Chi-Square:

Introduction to Nonparametric Stats

2

Chi-square

Parametric vs. nonparametric tests Hypotheses about Frequencies Two main uses:

Goodness of fit. 1 IV. Test of independence. 2 or more IVs.

Goodness-of-fit testBlind 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.

Goodness-of-fit (2)

E

EO 22 )(

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

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 k=4 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.

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

E

EO 22 )(

Use same formula. But now E is calculated by:

E=(rowtotal*columntotal)/grandtotal or equivalently:E=pctr*pctc*N, where pct means percentage.

Test of Independence (2)

Find expected values:

School tax increase

Ban EEO hiring prefs

Police Tax Increase

Total

Male (110*160)/340 = 51.76

(115*160)/340 = 54.12

(115*160)/340 = 54.12

160

Female (110*180)/340 = 58.24

(115*180)/340 = 60.88

(115*180)/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.

Test of Independence (3)

Find the value of chi-square:

E

EO 22 )(

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= 2

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.

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.

Effect Sizes for Contingencies - Phi

Nobt2

Type A Type B

Heart attack 25 10

No heart attack 5 40

80;56.302 Nobt

62.80

56.30

Phi

For 2x2 tables only

This is a strong relation. Anything larger than about .5 is unusual in psychology. Average is about .20. Data are hypothetical.

Contingency Coefficient

For 2-way tables other than 2x2, e.g., 3x2 or 4x3

2

2

obt

obt

NC

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

340;19.92 Nobt

16.19.9340

19.92

2

obt

obt

NC

This is a more typical result.

There is a significant association, but the association is not very strong.