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

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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

results:

A B C

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:

A B C

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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