MATH 203 Lecture Notes - Lecture 21: Contingency Table, Null Hypothesis, Complement Factor B
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MATH 203 Full Course Notes
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Two way analysis: h0: factor a and factor b levels are assigned independently. ^nij= ni . n. j n k r i=1. Ni. and n. j are the row |column totals for row i and column j respecively. If so, we would have evidence against h0 at the level: e. g. 1) 2) 2 * 4 coningency table -> (r= 2, c=4) so df = (1*3) = 3. 11000: is there a diference in probability of winning across cup sizes, null hypothesis: H0: probability of winning is the same for all cup sizes: we would consider the elements in the columns of the table as realizaions of three random sampling schemes with the same probability of success p. Esimated values: est. n21 = est. n. 1*est. p = 1500*0. 324 = 485, est. n22 = est. n. 2*est. p = 5000*0. 324 = 1618, est. n23 = est. n. 3*est. p = 3500*0. 324 = 1133, est. n24 = est. n. 4*est. p = 1000*0. 324 = 324.