ChiSquare Introduction to Nonparametric Stats Chisquare n n

Chi-Square: Introduction to Nonparametric Stats

Chi-square n n n Parametric vs. nonparametric tests Hypotheses about Frequencies Two main uses: q q Goodness of fit. 1 IV. Test of independence. 2 or more IVs.

Goodness-of-fit test Blind beer tasting study. Judges taste 4 beers and declare their favorite. 100 lucky judges. Results: Coors Frequency 15 Corona Miller Sam Adams Total 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) Where O is an observed frequency and E is an expected frequency under the null. Coors Corona Miller Sam Adams Total 15 45 30 10 100 Expected 25 25 O-E -10 20 5 -15 (O-E)2 100 400 25 225 16 1 9 Freq (O-E)2/E 4 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 110 115 340 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)/ (115*160)/ 160 340 = 51. 76 340 = 54. 12 Female (110*180)/ (115*180)/ 180 340 = 58. 24 340 = 60. 88 Total 110 115 340 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: School tax increase Ban EEO hiring prefs Police Tax Increase Male (40 -51. 76)2/51. 76 = 2. 19 2. 67 . 01 Female 2. 37 . 01 Total 1. 94 Total test value 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.

Effect Size n n n 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 For 2 x 2 tables only Phi Heart attack Type A Type B 25 10 No heart attack 5 40 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 2 x 2, e. g. , 3 x 2 or 4 x 3 School tax Ban EEO increase hiring prefs Police Tax Increase Total Male 40 65 55 160 Female 70 50 60 180 Total 110 115 340 This is a more typical result. There is a significant association, but the association is not very strong.