Chi-Square:

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Chi-Square

• Introduction to Nonparametric Stats

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Chi-square

• Parametric vs. nonparametric tests
• Hypotheses about Frequencies
• Two main uses
• Goodness of fit. 1 IV.
• Test of independence. 2 or more IVs.

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Goodness-of-fit test
Blind beer tasting study. Judges taste 4 beers
and declare their favorite. 100 lucky judges.
Results
Coors Corona Miller Sam Adams Total
Frequency 15 45 30 10 100
If no difference in taste (or all the same beer)
we expect about 25 people to choose each beer
(null hypothesis). There are 100 people and 4
choices (100/4 25).
We will test whether frequencies are equal across
beers.
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Goodness-of-fit (2)
Where O is an observed frequency and E is an
expected frequency under the null.
Coors Corona Miller Sam Adams Total
Freq 15 45 30 10 100
Expected 25 25 25 25
O-E -10 20 5 -15
(O-E)2 100 400 25 225
(O-E)2/E 4 16 1 9 30 test value
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Goodness-of-fit 3
Our test statistic was 30. The df for this test
are k-1, where k is the number of cells. In our
example k4 and df 3. Chi-square has a
distribution found in tables. For alpha.05 and
3 df, the critical value is 7.81, which is less
than 30. We reject the null hypothesis. People
can taste the difference among beers and have
favorites.
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Test of Independence (1)
Exit survey at polls. Voter preferences. Did
you vote yes for
School tax increase Ban EEO hiring prefs Police Tax Increase Total
Male 40 65 55 160
Female 70 50 60 180
Total 110 115 115 340
Use same formula. But now E is calculated by
E(rowtotalcolumntotal)/grandtotal or
equivalently EpctrpctcN, where pct means
percentage.
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Test of Independence (2)
Find expected values
School tax increase Ban EEO hiring prefs Police Tax Increase Total
Male (110160)/340 51.76 (115160)/340 54.12 (115160)/340 54.12 160
Female (110180)/340 58.24 (115180)/340 60.88 (115180)/340 60.88 180
Total 110 115 115 340
We use row and column totals to figure expected
cell frequencies under the null hypothesis that
all cell frequencies are proportional to their
row and column frequencies in the population.
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Test of Independence (3)
Find the value of chi-square
School tax increase Ban EEO hiring prefs Police Tax Increase Total
Male (40-51.76)2/51.76 2.67 2.19 .01
Female 2.37 1.94 .01 test value
Total 9.19
For the chi-square test of independence, the df
are (rows-1) times (cols-1) or for this example,
(2-1)(3-1) 2. From the chi-square table, we
find the critical value is 5.99 for an alpha of
.05, so we reject the null. Men and women have
different voting preferences.
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Effect Size

• Effect size index of magnitude of relations
• Statistical Significance probability of outcome
• Significant results when large magnitude or large
  sample size. Can have trivial magnitude but still
  significant results, so you want an effect size.

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Effect Sizes for Contingencies - Phi
For 2x2 tables only
Type A Type B
Heart attack 25 10
No heart attack 5 40
Phi
This is a strong relation. Anything larger than
about .5 is unusual in psychology. Average is
about .20. Data are hypothetical.
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Contingency Coefficient
For 2-way tables other than 2x2, e.g., 3x2 or 4x3
School tax increase Ban EEO hiring prefs Police Tax Increase Total
Male 40 65 55 160
Female 70 50 60 180
Total 110 115 115 340
This is a more typical result. There is a
significant association, but the association is
not very strong.