Chapter 10--Hypothesis Testing--Categorical Data

# We wish to test if the same probability distribution

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Unformatted text preview: .05? Chapter 10: Hypothesis Testing: Categorical Data Stat 491: Biostatistics Introduction Two-Sample Test for Binomial Proportions McNemar’s Test Estimation of Sample Size and Power R × C Contingency Tables Chi-Square Goodness-of-Fit Test The Kappa Statistic Test for Homogeneity of Proportions Consider the following data from a case-control study. Status Case Control Total <= 20 320 1422 1742 Age at ﬁrst birth 20 − 24 25 − 29 30 − 34 1206 1011 463 4432 2893 1092 5638 3904 1555 ≥ 35 220 406 626 Total 3220 10245 13465 Let p1 , p2 , p3 , p4 and p5 be the probability distribution of age at ﬁrst birth < 20, 20 − 24, 24 − 29, 30 − 34 and ≥ 35 given that a person does not have cancer (is a control). We wish to test if the same probability distribution holds for cases as well. Chapter 10: Hypothesis Testing: Categorical Data Stat 491: Biostatistics Introduction Two-Sample Test for Binomial Proportions McNemar’s Test Estimation of Sample Size and Power R × C Contingency Tables Chi-Square Goodness-of-Fit Test The Kappa Statistic Test for Homogeneity of Proportions Cont’d... Cases Control The data can be graphically displayed as 0.0 0.2 0.4 0.6 0.8 1.0 Are the two stacked bars signiﬁcantly diﬀerent? Lower ages appear more prevalent in the controls and higher ages in the cases. Chapter 10: Hypothesis Testing: Categorical Data Stat 491: Biostatistics Introduction Two-Sample Test for Binomial Proportions McNemar’s Test Estimation of Sample Size and Power R × C Contingency Tables Chi-Square Goodness-of-Fit Test The Kappa Statistic Test for Homogeneity of Proportions Cont’d... If the the probability distributions are homogeneous, then we can say that there is no association between case-control status and age at ﬁrst birth. Let n = n1 + n2 be the total number of subjects and Mj be the total number of subjects in the i th age-at-ﬁrst-birth group. If there is no association, the expected frequency in the (i , j )th cell is ni Eij = Mj × . n We construct the expected t...
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