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A STEPWISE MAXIMUM MODULUS TEST WITH UNEQUAL VARIANCES

D. Bristol and N. Stouffer

Purdue Pharma LP, Princeton NJ

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The maximum modulus test (RE) for simultaneously comparing the means of normal distributions to a known constant is usually presented with the

assumption of equal variances.

WHY?WHY?

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A competing procedure (RI) for use when the unknown variances are not assumed to be equal is

presented.

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Let Xij~ N(i, i2),j=1,...,ni, i=1,...,k,

where 12,…, k

2 are unknown.

Test H0: 1=…=k = 0 (known) against H1:i0 for at least one i.

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Let and si2 denote the sample

mean and sample variance for the random sample of size ni from the distribution with mean i (referred to as Treatment i).

iX

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Equal Variances

Assume 12=…=k

2 =2, where the common variance 2 is unknown.

Tukey (1952, 1953)

(see Tukey (1993))

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Let s2= denote the pooled estimate of the variance 2. Further, let ti=ni

1/2( -0)/s, i=1,...,k. Then ti has a noncentral t-distribution with noncentrality parameter i=ni

1/2(i-0)/ and =i=(ni-1) df, i=1,...,k.

2

1

/k

ii

s k

iX

Equal Variances

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Test Procedure RE:

Reject H0 in favor of H1 if

max{ | ti |; i=1,..,k} CE.

The subscript ‘E’ indicates that the equality of the variances is assumed.

The Maximum Modulus Test For Equal Variances

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Equal Variances

CE, 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.

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Unequal Variances

Chakraborti (1991) considered sampling one treatment at a time (stepwise sampling) and then testing H0i: i=0 against H1i: i0 after the observations are obtained from Treatment i and prior to sampling from subsequent treatments.

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Let Ti=ni1/2( -0)/si, i=1,...,k.

Ti has a noncentral t-distribution with noncentrality parameter i= ni

1/2(i-0)/i and i=ni-1 degrees of freedom, i=1,...,k. Here the maximum modulus test is

Test Procedure RI:

Reject H0 in favor of H1

if max{ | Ti |; i=1,..,k} CI.

iX

The Maximum Modulus Test For Unequal Variances

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Unequal VariancesValues of CI, 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.

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The lack of an assumption of equal variances allows estimation of each variance using data from the respective treatment. These

estimates, and thus Ti, i=1,..,k, are independent.

The subscript ‘I’ refers to this independence.

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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.

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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.

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Power Comparison

Let PE denote the estimated power for RE and let PI denote the

estimated power for RI.

Values of PI were compared to the exact results and the differences

were all <0.1%.

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(12,2

2,32)=

(4,4,4)(1

2,22,3

2)=(6,6,6)

n PI PE PI PE10 38.18 44.70 26.80 31.45

20 75.53 78.97 56.57 60.30

30 92.90 93.96 77.99 80.14

40 98.32 98.59 90.22 91.19

50 99.67 99.72 96.11 96.52

Power Comparison (1,…, k)=(1,…,1)

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Power ComparisonEqual Variances

0102030405060708090

100

0 10 20 30 40 50

Number of Observations

Pow

er (P

erce

nt)

PI

PE

1 = 2 = 3 =4 =1; 21 = 2

2 = 23 =2

4 =6

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(12,2

2,32)=

(3,6,6)(1

2,22,3

2,42)=

(3,3,6,6)n PI PE PI PE10 35.05 36.45 41.27 43.72

20 71.29 68.11 81.22 78.35

30 90.15 87.32 96.00 94.19

40 97.23 95.81 99.38 98.84

50 99.33 98.81 99.93 99.82

Power Comparison (1,…, k)=(1,…,1)

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(12,2

2,32)=

(4,4,9)(1

2,22,3

2,42)=

(4,4,9,9)n PI PE PI PE10 32.42 32.50 31.05 31.75

20 66.95 61.73 66.24 60.24

30 87.27 82.17 87.27 80.87

40 95.85 92.96 96.06 92.34

50 98.82 97.60 98.96 97.39

Power Comparison(1,…, k)=(1,…,1)

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Power ComparisonUnequal Variances

0

10

20

30

40

50

60

70

80

90

100

0 10 20 30 40 50

Number of Observations

Pow

er (P

erce

nt)

PI

PE

1 = 2 = 3 = 4 =1; 21= 4; 2

2=4; 23=9; 2

4=9

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Power Comparison(1,…, k)=(0,…,0)

(12,2

2,32)=

(1,1,1)(1

2,22,3

2,42)=

(1,1,1,1)n PI PE PI PE10 4.94 5.00 4.95 4.99

20 4.96 4.98 4.98 4.97

30 4.92 4.93 4.96 4.96

40 4.93 4.97 4.91 4.97

50 4.98 4.98 4.99 4.99

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Power Comparison(1,…, k)=(0,…,0)

(12,2

2,32)=

(1,2,3)(1

2,22,3

2,42)=

(1,2,3,4)n PI PE PI PE10 4.94 6.75 4.95 7.45

20 4.96 6.82 4.98 7.51

30 4.92 6.74 4.96 7.52

40 4.93 6.77 4.91 7.57

50 4.98 6.79 4.99 7.58

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Power Comparison(1,…, k)=(0,…,0)

(12,2

2,32)=

(4,4,9)(1

2,22,3

2,42)=

(4,4,9,9)n PI PE PI PE10 4.94 6.57 4.95 6.87

20 4.96 6.61 4.98 6.91

30 4.92 6.59 4.96 6.92

40 4.93 6.60 4.91 6.96

50 4.98 6.66 4.99 7.01

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Equal Unequal

Not True PE > PI

(Small Difference)

PI >PE

(Usually True)

True PI =

PE =

PI =

PE >

Variances

H0

Conclusion

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Recommendation:Use RI

•Preserves Type I error rate

•Unequal Variances: Gain (or small loss) in power

•Equal Variances: Small loss in power