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PROBMC |
Category: | Probability |
Syntax |
PROBMC(distribution, q, prob, df, nparms<, parameters>) |
Distribution | Argument |
---|---|
One-sided Dunnett |
'DUNNETT1' |
Two-sided Dunnett |
'DUNNETT2' |
Maximum Modulus |
'MAXMOD' |
Studentized Range |
'RANGE' |
Williams |
'WILLIAMS' |
Restriction: | Either q or prob can be specified, but not both. |
Restriction: | Either prob or q can be specified, but not both. |
Note: A missing value is interpreted as an infinite
value.
Note: For DUNNETT1 and DUNNETT2, the control group is not
counted.
Details |
The PROBMC function returns the probability or the quantile from various distributions with finite and infinite degrees of freedom for the the variance estimate.
The prob argument is the probability that the random variable is less than q. Therefore, p-values can be computed as 1- prob. For example, to compute the critical value for a 5% significance level, set prob= 0.95. The precision of the computed probability is O(10^{--8}) (absolute error); the precision of computed quantile is O(10^{--5}).
Note: The studentized range is not computed for finite degrees of freedom and unequal
sample sizes.
Note: Williams' test is computed
only for equal sample sizes.
The equations listed here define expressions used in equations that relate the probability, prob, and the quantile, q, for different distributions and different situations within each distribution. For these equations, let be the degrees of freedom, df.
Note:
The studentized range is not computed for finite degrees
of freedom and unequal sample sizes.
PROBMC computes the probabilities or quantiles from the distribution defined in Williams (1971, 1972) (See References). It arises when you compare the dose treatment means with a control mean to determine the lowest effective dose of treatment.
Note: Williams' Test is computed only for equal sample
sizes.
Let X_{1}, X_{2}, ..., X_{k} be identical independent N(0,1) random variables. Let Y_{k} denote their average given by
It is required to compute the distribution of
where
Y_{k} | is as defined previously |
Z | is a N(0,1) independent random variable |
S | is such that ½S^{2} is a ^{2} variable with degrees of freedom. |
As described in Williams (1971) (See References), the full computation is extremely lengthy and is carried out in three stages.
You can write a recursive expression for the probability of Y_{k} > d, with d being any real number.
To compute this probability, start from a N(0,1) density function
and recursively compute the convolution
From this sequential convolution, it is possible to compute all the elements of the recursive equation for , shown previously.
The third stage is not needed when = ∞. Due to the complexity of the operations, this lengthy algorithm is replaced by a much faster one when k 15 for both finite and infinite degrees of freedom . For k 16, the lengthy computation is carried out. It is extremely expensive and very slow due to the complexity of the algorithm.
Comparisons |
The MEANS statement in the GLM Procedure of SAS/STAT Software computes the following tests:
Examples |
This example shows how to use PROBMC in a DO loop to compute probabilities:
data probs; array par{5}; par{1}=.5; par{2}=.51; par{3}=.55; par{4}=.45; par{5}=.2; df=40; q=1; do test="dunnett1","dunnett2", "maxmod"; prob=probmc(test, q, ., df, 5, of par1-par5); put test $10. df q e18.13 prob e18.13; end; run;Probabilities from PROBMC shows the results of this DATA step that are printed to the SAS log.
DUNNETT1 40 1.00000000000E+00 4.82992188740E-01 DUNNETT2 40 1.00000000000E+00 1.64023099613E-01 MAXMOD 40 1.00000000000E+00 8.02784203408E-01 |
This example shows how to compare group means to find where the significant differences lie. The data for this example is taken from a paper by Duncan (1955) (See References) and can also be found in Hochberg and Tamhane (1987) (See References). The group means are
49.6 | |
71.2 | |
67.6 | |
61.5 | |
71.3 | |
58.1 | |
61.0 |
data duncan; array tr{7}$; array mu{7}; n=7; do i=1 to n; input tr{i} $1. mu{i}; end; input df s alpha; prob= 1-alpha; /* compute the interval */ x = probmc("RANGE", ., prob, df, 7); w = x * s / sqrt(6); /* compare the means */ do i = 1 to n; do j = i + 1 to n; dmean = abs(mu{i} - mu{j}); if dmean >= w then do; put tr{i} tr{j} dmean; end; end; end; datalines; A 49.6 B 71.2 C 67.6 D 61.5 E 71.3 F 58.1 G 61.0 30 8.924 .05 ;Group Differences shows the results of this DATA step that are printed to the SAS log.
