BIOL 570 1nd Edition Lecture 13 Outline of Last Lecture I. Chi square goodness of fit testsII. assumptions of testsOutline of Current Lecture II. Two categories: binomial test vs. chi square testIII. chi square test of associationIV. odds ratiosCurrent LectureTwo categories- Goodness of Fit1. binomial test (give exact P-value)2. x2 goodness of fit testExample: Hitchhikers Thumb HO: proportion of hitchhikers thumb = 0.25HA: p ≠ .25N= 8 students, 4HHT (x=4)1) binomial test is valid: test statistic = 42) x2 GOG invalid! – violates assumptions (No more than 20% of Expected < 5)Observed ExpectedTrait (HHT) 4 2No trait 4 6 Total: 8 8These 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.n=800, x=4001) Binomial test- hard to do without a computer!Pr(x=400) = 800! (400!400!) (.25)400(.75)4002)Obs ExpTrait (HHT) 400 200No trait 400 600- Use binomial test for exact P-value- Use x2 GOF when there is a large sample sizex2 GOF – 1 variablex2 test of association – 2 categorical variables= contingency table testStep 1: Is there any correlation between health status and diet (oil intake)? – 2 variablesStep 2:HO: health status is independent of oil intakeHA: health status is dependent Step 3: (obtaining random sample)Step 4: (creating contingency table) – all observed numbersLow Med High Total: Control 1368 1377 1409 4154Colon C 398 397 430 1225Rectal C 250 241 237 728 Total: 2016 2015 2076 Grand total: 6107** Grand total = sample sizeExpected # for any cell= (row total x column total)/ grand total= (728 x 2076)/ 6107 = 247.47Pr (high) = 2076/6107Pr (Rectal C) = 728/6107Two events (A + B) are independent:If P (A and B) = Pr (A) x Pr (B)Pr (high) x Pr (RC) x 6107 ←joint probabilityx2 = (1368 – 137.29)2/ 137.29 +…..+ for each cellx2 = 1.532Degrees of Freedom = (number of rows - 1) x (number of columns - 1) Conclusion: independentDo Not Reject HOOdd ratio – 2 variablesShotNo Yes Total: FluYes 15 3 18No 13 10 23 Total: 28 13 41Success = No flup-hat = probability of successesHave flu = 1 – phatNo shotp-hat no flu = 13/28 = .464p-hat flu = 15/28 = .536Shotp-hat no flu = 10/13 = .729p-hat flu = 3/13 = .231Odds of successes: O-hat = p-hatc 1-phatO-hat no shot= .464/.536 = .866O flu shot = .729/.231 = 3.33Odds RatioOR-hat = O-hat group 1 = shot = 3.33/.866 = 3.845 O-hat group 2 no shotOR-hat = 1: condition equally likelyOR-hat > 1: more likely in first groupOR-hat < 1: less “ ”95% CI for OR (.66-1.12) – cancer exampleIncludes 1, no