We had the following data, and rejected the null hypothesis of equal color distribution.
|
Color |
ObservedCount |
ExpectedCount |
fi–fi |
|
Blue |
14 |
29.5 |
- |
|
Brown |
67 |
29.5 |
- |
|
Green |
12 |
29.5 |
- |
|
|
38 |
29.5 |
- |
|
Red |
15 |
29.5 |
- |
|
Yellow |
31 |
29.5 |
- |
Now let’s take this data apart and see why we rejected that null hypothesis. Recall that for our original hypothesis test, we had an observed c2 statistic of 75.85 with 5 degrees of freedom.
Based on the deviations you calculate, decide which, if any, colors appear to behaving basically the same. (That is, are there any colors that seem to be present in roughly the same quantities?)
Examining just the data in each of those subgroups, evaluate if those colors really are basically equally likely. List your null and alternative hypotheses for each subtest, as well as the degrees of freedom.
Now “glue” the table back together. What should the proportions be of each subgroup? This will drive how you find the expected counts for each. Show that the result of this subtest, which should have its null hypothesis rejected, when added to the your other subtests, adds up to roughly your original c2 test. Also show that your degrees of freedom add up correctly.
A few answers to shoot for: I broke the M&M colors into the groups
(Blue-Green-Red), (Yellow-Orange) and (Brown). For (Blue-Green-Red), I
got expected counts of 13.67 for each color (not 29.5!), and c22 = 0.34.
For (Yellow-Orange) I got expected counts of 34.5 for each color, and c21 =
0.71. When I put the groups back, I got expected counts of 88.5 for
(Blue-Green-Red), 59 for (Yellow-Orange), and 29.5 for (Brown), with a test
statistic of 74.858. Note that 0.34 + 0.71 + 74.86 = 75.91, which is very
close to our original test statistic. Also 2 + 1 + 2 = 5 = original
number of degrees of freedom.