CJUS-K300 1nd EditionExam # 3 Study Guide Lectures: 11 - 14Lecture 11 (October 8) 2 Tailed TestsNull hypothesis: Ho - There’s no difference between the two variables. Research hypothesis (2-tailed/non-directional): H1 - There is a difference between the two variables. But not providing information on what the difference is. Starting off with an example: Domestic Violence Attitude ScalePossible range of scores: 1-50 where higher scores indicate officers taking domestic violence more seriously For the population of police officers: μ = 20For a sample of 75 officers: x = 29, s=4Ho : The sample of officers who have been through the 2 day training program come from a population of officers whose mean score on the Domestic Violence Attitude Scale is 20.H1: The sample of officers who have been through the 2 day training program come from a population of officers whose mean score on the Domestic Violence Attitude Scale is not 20.We typically express these in symbol form:- Ho: µ = 20- H1: μ ≠ 20Steps to test a null hypothesis (2 tailed) 1. State your null and research hypotheses2. Decide on your alpha level (.10, .05, or .01)3. Look up your critical value (e.g., zcrit) (draw a graph/picture it’s easier to visualize)4. Calculate your zobt5. Compare: Is –zcrit ≤ zobt ≤ +zcrit? (refer to your graph/picture)6. State your conclusionCritical regions for alpha .05 = 1.96Equation: Zobt=x−μs√ nIs -1.96 ≤ Zobt ≤ +1.96? or in other words is the Zobt outside of the critical zone? No.- Reject the null hypothesis and find in favor of the research hypothesis Since zobt is in the critical region, we reject the Ho and find instead in favor of the H1.Conclusion: We are 95% certain that the sample of officers who have been through the 2 day training program come from a population whose mean score on the Domestic Violence Attitude Scale is not 20Lecture 12 (October 13) 1 Tailed TestsWhy? Reasons to specify a directional hypothesis:1. It fits the theory2. It gives a mathematical advantageIn a two tailed test both the tails or extremities are included, 2.5% in each tail. For a one tailed test we combine the tails making it 5% in one of the tails. How do we know which tail to use? Asa good rule of thumb you can follow the sign of the null hypothesis. > Means right hand tail. < Means left hand tail (usually)Steps in testing a null hypothesis with a 1 tailed test (so similar to the 2 tailed test)1. State your null & research hypotheses2. Decide on your alpha level (usually given to you)3. Look up the critical value (Zcrit and draw a graph if you wish)4. Calculate Zobt=x−μs√ n5. Compare Zobt and Zcrit6. State your conclusion in wordsUsing the same example from last class with 2 tailed tests, we will look again at the Violence Attitude Scale.Possible range 1-50 (higher score = taking domestic violence more seriously)Population of police officers μ = 20Sample of 75 officers x = 20.8, s = 4 Ho=20 H1>20 H1 interpreted: The sample of officers who have been through the 2 day training program come from a population of officers whose mean score on the Domestic Violence Attitude Scales is GREATER than 20. Alpha: .05 corresponds to 1.65Zobt: 1.74 Reject the null Another exampleDo teenage shoplifters steal less valuable items than the general population of shoplifters? You can find hints in the research question about the direction of the testPopulation of all shoplifters the mean value of stolen items $83 so μ = 83For a sample of 97 shoplifters ages 13-17 years, the mean value of stolen items was $81 so X=81, s=11, alpha =.05 (look up alpha in the 1-tailed test column) = 1.65Zobt = 1.80Lecture 13 (October 15) Test of Best FitTime of day Observed frequencyExpected frequency12am-6am 20 256am-12pm 7 2512pm-6pm 31 256pm=12am 42 25Total 100 100Ho: x2 = 0 (In the population of all ambulance calls at Green College, the calls are distributed evenly across the 4 time of day categories)Hi: x2 > 0 (In the population of all ambulance calls at Green College, the calls are not distributed evenly across the 4 time of day categories)alpha = .05 Degrees of freedom = K-1 (K most likely stands for categories in german) d.f = 4-1 = 3 at an alpha of .05 X^2crit = 7.815Xobt2=∑(ƒo− ƒe)2ƒe This is the equation in theory and we have to understand this equation inorder to understand what our answer means but we will be using a somewhat more simple equation to work with day to day. Xobt2= 26.96Time of dayƒoƒeƒo−ƒe(ƒo−ƒe)2(ƒo−ƒe)ƒe212am -6am 20 25 -5 25 16am-Noon 7 25 -18 324 12.96Noon-6pm 31 25 7 49 1.446pm-12am 42 25 17 289 11.56Total 100 26.96Is 26.96>7.815? Yes so we Reject HoConclusion: We are 95% certain that, in the population of all Green College ambulance calls, theambulance calls are not evenly distributed across the 4 times of day categories. (You will absolutely need to know how to write this out.)Chi-Squared Computational Formula (aka the formula we will actually be using)Xobt2=∑ƒo2ƒe−n Using the same example with this equation we would do:20225 + 7225+31225 + 42225−100=26.96Lecture 14 (October 20) Test of IndependenceExpected 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=84Testing 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