P(t) = 1 -e-AI = 1 - S(t) _
The conditional probability that the individual will develop CVD in the forthcoming interval of unit length, given that the event has not yet occurred prior to that interval, is constant and equal to A.. In other words the hazard function for this model is constant and equal to A.. This assumption of constant success rate may not be appropriate for certain diseases or for large groups of individuals, but may be realistic for certain age-, sex- or race-specific groups. For example, the risk of CVD does not change dramatically for white males, in the USA., during the years 40-45. If J..L denotes the mean survival time, then A. = 1/J..L.
If we consider the information on follow-up of the five individuals discussed in an earlier example, we note that two of these persons had developed CVD, by the end of the study.
This type of observation, or in fact any observation of follow-up which is terminated before the event has occurred, is said to be censored. Under the assumption of exponential survival the maximum likelihood estimate of the hazard function is
~=d/F
(see Gross and Clark26 Chapter 3) where d is the number of events and F is the total follow-up time. In the previous example, d = 2, F = 20 and ~ = 0.1. Thus, the incidence density is the estimate of the hazard under exponential survival. This result will provide the basis for development of formulae to determine necessary sample size. In many epidemiologic studies the hazard rate may change considerably with age although the assumption of a constant hazard may be approximately true during the time period considered in most studies.
Suppose we are planning an experiment which involves observing experimental subjects from the time they receive a "treatment" until a "success" occurs. (For example, the experimental subject may be a human, the treatment may be an analgesic, and success might be the end of a headache.) The subject is observed until "success" occurs; thus there is no censoring. Let tJ, ... ,tn represent the observed success times for then subjects.
In this case ~ = 1/t, where
n t=(1/n)Lti.
~1
It follows from the theory of maximum likelihood estimation that where n is sufficiently large, ~is normally distributed with mean A. and variance A.2/n. This information may be used in the same way that similar information was used to develop sample size formulae for estimation and tests about proportions. The sample size which is necessary to estimate A. to within E of its true value with probability (1-a) is given by the formula
(15)
from I ~-A.I = z1--«12
[AtJ
n] withe= 1~-AJ/A. • These values of n may be looked up directly in Table 15.Example 1.6.1
How many people should be followed-up to estimate the hazard (incidence rate) to within 10% of the true value with 95% confidence?
Statistical Methods for Sample Size Determination Solution
Using formula (15) with e = 0.1 0 and z = 1.960 it follows that n = (1.960/0.1 )2 =384.16,
indicating that we would need to follow-up 385 individuals until the "success" time of each individual was known. This value may be found by using Table 15, at the intersection of the row corresponding to e=0.1 0 and the column corresponding to the 95% confidence level.
31
In order to test a hypothesis about the hazard (incidence rate), we must consider formulation of the alternative hypothesis. The incidence rate expresses risk on the basis of events per so many persons per time unit. We may wish to state the null and alternative hypotheses directly. For example, H0: A= 0.1 versus Ha: A= 0.05. Under Ha we have 5 events per 100 persons per time unit or 1/2 the incidence rate under Ho.
Alternatively, the null and alternative hypothesis may be expressed in terms of the mean length of time to failure, Jl, through the relationship Jl = 1/A. For example, when A = 0.1, Jl = 1/0.1 = 10. In the alternative we might express our belief that the mean length of time to failure will be 15 (this would correspond to A = 1/15 = 0.06667). Since the expression ofthe null and alternative hypothesis in terms of Jl or A is equally satisfactory, the decision should be based on whichever is most biologically meaningful for the problem at hand. This will depend on the available data. If incidence data are available, 'Ao and 'Aa should be specified directly; otherwise A.o and Aa should be specified indirectly through Jlo and lla· The test statistic for H0: A='Ao is
z = (~-'Ao)(,ln)/'J...o
which is distributed N(0,1) under the null hypothesis. To obtain an expression for the necessary sample size which will detect the two-sided alternative Ha: A ::;:. Ao with stated power, we must find that value of n such that
Using a strategy similar to the one used in developing the formula for the sample size for hypothesis testing for a single population proportion, Fig. 3 and equation (4), gives:
Solving for n, it follows that
(z1-w2Ao + z1-~)2
n= (16)
('J...o-J...a)2
Tables 16a-16i provide sample sizes based on formula (16). The appropriate sample size is located in the table, for specified a and ~ levels, at the intersection of the row representing 'Aa and the column representing
Ao·
Example 1.6.2
Suppose it is widely reported that the hazard due to a certain chemical exposure in a particular industry had always been 0.20, but, recently with the introduction of new production techniques the hazard has been changed by 25%. How many people should be followed-up to test H0 : A.o = 0.20 versus 'Aa = 0.15 or 0.25 at the 5% level of significance and with 80% power?
