General Contingency TableVariable 1Tests of IndependenceContingency-Table MethodSame Example as last class: Suppose we sample 1000 students in the SPHTM and 400 are male. Suppose we sample 600 students in the SOM and 300 are male. Use a two-tailed test to test if the proportions of students who are male are different in the two schools. Conduct the test at the 5% level of significance.Male Female TotalSPHTM 40 60 100SOM 30 30 60Total 70 90 160Each population is a row in the table. Each gender is a column. Each person falls into oneof the cells.Note: Number of students at SPHTM is a row total called a row marginal. Note: There are a total of 70 men. This is called acolumn marginal.Note: There are a total of 160 people in the study. This is called the grand total. The above table is a 2 x 2 contingency table. There are two categories each at two levels. Itis arbitrary which variable is the row variableand which is the column variable. Each cell isthe number of subjects with that particular value for each variable. Note: These are the Observed values. These are sometimes labeled O11, O12, O21, O22 where Oij stands for the ith row and the jth column. For this example, we will be conducting a test for homogeneity of two binomial proportions. Is the proportion of Males the same for each population?Expected Table—We need to calculate the numberof subjects in each cell which we would expectif there was no relationship between gender and school; i.e., p1 = p2 = p.Where p1 = Pr(M | SPHTM)and p2 = Pr(M | SOM)To generalize: Let n1 = sample size of 1st sample n2 = sample size of 2nd sample X1 = number exposed (with characteristic)in the 1st sample X2 = number exposed (with characteristic)in the 2nd sampleGeneral Contingency TableVariable 1Population 1 2 Total1 X1n1 – X1n12 X2n2 – X2n2Total X1 + X2n1 + n2 –(X1 + X2)n1 + n2If H0 is true, best estimate of p is2121212211nnXXnnpnpnp orˆˆˆUnder H0, for (1, 1) cell we expect 212111nnXXnpnˆ in that cellThis is the row total * column total / grand totalSimilarly, in (2, 1) cell we expect 212122nnXXnpnˆWe usually call the 4 expected values E11, E12, E21, and E22ExampleExpected TableMale Female TotalSPHTM 43.75 56.25 100SOM 26.25 33.75 60Total 70 90 16075.431607010011*ENote: Row totals are the same for Expected and ObservedNote: Column totals are the same for Expected and ObservedNote: Grand total is the same for Expected and ObservedIf H0 were true, the Expected would be the same as the Observed. How different do they have to be to reject H0?Test Statistic i,jijijijEEOX22This test statistic follows a 2 (chi-square) distribution. 2 is a family of distributions dependent on a parameter called degrees of freedom. The shape is skewed. The mean = k,the number of degrees of freedom and the variance = 2k. The mode is k – 2 unless k = 1. Then the mode = 0 if k = 1. The 2 distribution is derived from the Gaussian distribution.2222212122χσμXσμXχσμXzσμXzSee Table 6 (page 824). We will be interested in the upper values. For the upper 5%, use the column marked .95. For a 2 x 2 contingency table, degrees of freedom = 1. This is (number of rows – 1) * (number of columns – 1).Therefore, i,jijijij~χEEOX2122This is a large sample approximation and holds as long as no Expected Value < 5.For the example,       52381.141667.53571.25.32143.75.3375.333025.2625.263025.5625.566075.4375.434022222XCritical Value: 21 = 3.84p-value: 0.10 < p < 0.25Continuity Correction—A more accurate p-value can be obtained with a correction. We will use E.E||O250This is called Yates Correction. H0: p1 = p2H1: p1  p2       0.50 p 0.251446.13130.04024.0*1878.02414.033.750.5|33.7530|26.250.5|26.2530|56.250.5|56.2560|43.750.5|43.7540|E0.5|EO|X222222Note: This test is identical to the 2-sample binomial test. We get the same p-value and the same decision about accepting or rejecting H0. Short Computational Formula for 2 x 2 OnlyVariablePopulation 1 2 Total1 a b a+b2 c d c+dTotal a+c b+d n1.144690*70*60*100160(520)90*70*60*1002160|30*6030*40|160d)c)(bd)(ab)(c(a2n|c*bd*a|nX2222Tests of IndependenceSometimes we want to examine the relationship between two measures for the same individual. For example, obese or non-obese vs hypertensive or normal. This is called a test of independence or test of association between two characteristics. The same test procedure is used. One characteristic is therow variable and the other is the column variable. The two characteristics are independent if the distribution of one characteristic is the same no matter what the distribution of the other characteristic.Example: A sample of 500 college students participated in a study designed to evaluate the level of college students’ knowledge of a certain group of common diseases. The following table shows the students classified by major field of study and level of knowledgeof the group of diseases:Knowledge of DiseasesMajor Good Poor TotalPremedical 31 91 122Other 19 359 378Total 50 450 500 Do these data suggest that there is a relationship between knowledge of the group of diseases and major field of study of the college students from which the present sample was drawn? Test at the 5% level of significance.H0: Knowledge of the group of diseases and majorfield of study are independentH1: Knowledge of the group of diseases and majorfield of study are not independentTest Statistic: i,jijijijEEOX2250.X2 follows a χ2 distribution with 1 degree of freedom. Reject H0 if X2 > χ21, .95 = 3.84. Expected TableKnowledge of DiseasesMajor Good Poor TotalPremedical 12.2 109.8 122Other 37.8 340.2 378Total 50 450 500E11 = 122 * 50 / 500 = 12.23440980868053452723405023403598375083719810950810991212502123122222......).|.(|.).|.(|.).|.(|.).|.(|XReject H0. p < 0.001. The two variables, knowledge of the group of diseases and major field are not independent.Tests of Independence:1. A single sample is selected from a population and the subjects or objects are