CJUS-K300 1nd Edition Lecture 14Outline of Last Lecture I. Chi Square Goodness of Fit TestOutline of Current Lecture II. Chi Square Test of IndependenceCurrent LectureExpected Frequency Table LablesWhite Non White Row MarginalsNon-Capital Cell A 118 Cell B 70 118Capital Cell C 97 Cell D 17 114Column Marginals 215 87 302Independent Events: P(A and B) = P(A) x P(B)P(non-cap) = 188/302 = .622P(white victim) = 215/302 = .712P(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 wayRow marginal x column marginal and dividing by total nfe=RM x CMn So for Cell A fe=118 x 215302=84These 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 hypothesisAlpha of .05d.f = (Rows-1)(Columns-1) = (2-1)(2-1) = 1 x 1 = 1 Chi square crit = 3.841 Observed 118, 70, 97, 17Expected 134, 54, 81, 33N=302 Xobt2=∑ƒo2ƒe−n 103.9 + 90.7 + 116.2 + 8.75 = 319.55- 302 = 17.55So we reject HoConclusion: 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 RMNon-capital 70 26 96Capital 42 20 62CM 112 46 N = 158fo - 70, 26, 42, 20fe - 68, 27.9, 43.9, 18Xobt – 72, 24.2, 40.2, 22.2 = 158.6 –n = .6df = 1, alpha = .01, chi square crit = 6.635Fail to reject the null, Conclusion: We are 99% confident that they are independent eventsSteps1. Determine the expected frequencies2. State the null and research hypotheses3. Determine d.f and, using an alpha of .01, look up chi square crit4. Calculate Chi square obt5. Reject of fail to reject the null6. State your conclusion in wordsIndependent variable – gender, Dependent – type of
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