE is an estimate of the population variance
based on the deviation of scores around their
respective treatment means. It is a weighted
average of the treatment variances (see p.80)
2. MST is also an estimate of the population
variance if the null hypothesis is true. It is based
upon the deviations of group means about the grand
mean. Since it is influenced by treatment effects, it is
only an estimate of the same population variance if
the treatment effects are zero; i.e., when the null
hypothesis is true.
3. It turns out that if the null hypothesis is true, the
ratio of these two variance estimates is distributed
as an F-distribution:
F = MST / MSE
4. Since under the null hypothesis the two mean
squares are estimating the same population value,
this ratio should be close to 1 when the null is true.
The observed value of F is compared to the
sampling distribution of F to get a p-value (or
empirical p-value via permutation test) to look for
departures from the null hypothesis.
5. If the observed F ratio is "large", then perhaps
the assumption of the null hypothesis of no
treatment effect is false, and we should reject the
null.
• See section 3.1.3 on p. 83 for an alternative
statistic to use for the permutation test – the
author calls it SSX where
SSX 
k
n X
i 1
i
2
i
• The use of SSX in place of F is justified in
permutation tests of this type…
• HW for Thursday: Read section 3.1 and 3.2
about the Kruskal-Wallis test… Do problem #2
on page 105 – include the use of the Kruskal
– Wallis test on this data. Write up a solution
and hand it in as part of the mid-term exam.
This will be the last question on the take-home
part of the midterm.