Solution
Using formula (16) with A.o = 0.20; A,.= 0.15 or 0.25; z1_w2 = 1.960 and z1_~ = 0.842 it follows that
or
n = (1.960 X 0.20 + 0.842 X 0.15)2/(0.05)2
=107.45
n = (1.960 X 0.20 + 0.842 X 0.25)2/(0.05)2
= 145.20
Therefore 146 people would have to be followed-up. These results could have been obtained from Table 16e by taking the larger of the two numbers given in the column headed 0.20 and rows headed 0.15 and 0.25.
In the above example, under the null hypothesis the average time to follow-up is J.!o=l/A.0=5 "years" and under the alternative it is lla=l/A,.=4 or 6.7 "years" depending on whether \>A.0 or \ <A.0• Thus the study could take up to nearly 7 "years" to complete if the unit of time for the study is in years. This points out the problem of not allowing for censoring. We need few subjects but it requires a long time to complete. Modifications in the study design based on control of the follow-up period will now be presented for the two-sample problem.
Typically, an investigator will be interested in comparing the incidence rate in two populations. In this situation the goal is usually to test the null hypothesis H0 : A.1 = A.z (or, H0: A.1-A.2=0) rather than to estimate the difference with stated precision. Hence sample size formulae will be developed only for the hypothesis testing situation.
Consider the situation in which each subject is followed until the event of interest is observed. The null hypothesis is stated as H0: A.1-A.2 = 0 and the two-sided alternative is Ha: A.1-A.2 -:t 0, where both A.1 and A.2 are specified under Ha. The methodology for selecting the values of A.1 and A.2 is identical to that used in the single-sample case. The test statistic is
where
X
= (~1
+ ~i)/2 assuming equal group sizes. We must find that value of n such thatProb{ z > z1
.a12l
Ha} = 1-~.Using the same method as was used in developing formula (7) for the sample size for hypothesis testing for two population proportions, and Fig. 4 (replacing P1-P2 with A.1-A.z) gives:
Then, solving for n it follows that
{z1-a!2,1[2A. ] -2 + Z1-~,I[A.14A.z2JF
n= (17)
(A.rA.z)2
Tables 17a-17i provide sample sizes based on formula (17). The appropriate sample size is located in the table, for specified a and ~ levels, at the intersection of the row representing A.z and the column representing A.1•
Statistical Methods for Sample Size Determination 33
If there are to be unequal sample sizes, n1 and n2, from the two populations, then formula (17) generalizes to:
n1=--- (17a)
with X=(~1 + !&2)/(1+k) where k = n2/n1. Tables 17a-17i cover the most common situation of k= 1.
Example 1.6.3
Suppose that a disease hazard due to a certain chemical exposure in a particular industry is believed to be roughly 0.1 0 and a competing industry uses techniques with a disease hazard believed to be roughly 0.05, how many people would be needed to be followed-up in each industrial exposure to test, at the 5% level of significance with a power of 80%, whether there is a difference in the disease hazards in the two industries?
Solution
Using formula (17) with X= (0.1+ 0.05)/2 = 0.075 yields
{
1.960~
2(0.075)2 +0.842J (0.1 )2+ (0.05)2 }
2
n = 2 =36.49
(0.1 -0 .05)
Hence, at least 37 subjects would have to be followed-up in each group until the event/failure occurs. This value may be found as the first entry in Table 17e.
An alternative strategy is to begin the study on a fixed date, allow patients to enroll in the study throughout the period, and terminate enrollment and follow up for "T" years. This controls the time duration of the study but we must worry about how to account for the censored observations which are bound to occur. The mathematical details of the necessary modifications to formula (17) are sketched in the papers by Donneri2 and Lachin37 and developed more fully in Gross and Clark26. The modification of formula (17) requires that we evaluate
f(A.) = ~vs T/(
n -
1 + e-A.T) and use the following formula for n:n=
Example 1.6.4
{z1-aa'i'[2f(X)l + z
1
_~ 'i'[f(A.1)+f(A,z)]}2 (A.1-A,z)2Consider the data in Example I. 6.3 with the additional limitation that the study will terminate in 5 years. We wish to test H0 : A.1 = A.2 = 0.1 versus the alternative that Ha: A.1 = 0.1, A.2= 0.05 with a = 0.05, ~ = 0.2. How many people should be followed up?
