Date: Aug 13, 2012 11:32 AM
Author: David Bee
Subject: [ap-stat] A Closing Lesson

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In a summer course taught by me at a local college that concluded
last Thursday, the closing lesson was all about one problem,
which was this (thanks to Daren[?], or whoever else provided such
in the past):

Consider three types of popcorn, A, B, and C. Four containers of 500
kernels each are randomly selected and popped for five minutes,
one container at a time in random order. (Sufficient time is left
between popping to allow the popper to cool down.) After each five-
minute period the number of unpopped kernels is counted, with these
52 44 60
60 50 58
56 52 60
52 42 50
xbar 55.0 47.0 57.0
s 3.830 4.761 4.761

After looking at the stats, we formulated the hypotheses:

Ho: mu(A) = mu(B) = mu(C)
Ha: The means are not all equal.

Suppose alpha = 0.05.

After mentioning a couple of essential ANOVA assumptions (normal
distributions with common variances among treatments, and so we
assume the distributions of the numbers of unpopped kernels are
normal with common mean and common variance) and a rule of thumb
about whether the sample variances (s^2) indicate the common
variance assumption is not good, the two main ANOVA concepts were
introduced, namely estimates of the common population variance from
the variance of the samplemeans (variation among/between treatments)
and from pooling the variances (variation within treatments).

Such lead to 112.0 and 20.0 respectively, and so an F ratio of
112.0/20.0 = 5.6.

Then we considered the corresponding numberS of degrees of freedom:

Total DoF = 12 - 1 = 11, or Total = kn - 1 = (3)(4) - 1 = 11
DoF for Numerator = 3 - 1 = 2, or DoF Num = k - 1 = 3 - 1 = 2
DoF for Denominator = 11 - 2 = 9, or DoF Den = kn-1-(k-1)= k(n-1)

The students then found the P-Value using this "F Distribution",
resulting in
P-Value = P(F(2,9)>5.6) = 0.026 < alpha = 0.05

and so rejection of the null hypothesis at the 0.05 level.

But as there was still much time left in this 110-minute lesson,
the students were introduced to a classical nonparametric method,
namely the Kruskal-Wallis Test, which was introduced by K and W in
an article in JASA sixty years ago. (The article was titled simply
"Use of Ranks on One-Criterion Variance Analysis", 1952.)

The hypotheses could be stated in terms of medians or distributions,
but the method involves taking the 12 data values and ranking them
in order from smallest to largest, resulting in the following:

6.0 2 11.0
11.0 3.5 9
8 6.0 11.0
6.0 1 3.5
R 31.0 12.5 34.5

R is the sum of the ranks in each sample, with ties represented
by ordinary averages of the corresponding ranks.

The Kruskall-Wallis Test Stat, which is the rank version of the F
test stat above (and can be easily found in a book or on the Web),
came out to be 5.375, and such stats have approximately a chi-square
distribution with k-1 = 3-1 degrees of freedom. Thus, here, we have
P(Chi-Square > 5.375) = 0.068 > alpha = 0.05, and so now the null
hypothesis is not rejected. (BTW, checking a table of exact P-
Values for the Kruskal-Wallis Test for such small samples shows a
P-Value of about the same; also, although too many ties is a
problem, such wasn't the case here.)

Therefore, this course concluded with a contradictory pair of tests
for the same data set, which should be some good food for thought
for students who want to use statistical testing in the future sans
being attentive to doing such...

-- David Bee

PS: Thinking of the Olympics, it should not be difficult to try the
above parametric F test and nonparametric K-W test on comparing
the mean/median times of some comparable races --- and compare
the results...

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