Chi-Square Test Mon, Apr 19th, 2004

Chi-Square Test Mon, Apr 19th, 2004

Chi-Square ( 2) Are 2 categorical variables related (correlated) or independent of each other? Compares # in categories that would be expected by chance (E) to # in categories actually observed (O) Null hyp (Ho) – no relationship between 2 variables (they’re independent of ea.other) Alternate (Ha) – the 2 variables are related (not independent)

2 Formula 2 [(fo – fe)2] fe So we’ll compare observed and expected frequencies for each cell in the table

Example Is age ( 30 v. 30) related to preference for analog/digital watches? Digital Analog Undecided Under 30 90 40 10 Over 30 40 10 10

Step 1: Compute Marginals Marginals are the row and column totals: Digital Analog Undecided Under 30 90 40 10 140 Over 30 10 40 10 60 100 80 20

Step 2: Compute Expected Frequencies (fE) fE (Column marginal * Row marginal ) / N For people under 30: fe (digital) 100 * 140 / 200 70 fe (analog) 80*140 / 200 56 fe (undec) 20*140 / 200 14 Over 30: fe (digital) 100*60 / 200 30 fe (analog) 80*60 / 200 24 fe (undec) 20*60 / 200 6

Step 3: Compute X2 Find difference (residual) betw observed & expected for each cell (fo – fe) Square those differences Divide squared differences by fe Sum the results

Summary Fo Fe Fo-fe (fo-fe)2 (fo-fe)2 / fe 30 D 90 70 20 400 5.71 30 A 40 56 -16 256 4.57 30 U 10 14 -4 16 1.14 30 D 10 30 -20 400 13.33 30A 40 24 16 256 10.67 30 U 10 6 4 16 2.67

(cont.) Last step: Add up (fo-fe)2 / fe 2 5.71 4.57 1.14 13.33 10.67 2.67 38.09 Step 4: Compare to 2 critical with df (# columns – 1) (# rows – 1) Here df (2-1)(3-1) 2 df, .05, critical 5.99

Hypothesis Test If 2 observed 2 critical, reject Ho – Reject Ho conclude there is a relationship between the 2 variables – Here, 38.09 5.99, reject Ho, there is a relationship between age & watch preference

In SPSS Analyze Descriptive Stats Crosstabs – Choose whichever variable you’d like for ‘row variable’ and the other for ‘column variable’ – Click “Statistics” button, and check chi-squared option – Click “Cells” Button, choose “expected count”

SPSS (cont.) Output – look for “Pearson chi-sq” and “Asymp Sig” column gives significance value for chi-sq test: – If “Asymp Sig” value is .05 (alpha), reject Ho Note there is an option for clustered graphing, read this example in the lab