Categorical Data Analysis Chapter 3

Categorical Data Analysis Chapter 3

The actual coverage probability is usually a bit higher than the nominal level. Confidence intervals for association parameteres Consider the odds ratio in the 2x2 table, ˆθ = (n 11 n 22 )/(n 12 n 21 ) Unless n is vary large, its sampling distribution is highly skewed. The log transform converges more reapidly to normality. An estimated standard error for log ˆθ is ˆσ(log ˆθ) 1 = n 1 1 + 1 n 1 2 + 1 n 2 1 + 1 n 2 2 By the large-sample normaility of log ˆθ, the Wald confidence interval for log θ is log ˆθ ± z α/2ˆσ(log ˆθ) The Wald CI for θ is exp ( ) log ˆθ ± z α/2ˆσ(log ˆθ)

Wald CI for difference of proportions The estimated difference of proportions ˆπ 1 ˆπ 2 is unbiased for the true difference π 1 π 2 and has the standard error π 1 (1 π 1 ) σ(ˆπ 1 ˆπ 2 ) = + π 1(1 π 1 ) n 1 n 2 The estimate ˆσ(ˆπ 1 ˆπ 2 ) replaces π i by ˆπ i. Then ˆπ 1 ˆπ 2 ± z α/2 σ(ˆπ 1 ˆπ 2 ) is a Wald CI for π 1 π 2. Like the Wald interval for a single proportion, it usually has true coverage probability less than the nominal confidence level, especially when π 1 and π 2 are near 0 or 1.

Wald CI for relative risk Like the odds ratio, the log relative risk log r converges to normality faster on the log scale. An estimated standard error for log r is ˆσ(log r) = 1 ˆπ1 y 1 + 1 ˆπ 2 y 2 The Wald interval for r is ( ) exp log r ± z α/2ˆσ(log r). It tends to be somewhat conservative.

Example: Aspirin and Heart attacks revisited The proportions having fatal hear attacks were 18/11,034=0.00163 for placebo group and 5/11,037=0.00045 for aspirin group. The 95% CI for the log relative risk is log(0.00163/0.00045) ± 1.96(0.505) = log(1.34, 9.70) Despite the very large sample sizes, due to the very low rate of heart attach deaths, the estimated effect is imprecise. The Wald 95% CI for π 1 π 2 is 0.0012 ± 1.96(0.00043) = (0.0003, 0.002) The Wald 95% CI for odds ratio is log(3.62) ± 1.96(0.51) = log(1.33, 9.84)

Deriving standard errors with the delta method If n(t n θ) d N(0, σ 2 ) then n(g(tn ) g(θ) d N(0, [g (θ)] 2 σ 2 )

Score confidence interval for difference in proportions Consider testing H 0 : π 1 π 2 = 0 Let ˆπ 1 ( 0 ) and ˆπ 2 ( 0 ) denote the ML estimates of π 1 and π 2 subject to the constraint π 1 π 2 = 0. The score test statistic is z( 0 ) = (ˆπ 1 ˆπ 2 ) 0 ˆπ1 ( 0 )[1 ˆπ 1 ( 0 )] n 1 + ˆπ 2( 0 )[1 ˆπ 2 ( 0 )] n 2 The score CI is the set of 0 such that z( 0 ) < z α/2.

Score confidence interval for odds ratio Consider testing on the odds ratio H 0 : θ = θ 0 Let ˆµ ij (θ 0 ) be the unique expected frequeny estimates that have the same row and column margins as {n i j} and satisfy The set of θ 0 satisfying ˆµ 11 (θ 0 )ˆµ 22 (θ 0 ) ˆµ 12 (θ 0 )ˆµ 21 (θ 0 ) = θ 0 X 2 (θ 0 ) = (n ij ˆµ ij (θ 0 )) 2 /ˆµ ij (θ 0 ) < χ 2 1(α) form a 100(1 α)% score-test-based confidence interval.

Profile likelihood CI Consider testing on the odds ratio H 0 : θ = θ 0 The set of θ 0 satisfying G 2 (θ 0 ) = 2 i n ij log[n ij /ˆµ ij (θ 0 )] < χ 2 1(α) j form a 100(1 α)% likeli-ratio test-based CI.

