Ch 13: Chi-square tests Part 2: Nov 29, 2007

Chi-sq Test for Independence Deals with 2 nominal variables Create ‘contingency tables’ –Crosses the 2 variables of interest so that you can see the frequencies of their combinations and the totals: Species Learned Task? RatMonkeyHuman Yes2414 No28166

Independence Question is whether there is any relation between the 2 nominal variables –If no relation, the proportion of each species who learned the task is same as proportion who did not. The species & learning variables are independent

Hypotheses Null hyp is that the 2 variables are independent (no relation between them) Research hyp is that they are not independent (they are related…) Need to determine Expected Frequencies (E) for each cell. –Find E by assuming the 2 variables are independent

(cont.) If independent, proportions should be the same up & down the cells of each column. 1. Find each row’s percentage of the total. 2. For each cell, multiply its row’s % by it’s column total. Or… E = (R/N) (C) Where R is frequency observed in this cell’s row N is number of people total in the sample C is frequency observed in this cell’s column

Species Learn Task? Rat MonkeyHuman Yes2414 No28166 So, for the cell “Rat/Yes”, E = [(20/70) * 30] = 8.57 Find the E’s for the other cells… Total N = 70

Figuring Chi-sq Same formula here as for the 1 variable case (goodness-of-fit): Χ 2 = (O – E) 2 E Find chi-sq for last example… Df = (N columns – 1) (N rows – 1) Σ

Effect Size Can use chi-sq to find effect size (how strong is the association between the 2 variables?) –Phi coefficient for 2x2 contingency tables: Φ = sqrt (X 2 / N) Will range from 0-1 and can be considered a correlation coefficient. Cohen’s guidelines = around.10 small effect,.30 medium effect, >.50 large effect

Effect Size (cont.) If table is greater than 2x2, can’t use phi, use Cramer’s phi: –Same as phi, but divide by N times df of the smaller side of the table If 3 row x 4 column table, you’d go with the 2 df from the 3 rows side of the table. –Cramer’s Φ = sqrt [X 2 / (N)(df smaller )] –Cohen’s standards change for this – use Table 13-8 to interpret Cramer’s (depends on df)

SPSS: Chi-sq test of Independence (for 2 variables) Analyze  Descriptives  Crosstabs –Choose Row and Column (doesn’t matter which variable is entered where…) –On the ‘statistics’ button check box for ‘chi- square’  OK –See output from 2 nd box –1 st row gives “Pearson’s chi-sq” figure, df, and ‘sig’ value –Compare ‘sig’ to your alpha level (.05) and if ‘sig’ < alpha  reject null