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Chi-Square Test Mon, Apr 19 th, 2004. Chi-Square (  2 ) wAre 2 categorical variables related (correlated) or independent of each other? wCompares # in.

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Presentation on theme: "Chi-Square Test Mon, Apr 19 th, 2004. Chi-Square (  2 ) wAre 2 categorical variables related (correlated) or independent of each other? wCompares # in."— Presentation transcript:

1 Chi-Square Test Mon, Apr 19 th, 2004

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

3  2 Formula w  2 =  [(fo – fe) 2] fe So we’ll compare observed and expected frequencies for each cell in the table…

4 Example wIs age ( 30) related to preference for analog/digital watches? DigitalAnalogUndecided Under 30904010 Over 30104010

5 Step 1: Compute Marginals DigitalAnalogUndecided Under 30904010 Over 30104010 Marginals are the row and column totals: 140 60 100 80 20

6 Step 2: Compute Expected Frequencies (f E ) wf E = (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

7 Step 3: Compute X 2 wFind difference (residual) betw observed & expected for each cell (fo – fe) wSquare those differences wDivide squared differences by fe wSum the results

8 Summary FoFeFo-fe(fo-fe) 2 (fo-fe) 2 / fe <30 D 9070204005.71 <30 A 4056-162564.57 <30 U 1014-4161.14 >30 D 1030-2040013.33 >30A 40241625610.67 >30 U 1064162.67

9 (cont.) wLast step: Add up (fo-fe) 2 / fe w  2 = 5.71 + 4.57 + 1.14 + 13.33 + 10.67 + 2.67 = 38.09 wStep 4: Compare to  2 critical with df = (# columns – 1) (# rows – 1) wHere df = (2-1)(3-1) = 2 df,  =.05, critical = 5.99

10 Hypothesis Test wIf  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

11 In SPSS wAnalyze  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”

12 SPSS (cont.) wOutput – 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 wNote there is an option for clustered graphing, read this example in the lab


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