Solution
Using the formula for f(A.):
f(X = o.o75) = o.o339
(18)
and
f(A.t = 0.1) = 0.0469 f(~ = 0.05) = 0.0217
n = {1.960 v'[2(0.0339)] + 0.8416 v'[0.0469 + 0.0217]}2/(0.1-0.05)2
= 213.7
By restricting the study duration to 5 years we need many more subjects (214) in each group than was previously the case. The reason for this is that the average time to failure is 10 years under H0 and 20 years under Ha. . The survival rates are too high to be realistic for a 5 year study.
Example 1.6.5
Suppose that the average survival for patients suffering from a specific disease and receiving a standard treatment is 2 years but there is a new treatment which will receive approval for marketing if it can be demonstrated that it would increase the patients' survival, on average, by at least 1 year; how many subjects would a 5 -year study require?
Solution
In this example
"-1 =
0.5 , "-2 = 0.33 andX.
= 0.4167and
f
(X.
= 0.4167) = 0.2995 f (A.1 = 0.5) = 0.3950 f ( "-2 = 0.33) = 0.2164n = {1.645v'[2(0.2995)] + 0.842v'[0.3950 + 0.2164]}2/(0.5 - 0.33)2
=163.74
Thus 164 subjects would be needed in each group. If, however, the follow-up was uncensored 100 subjects would be needed in each group.
Examples 1.6.4 and 1.6.5 illustrate that the proposed length of the study cannot be chosen with total disregard to the survival times that are likely to be observed.
Using the results presented in Gross and Clark26, Lachin37 has extended formula (18) to cover the situation in which subjects are enrolled for T 1 years and the total duration of the study is T years ..
The following formula is used for n:
where
n=
Example 1.6.6
{z1-CXJ2v'[2g(X,)] + z
1
_~v'[g(A.t)+g(A.z)]}2 (A.t-A.z)2Suppose, in Example 1.6.5, that subjects are to be enrolled in the study for 2.5 years and then continue the follow-up for another 2.5 years, how many people would have to be included in the study?
Solution
With T 1 = 2.5, T = 5, "-1 = 0.5, A.2 = 0.3333 and
X.
= 0.4167 yields g(X.
= 0.4167) = 0.2224(19)
Statistical Methods for Sample Size Determination
g
P-1
= o.5ooo) = 0.2989 g (1 .. 2 = 0.3333) = o.1576 Then using formula (19)n = {1.645"[2(0.2224)] + 0.842 "[0.2989 + 0.1576]}2/(0.5-0.3333)2
= 99.88
35
Thus only 100 subjects would be needed in each group if the design allowed for 2.5 years of follow-up after an enrollment period of 2.5 years. The saving of subjects occurs because all subjects are being followed for at least 2.5 years, which is the average survival time under H0 and Ha. Thus we expect to observe many more failures/events.
The precision of the study depends on the expected number of events. The design which generates the greatest number of events over the shortest enrollment and follow-up period will require the fewest overall number of participants.
The development of formulae for sample size involving incidence rates was framed entirely in the domain of survival studies. Examples of these in health research include most clinical trials of cancer therapies. The results are easily interpreted in this context.
The extension to other types of studies comes from the observation that the measure, incidence density, or person-years incidence, is an estimate of a hazard function under exponential survival. This assumption will be approximately valid for relatively homogeneous groups of subjects.
Finally, the use of the incidence density ratio (IDR) as a measure of comparing two populations will, under certain conditions, be as an approximation to the relative risk. In the notation of this Chapter, the test of the hypothesis that A.1 = A.2 is fully equivalent to a test that the IDR = A.JIA.2= 1. Direct tests about the ratio would be based on ln(IDR) = ln(A.1)-ln(A.2). Sample sizes based on the distribution theory ofln(A.1)-ln(A.2) will not differ significantly from the sample sizes obtained from equations (17), (18), and (19), which are likely to be accurate enough for most purposes.
associated methods for inference concerning the population proportion can be extended for application to continuous random variables and parameters such as population means and totals. In this Chapter we present some of the formulae necessary to determine sample sizes for estimating and testing hypotheses about the population mean.