A B 21.6 A C 18 A E 21.7 |
This example shows how to compute 95% one-sided and two-sided confidence intervals of Dunnett's test. This example and the data come from Dunnett (1955) (See References) and can also be found in Hochberg and Tamhane (1987) (See References). The data are blood count measurements on three groups of animals. As shown in the following table, the third group serves as the control, while the first two groups were treated with different drugs. The numbers of animals in these three groups are unequal.
Treatment Group: | Drug A | Drug B | Control |
---|---|---|---|
9.76 | 12.80 | 7.40 | |
8.80 | 9.68 | 8.50 | |
7.68 | 12.16 | 7.20 | |
9.36 | 9.20 | 8.24 | |
10.55 | 9.84 | ||
8.32 | |||
Group Mean | 8.90 | 10.88 | 8.25 |
n | 4 | 5 | 6 |
The mean square error s^{2} = 1.3805 (s = 1.175) with = 12.
data a; array drug{3}$; array count{3}; array mu{3}; array lambda{2}; array delta{2}; array left{2}; array right{2}; /* input the table */ do i = 1 to 3; input drug{i} count{i} mu{i}; end; /* input the alpha level, */ /* the degrees of freedom, */ /* and the mean square error */ input alpha df s; /* from the sample size, */ /* compute the lambdas */ do i = 1 to 2; lambda{i} = sqrt(count{i}/ (count{i} + count{3})); end; /* run the one-sided Dunnett's test */ test="dunnett1"; x = probmc(test, ., 1 - alpha, df, 2, of lambda1-lambda2); do i = 1 to 2; delta{i} = x * s * sqrt(1/count{i} + 1/count{3}); left{i} = mu{i} - mu{3} - delta{i}; end; put test $10. x left{1} left{2}; /* run the two-sided Dunnett's test */ test="dunnett2"; x = probmc(test, ., 1 - alpha, df, 2, of lambda1-lambda2); do i=1 to 2; delta{i} = x * s * sqrt(1/count{i} + 1/count{3}); left{i} = mu{i} - mu{3} - delta{i}; right{i} = mu{i} - mu{3} + delta{i}; end; put test $10. left{1} right{1}; put test $10. left{2} right{2}; datalines; A 4 8.90 B 5 10.88 C 6 8.25 0.05 12 1.175 ;Confidence Intervals shows the results of this DATA step that are printed to the SAS log.
DUNNETT1 2.1210786586 -0.958751705 1.1208571303 DUNNETT2 -1.256411895 2.5564118953 DUNNETT2 0.8416271203 4.4183728797 |
Treatment | Mean |
---|---|
X_{0} | 10.4 |
X_{1} | 9.9 |
X_{2} | 10.0 |
X_{3} | 10.6 |
X_{4} | 11.4 |
X_{5} | 11.9 |
X_{6} | 11.7 |
The mean square, with (7 - 1)(8 - 1) = 42 degrees of freedom, is s^{2} = 1.16.
Determine the maximum likelihood estimates M_{i} through the averaging process.
The maximum likelihood estimates under the alternative hypothesis are
M_{0} = M_{1} = M_{2} = X_{0,1,2} = 10.1 | |
M_{3} = X_{3} = 10.6 | |
M_{4} = X_{4} = 11.4 | |
M_{5} = M_{6} = X_{5,6} = 11.8 |
Now compute , and the probability that corresponds to k = 6, = 42, and t = 2.60 is .9924467341, which shows strong evidence that there is a response to the substance. You can also compute the quantiles for the upper 5% and 1% tails, as shown in the following table.
SAS Statements | Results |
---|---|
prob=probmc("williams",2.6,.,42,6); |
0.99244673 |
quant5=probmc("williams",.,.95,42,6); |
1.80654052 |
quant1=probmc("williams",.,.99,42,6); |
2.49087829 |
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