CJUS K300 1nd Edition Lecture 14 Outline of Last Lecture I Chi Square Goodness of Fit Test Outline of Current Lecture II Chi Square Test of Independence Current Lecture Expected Frequency Table Lables White Non White Row Marginals Non Capital Cell A 118 Cell B 70 118 Capital Cell C 97 Cell D 17 114 Column Marginals 215 87 302 Independent Events P A and B P A x P B P non cap 188 302 622 P white victim 215 302 712 P non cap and white victim 622 X 712 443 443 X 302 134 what you put in the cell This is how we COULD get the expected frequencies but there is an easier way Row marginal x column marginal and dividing by total n fe RM x CM n So for Cell A f e 118 x 215 84 302 These notes represent a detailed interpretation of the professor s lecture GradeBuddy is best used as a supplement to your own notes not as a substitute Testing the hypothesis Alpha of 05 d f Rows 1 Columns 1 2 1 2 1 1 x 1 1 Chi square crit 3 841 Observed 118 70 97 17 Expected 134 54 81 33 N 302 X 2obt 2o n e 103 9 90 7 116 2 8 75 319 55 302 17 55 So we reject Ho Conclusion We are 95 certain that in the population race of victim and whether or not a homicide case is charges as a noncapital or capital offense are NOT independent Are gender of victim and how a case is charged independent Male Female RM Non capital 70 26 96 Capital 42 20 62 CM 112 46 N 158 f o 70 26 42 20 f e 68 27 9 43 9 18 X obt 72 24 2 40 2 22 2 158 6 n 6 df 1 alpha 01 chi square crit 6 635 Fail to reject the null Conclusion We are 99 confident that they are independent events Steps 1 2 3 4 5 6 Determine the expected frequencies State the null and research hypotheses Determine d f and using an alpha of 01 look up chi square crit Calculate Chi square obt Reject of fail to reject the null State your conclusion in words Independent variable gender Dependent type of charge
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