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Chapter 212
Non-Inferiority Tests for the Odds Ratio of
Two Proportions
Introduction
This module provides power analysis and sample size calculation for non-inferiority tests of the odds ratio in
two-sample designs in which the outcome is binary. Users may choose between two popular test statistics
commonly used for running the hypothesis test.
The power calculations assume that independent, random samples are drawn from two populations.
Example
A non-inferiority test example will set the stage for the discussion of the terminology that follows. Suppose
that the current treatment for a disease works 70% of the time. Unfortunately, this treatment is expensive
and occasionally exhibits serious side-effects. A promising new treatment has been developed to the point
where it can be tested. One of the first questions that must be answered is whether the new treatment is as
good as the current treatment. In other words, do at least 70% of treated subjects respond to the new
treatment?
Because of the many benefits of the new treatment, clinicians are willing to adopt the new treatment even if
it is slightly less effective than the current treatment. They must determine, however, how much less
effective the new treatment can be and still be adopted. Should it be adopted if 69% respond? 68%? 65%?
60%? There is a percentage below 70% at which the ratio of the two treatments is no longer considered
ignorable. After thoughtful discussion with several clinicians, it was decided that if the odds ratio is no less
than 0.9, the new treatment would be adopted. This odds ratio is called the margin of non-inferiority. The
margin of non-inferiority in this example is 0.9.
The developers must design an experiment to test the hypothesis that the odds ratio of the new treatment
to the standard is at least 0.9. The statistical hypothesis to be tested is
:
0.9
versus
:
> 0.9
Notice that when the null hypothesis is rejected, the conclusion is that the odds ratio is at least 0.9. Note
that even though the response rate of the current treatment is 0.70, the hypothesis test is about an odds
ratio of 0.9. Also notice that a rejection of the null hypothesis results in the conclusion of interest.
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Technical Details
The details of sample size calculation for the two-sample design for binary outcomes are presented in the
chapter “Tests for Two Proportions,” and they will not be duplicated here. Instead, this chapter only
discusses those changes necessary for non-inferiority tests.
This procedure has the capability for calculating power based on large sample (normal approximation)
results and based on the enumeration of all possible values in the binomial distribution.
Suppose you have two populations from which dichotomous (binary) responses will be recorded. Assume
without loss of generality that the higher proportions are better. The probability (or risk) of cure in
population 1 (the treatment group) is
and in population 2 (the reference group) is
. Random samples of
and
individuals are obtained from these two populations. The data from these samples can be
displayed in a 2-by-2 contingency table as follows
Group Success Failure Total
Treatment
Control
Totals
The binomial proportions,
and
, are estimated from these data using the formulae
=
=
and
=
=
Define
=
(
1
)
. Let
.
represent the group 1 proportion tested by the null hypothesis,
. The
power of a test is computed at a specific value of the proportion which we will call
.
. Let
represent the
smallest odds ratio (margin of non-inferiority) between the two proportions that still results in the
conclusion that the new treatment is not inferior to the current treatment. For a non-inferiority test,
< 1
The set of statistical hypotheses that are tested is
:
versus
:
>
which can be rearranged to give
:
versus
:
>
There are three common methods of specifying the margin of non-inferiority. The most direct is to simply
give values for
and
.
. However, it is often more meaningful to give
and then specify
.
implicitly by
specifying the difference, ratio, or odds ratio. Mathematically, the definitions of these parameterizations are
Parameter Computation Hypotheses
Difference
=
.
:
versus
:
>
Ratio
=
.
:
versus
:
>
Odds Ratio
=
.
:
versus
:
>
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Odds Ratio
The odds ratio, =
=
(
(
1
)
) (
(
1
)
)
, gives the relative change in the odds of the
response. Testing non-inferiority uses the formulation
:
versus
:
>
or equivalently
:
versus
: >
.
For non-inferiority tests with higher proportions better,
< 1. For non-inferiority tests with higher
proportions worse,
> 1.
A Note on Setting the Significance Level, Alpha
Setting the significance level has always been somewhat arbitrary. For planning purposes, the standard has
become to set alpha to 0.05 for two-sided tests. Almost universally, when someone states that a result is
statistically significant, they mean statistically significant at the 0.05 level.
Although 0.05 may be the standard for two-sided tests, it is not always the standard for one-sided tests,
such as non-inferiority tests. Statisticians often recommend that the alpha level for one-sided tests be set at
0.025 since this is the amount put in each tail of a two-sided test.
Power Calculation
The power for a test statistic that is based on the normal approximation can be computed exactly using two
binomial distributions. The following steps are taken to compute the power of these tests.
1. Find the critical value using the standard normal distribution. The critical value,
, is that value
of z that leaves exactly the target value of alpha in the appropriate tail of the normal distribution.
2. Compute the value of the test statistic,
, for every combination of
and
. Note that
ranges
from 0 to
, and
ranges from 0 to
. A small value (around 0.0001) can be added to the zero-cell
counts to avoid numerical problems that occur when the cell value is zero.
3. If
>
, the combination is in the rejection region. Call all combinations of
and
that
lead to a rejection the set A.
4. Compute the power for given values of
.
and
as
1 =
.
.
.
5. Compute the actual value of alpha achieved by the design by substituting
.
for
.
to obtain
=
.
.
.
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Asymptotic Approximations
When the values of
and
are large (say over 200), these formulas often take a long time to evaluate. In
this case, a large sample approximation can be used. The large sample approximation is made by replacing
the values of
and
in the z statistic with the corresponding values of
.
and
, and then computing
the results based on the normal distribution.
