ELSEVIER
LETTERS TO THE EDITOR NUMBER OF PHONE LINES FOR TELEPHONE
RANDOMIZATION
Many multi-institution clinical trials register and randomize patients by telephone, by speaking directly with the trial registrar, responding to a menu of prerecorded questions by touch-tone, or logging onto a central computer by modem. For smooth conduct of the trial, this process should not be burdensome to the caller. Frequent busy signals would be particularly problematic in trials with urgent treatment, for example, trials for patients in acute trauma. The probability of a busy signal depends on the arrival rate of incoming calls, the length of each call, and the number of available phone lines. Only the last of these parameters may be under the control of the coordinating center. This letter uses queuing theory to provide a rational basis for choosing the number of phone lines. Suppose K phone lines are available and calls arrive randomly according to a Poisson process, with rate X per minute. If each call lasts T minutes, a busy signal will occur if K more calls arrive before T minutes have elapsed. Interarrival times have independent exponential distributions with rate A. The waiting time W until the Kth call is gamma distributed with rate parameter X and shape parameter K, since this random variable is the sum of K independent and identically distributed exponentials [l]. Therefore, the probability b of a busy signal with K phone lines can be calculated from the cumulative distribution function (cdf) of the gamma distribution:
or from the cdf of a chi square random variable [l]: b = Pr(V<2hT),
where V m $(2K)
The expected number of additional calls that arrive during any given call is XT and the expected number of busy signals per year is bX X&l X24X365. Table 1 shows the minimum number of phone lines required to ensure a specified probability of a busy signal as a function of the expected number of additional calls that arrive during any given call. The number of phone lines
Table 1
Minimum
Number of Phone Lines Required
b
0.01
0.1
1
5
10
2
3
6
14
22
co.01
1
2
5
12
19
co.1
1
1
3
9
15
Controlled Clinical Trials 18:38&382 (1997) Published by Elsevier Science Inc. 1997 655 Avenue of the Americas, New York, NY 10010
0197-2456/97/$0.00
Letters to the Editor
381
required is more sensitive to the rate of incoming calls, A, or the length of a call, T, than to the desired probability of a busy signal, b. When the rate of incoming calls or the length of each call is not fixed, it is useful to consider a variety of scenarios. Calculating the probability of a busy signal during the time when the rate of calls is highest yields a worst-case scenario. If, for example, call lengths are expected to vary uniformly between five and ten minutes, a quick approximation can be obtained by averaging the results of calculations under the assumptions of five and ten minutes. We hope that the formulas given above provide clinical trialists with a useful tool for anticipating needs, managing resources, and justifying expenses. Leslie A. Kalish, ScD Henry A. Feldman, PhD New England Research Institutes Watertown, Massachusetts PII !30197-2456(97)00007-X
REFERENCE 1. Bain L. Gamma distribution. In: Kotz S, Johnson NL, eds. Encyclopedia ofStatistical Sciences, Vol. 3. New York: John Wiley and Sons; 1982~292-298.
THE ODDS
RATIO
The odds ratio (OR) has a long history of applications in epidemiology and biostatistics. Its recent use to support the FDA licensing application of a diagnostic product led to developing a simple way of presenting the odds ratio to nonstatisticians and nonepidemiologists. Suppose one had a gold standard (truth) (G) and a diagnostic test with results as given in the illustrative table below.
Gold Standard Results Diagnostic Test Results Positive Negative Total
Positive
Negative
Total
60 20 80
10 75 85
70 95 165
The OR is given by 60X75/10X20 = 22.5. The Positive Predictive Value (PPV) is 60/70 = 0.857 and the Negative Predictive Value (NPV) is 75/95 = 0.789. The odds for the gold standard’s being positive when the test is positive is thus (60/ 70)/(10/70) = 6. Similarly, the odds for the gold standard’s being negative when the test is negative is (75/95)/(20/95) = 3.75. The product of these is the OR. Mathematically, the result is