# Lecture12 - 6 June 2003 Biostatistics 6650-L12 1 Todays...

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6 June 2003 Biostatistics 6650--L12 1

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6 June 2003 Biostatistics 6650--L12 2 Today’s Schedule Contingency tables 2x2 tables RxC tables 1-way tables(Goodness of fit) McNemar’s test for paired(correlated) proportions
6 June 2003 Biostatistics 6650--L12 3 2x2 Tables Recall the test for comparing two independent binomial proportions, p 1 and p 2 , using a normal approximation. Ho: p 1 = p 2 , Ha: p 1 = p 2 not using correction factors 1 2 obs 1 2 i i i i 1 2 1 2 1 2 ˆ ˆ p -p Z = , 1 1 ˆ ˆ pq( + ) n n ˆ p =x /n , i=1,2, x =number of successes in group i ˆ p=(x +x )/(n +n ), ˆ ˆ ˆ ˆ n pq 5 and n pq 5

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6 June 2003 Biostatistics 6650--L12 4 2x2 Tables Another way to test the same hypothesis is by considering a 2x2 contingency table “Success” “Failure” Population 1 Population 2 This can be extended to test the equality of K proportions Ho: p 1 =p 2 =…=p K We can evaluate two types of hypotheses using 2x2 tables 1. Ho: homogeneity (all proportions equal) 2. Ho: independence of rows and columns
6 June 2003 Biostatistics 6650--L12 5 2x2 Tables Now fill in the table with observed frequencies “Success” “Failure” marginal totals Population 1 x 1 n 1 -x 1 n 1 Population 2 x 2 n 2 -x 2 n 2 x 1 +x 2 N-(x 1 +x 2 ) N=n 1 +n 2 =Grand total marginal totals Under the null hypothesis, p 1 =p 2 =p, so we combine both groups to estimate the overall proportion of successes: 1 2 1 2 (x + x ) ˆ p= n + n

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6 June 2003 Biostatistics 6650--L12 6 2x2 Tables Acute Renal Failure example(ARF): Hospitalized ARF patients requiring dialysis. Randomized trial comparing type of dialyzer membrane (cellulose vs polysulfone). Outcome is proportion(p) with recovery of renal function within 30 days. Two independent treatment groups with fixed n1 and n2 Recovery No recovery • Cellulose x 1 =19 n 1 -x 1 = 14 33=n 1 , =19/33=.58 Polysulfone x 2 =13 n 2 -x 2 = 20 33=n 2 , =13/33=.39 32 34 66=N Ho: p 1 =p 2 =p, Homogeneity hypothesis Row totals fixed. Column totals random. Mayo Proc 2000, 75:1141-47 1 ˆ p 2 ˆ p
6 June 2003 Biostatistics 6650--L12 7 2x2 Tables Association of Gleason score and lymph node status in N=485 1990 Mayo prostatectomies--L4 N is fixed, n1 and n2 not fixed. Cross classify two factors. Node - Node + Gleason<7 340 31 371 Gleason 7+ 85 29 114 425 60 485=N Ho: row factor is independent of column factor I.e. that Pr(Gleason 7 and node +)=Pr(Gleason 7)Pr(Node +) Ha: is that the two factors are associated

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6 June 2003 Biostatistics 6650--L12 8 2x2 Tables Association of Gleason score and lymph node status in N=485 1990 Mayo prostatectomies--L4 Expected cell proportions under independence model: p ij =p i. p .j Node - Node + Gleason<7 .67 .09 .76 Gleason 7+ .21 .03 =.12(.24) .24 =114/485 .88 .12 1.00
6 June 2003 Biostatistics 6650--L12 9 2x2 Tables: Notation It turns out that the statistical test used is identical for both the homogeneity and independence hypotheses The test statistic, X 2 obs , is based on observed and expected frequencies or percentages Let O ij = observed number of subjects in the i-th row and j-th column of the table.

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