ADM 2304 Lecture Notes - Lecture 10: Contingency Table, Null Hypothesis, Test Statistic
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> could be displayed as a two-way table: Contingency table (or two-way table, cross tabulation, crosstab): > assume that there is no relationship between row and column and compare the observed frequencies with what one would expect if the two variables were independent. > h0: row and column variables are independent. Key idea: if you have independence between the row and column variables then: > pr(in ith row and jth column) = pr(in ith row) pr(in jth col) > pr(ses = low & ss = current) = pr(ses = low) pr(ss = current) Therefore, if you have independence between the row and column variables then, the expected frequency in the (i,j)th cell is: > we want to compare eij and oij. If we have independence they should be fairly similar. >not surprisingly, the chi-square test statistic follows a chi-square distribution with d. f. As per usual, we will reject h0 if: = (3 1)(3 1) = 4 p-value :