Example: Aspirin and heart attacks profile LRT CI for odds ratio: (1.44, 2.34) score CI for difference in proportion: (0.0047, 0.0108) score CI for relative risk (1.43, 2.30) score CI for odds ratio: (1.44, 2.33)

Testing independence in two-way contingency tables Consider the hypothesis of statistical independence H 0 : π ij = π i+ π +j for all i and j The Pearson X 2 test statistic is X 2 = (n ij ˆµ ij ) 2 ˆµ i j ij where ˆµ ij = nˆπ i+ˆπ +j = n i+ n +j /n which is the expected cell count under independence. Under H 0, X 2 follows an asymptotic chi-square distribution with. df = (IJ 1) (I 1) (J 1) = (I 1)(J 1) The likelihood-ratio test produces a different statistic: G 2 = 2 n ij log(n ij /ˆµ ij ) i j which follows the same asymptotic distribution as X 2

Adequacy of Chi-Squared approximations When there are independent multinomial samples, independence between the row and column corresponds to homogeneity of each outcome probability among the rows or columns with fixed margin. The limiting chi-squared results still hold. The convergence of the actual sampling distribution of X 2 or G 2 to the chi-squared distribution applies as n grows for a fixed number of cells. The adequacy of the approximation depends on both n and the number of cells. The size of n/ij that produces adequate approximations for X 2 tends to decrease as IJ increases.

Adequacy of Chi-Squared approximations Research has shown that X 2 performs adequately with smaller n and more sparse tables than G 2. The distribution of G 2 is usually poorly approximated by chi-squared when n/ij < 5. Chi-squared approximations for both tend to be poor for tables containing both very small and moderately large µ ij. Small-sample methods are available whenever it is doubtful.

Example: Education and belief in God X 2 = 76.1, G 2 = 73.2 with df = (3 1)(6 1) = 10. The P-values are < 0.0001. These statistics provide extremely strong evidence of an association.

Following-up Chi-squared tests When a test of independence has a small p-value, what does it say about the strength of the association? Not much, the smaller the p-value, the stronger the evidence that AN association exists It does not tell you that the association is very strong To understand more about assoication, do 1) a residual analysis 2) Consider partitioning the Chi-square statistics into independent pieces to examine association in subtables

Residuals Pearson residuals e ij = n ij ˆµ ij ˆµij Pearson residuals have asymptotic variances less than 1, averaging [(I 1)(J 1)]/IJ Standardized residuals r ij = n ij ˆµ ij ˆµ ij (1 p i+ )(1 p +j ) In 2x2 tables, df = 1 and r 11 = r 12 = r 21 = r 22, and any r 2 ij = X 2 A standardized residual that exceeds about 2 or 3 in absolute value indicates lack of fit of H 0.

Example: Education and Belief in God revisited n 36 = 293, ˆµ 36 = 358.8, p 3+ = 581/2000 = 0.2905, p +6 = 1235/2000 = 0.6175. r 36 = (293 358.8)/ 358.8(1 0.2905)(1 0.6175) = 6.7 We can infer that in the population in 2008, fewer people at the highest level of education would have responded know God exists than if the variables were truly independent.

Example: Mosaic plot

Partitioning Chi-Squared After rejecting independence, the next question could be Are there individual comparisons more significant than others? Partitioning may show the association is largly dependent on certain categories or groupings of categories For IXJ tables, one way to partition G 2 to the G 2 of the (I 1)(J 1) separate 2x2 tables is The G 2 s of the (I 1)(J 1) tables are independent.

Example: Origin of Schizophrenia Here G 2 = 23.04 with df = 4. To understand the association better, we partition G 2 into 4 independent components.

Example the order of the above tables, G 2 = 0.29, 1.36, 12.95, 8.43 respectively. In The psychoanalytic school seems more likely than the other schools to ascribe the origins of schizophrenia as being a combination Of those who chose either the biogenic or environmental origin, members of the psychoanalytic school where somewhat more likely than the other schools to choose the environmental origin.

Partitioning For G 2, exact partitioning occurs Pearson X 2 does not have this property, but since X 2 and G 2 are asymptotically equivalent, X 2 can be used for subtables too The selection of subtables is not unique. To initiate the process, you can use your residual analysis to identify the most extreme cells and begin there Association measures such as odds ratio, relative risk, difference of proportions, and association factors and their CI can also be used describe strength of association in subtables.

Rules for partitioning The df for the subtables must sum to the df for the full table Each cell count in the full table must be a cell count in one and only one subtable Each marginal total of the full table must be a marginal total for one and only one subtable