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1 A STEPWISE MAXIMUM MODULUS TEST WITH UNEQUAL VARIANCES D. Bristol and N. Stouffer Purdue Pharma LP, Princeton NJ.

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Presentation on theme: "1 A STEPWISE MAXIMUM MODULUS TEST WITH UNEQUAL VARIANCES D. Bristol and N. Stouffer Purdue Pharma LP, Princeton NJ."— Presentation transcript:

1 1 A STEPWISE MAXIMUM MODULUS TEST WITH UNEQUAL VARIANCES D. Bristol and N. Stouffer Purdue Pharma LP, Princeton NJ

2 2 The maximum modulus test (R E ) for simultaneously comparing the means of normal distributions to a known constant is usually presented with the assumption of equal variances. WHY?

3 3 A competing procedure (R I ) for use when the unknown variances are not assumed to be equal is presented.

4 4 Let X ij ~ N( i, i 2 ),j=1,...,n i, i=1,...,k, where 1 2,…, k 2 are unknown. Test H 0 : 1 =…= k = 0 (known) against H 1 : i 0 for at least one i.

5 5 Let and s i 2 denote the sample mean and sample variance for the random sample of size n i from the distribution with mean i (referred to as Treatment i).

6 6 Equal Variances Assume 1 2 =…= k 2 = 2, where the common variance 2 is unknown. Tukey (1952, 1953) (see Tukey (1993))

7 7 Let s 2 = denote the pooled estimate of the variance 2. Further, let t i =n i 1/2 ( - 0 )/s, i=1,...,k. Then t i has a noncentral t-distribution with noncentrality parameter i =n i 1/2 ( i - 0 )/ and = i = (n i -1) df, i=1,...,k. Equal Variances

8 8 Test Procedure R E : Reject H 0 in favor of H 1 if max{ | t i |; i=1,..,k} C E. The subscript E indicates that the equality of the variances is assumed. The Maximum Modulus Test For Equal Variances

9 9 Equal Variances C E, chosen to have an -level test, is available from Miller (1981) - Pillai and Ramachandran (1954) - Hahn and Hendrickson (1971) See also Dunnett (1955) and Bechhofer and Dunnett (1988). SAS function PROBMC.

10 10 Unequal Variances Chakraborti (1991) considered sampling one treatment at a time (stepwise sampling) and then testing H 0i : i = 0 against H 1i : i 0 after the observations are obtained from Treatment i and prior to sampling from subsequent treatments.

11 11 Let T i =n i 1/2 ( - 0 )/s i, i=1,...,k. T i has a noncentral t-distribution with noncentrality parameter i = n i 1/2 ( i - 0 )/ i and i =n i -1 degrees of freedom, i=1,...,k. Here the maximum modulus test is Test Procedure R I : Reject H 0 in favor of H 1 if max{ | T i |; i=1,..,k} C I. The Maximum Modulus Test For Unequal Variances

12 12 Unequal Variances Values of C I, chosen to have an -level test, are given in Chakraborti (1991), with emphasis on small unequal sample sizes. According to SAS (1992), PROBMC can be used for this situation; however, PROBMC can only be used when the unknown variances are assumed to be equal.

13 13 The lack of an assumption of equal variances allows estimation of each variance using data from the respective treatment. These estimates, and thus T i, i=1,..,k, are independent. The subscript I refers to this independence.

14 14 Power Comparison The assumption of equal variances should provide desirable properties when it is true. It is also important to examine the consequences of making this assumption when it is not true and not making this assumption when it is true.

15 15 Power Comparison Comparison of the power of the two procedures is made using various simulations for =0.05. For each entry, 500,000 simulations were performed. Without loss of generality, 0 =0.

16 16 Power Comparison Let P E denote the estimated power for R E and let P I denote the estimated power for R I. Values of P I were compared to the exact results and the differences were all <0.1%.

17 17 Power Comparison ( 1,…, k )=(1,…,1)

18 18 Power Comparison Equal Variances 1 = 2 = 3 = 4 =1; 2 1 = 2 2 = 2 3 = 2 4 =6

19 19 Power Comparison ( 1,…, k )=(1,…,1)

20 20 Power Comparison ( 1,…, k )=(1,…,1)

21 21 Power Comparison Unequal Variances 1 = 2 = 3 = 4 =1; 2 1 = 4; 2 2 =4; 2 3 =9; 2 4 =9

22 22 Power Comparison ( 1,…, k )=(0,…,0)

23 23 Power Comparison ( 1,…, k )=(0,…,0)

24 24 Power Comparison ( 1,…, k )=(0,…,0)

25 25 EqualUnequal Not True P E > P I (Small Difference) P I >P E (Usually True) True P I = P E = P I = P E > Variances H0H0 Conclusion

26 26 Recommendation: Use R I Preserves Type I error rate Unequal Variances: Gain (or small loss) in power Equal Variances: Small loss in power


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