Test Statistics
Two test statistics have been proposed for testing whether the odds ratio is different from a specified value.
The main difference between the test statistics is in the formula used to compute the standard error used in
the denominator. These tests are both likelihood score tests.
In power calculations, the values of
and
are not known. The corresponding values of
.
and
may
be reasonable substitutes.
Following is a list of the test statistics available in PASS. The availability of several test statistics begs the
question of which test statistic one should use. The answer is simple: one should use the test statistic that
will be used to analyze the data. You may choose a method because it is a standard in your industry,
because it seems to have better statistical properties, or because your statistical package calculates it.
Whatever your reasons for selecting a certain test statistic, you should use the same test statistic when
doing the analysis after the data have been collected.
Miettinen and Nurminen’s Likelihood Score Test
Miettinen and Nurminen (1985) proposed a test statistic for testing whether the odds ratio is equal to a
specified value,
. Because the approach they used with the difference and ratio does not easily extend to
the odds ratio, they used a score statistic approach for the odds ratio. The regular MLE’s are
and
. The
constrained MLE’s are
and
. These estimates are constrained so that
=
. A correction factor of
N/(N-1) is applied to make the variance estimate less biased. The significance level of the test statistic is
based on the asymptotic normality of the score statistic.
The formula for computing the test statistic is
=
(
)
(
)
1
+
1
1
where
=
1 +
(
1
)
=
+
4
2
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=
(
1
)
,
=
+
(
1
)
,
=
Farrington and Manning’s Likelihood Score Test
Farrington and Manning (1990) indicate that the Miettinen and Nurminen statistic may be modified by
removing the factor N/(N-1).
The formula for computing this test statistic is
=
(
)
(
)
1
+
1
where the estimates,
and
, are computed as in the corresponding test of Miettinen and Nurminen
(1985) given above.
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Example 1 – Finding Power
A study is being designed to establish the non-inferiority of a new treatment compared to the current
treatment. Historically, the current treatment has enjoyed a 62.5% cure rate. The new treatment reduces
the seriousness of certain side effects that occur with the current treatment. Thus, the new treatment will be
adopted even if it is slightly less effective than the current treatment. The researchers will recommend
adoption of the odds ratio of the new treatment to the old treatment is at least 0.80.
The researchers plan to use the Farrington and Manning likelihood score test statistic to analyze the data
that will be (or has been) obtained. They want to study the power of the Farrington and Manning test at
group sample sizes ranging from 50 to 500 when the actual odds ratio is 1. The significance level will be
0.05.
Setup
If the procedure window is not already open, use the PASS Home window to open it. The parameters for this
example are listed below and are stored in the Example 1 settings file. To load these settings to the
procedure window, click Open Example Settings File in the Help Center or File menu.
Design Tab
_____________ _______________________________________
Solve For ....................................................... Power
Power Calculation Method ............................. Normal Approximation
Higher Proportions Are .................................. Better (H1: OR > OR0)
Test Type ....................................................... Likelihood Score (Farr. & Mann.)
Alpha.............................................................. 0.05
Group Allocation ............................................ Equal (N1 = N2)
Sample Size Per Group ................................. 50 to 500 by 50
OR0 (Non-Inferiority Odds Ratio) ................... 0.80
OR1 (Actual Odds Ratio) ............................... 1.0
P2 (Group 2 Proportion) ................................. 0.625
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Output
Click the Calculate button to perform the calculations and generate the following output.
Numeric Reports
Numeric Results
───────────────────────────────────────────────────────────────────────────────────────────────
Solve For: Power
Groups: 1 = Treatment, 2 = Reference
Test Statistic: Farrington & Manning Likelihood Score Test
Hypotheses: H0: OR ≤ OR0 vs. H1: OR > OR0
───────────────────────────────────────────────────────────────────────────────────────────────
Proportions Odds Ratio
Sample Size ─────────────────────────── ──────────────────
────────────── Non-Inferiority Actual Reference Non-Inferiority Actual
Power* N1 N2 N P1.0 P1.1 P2 OR0 OR1 Alpha
──────────────────────────────────────────────────────────────────────────────────────────────────────────────────
0.13427 50 50 100 0.5714 0.625 0.625 0.8 1 0.05
0.18885 100 100 200 0.5714 0.625 0.625 0.8 1 0.05
0.23884 150 150 300 0.5714 0.625 0.625 0.8 1 0.05
0.28606 200 200 400 0.5714 0.625 0.625 0.8 1 0.05
0.33101 250 250 500 0.5714 0.625 0.625 0.8 1 0.05
0.37390 300 300 600 0.5714 0.625 0.625 0.8 1 0.05
0.41477 350 350 700 0.5714 0.625 0.625 0.8 1 0.05
0.45368 400 400 800 0.5714 0.625 0.625 0.8 1 0.05
0.49064 450 450 900 0.5714 0.625 0.625 0.8 1 0.05
0.52568 500 500 1000 0.5714 0.625 0.625 0.8 1 0.05
───────────────────────────────────────────────────────────────────────────────────────────────
* Power was computed using the normal approximation method.
Power The probability of rejecting a false null hypothesis when the alternative hypothesis is true.
N1 and N2 The number of items sampled from each population.
N The total sample size. N = N1 + N2.
P1 The proportion for group 1, which is the treatment or experimental group.
P1.0 The smallest group 1 proportion that still yields a non-inferiority conclusion. P1.0 = P1|H0.
P1.1 The proportion for group 1 under the alternative hypothesis at which power and sample size calculations are
made. P1.1 = P1|H1.
P2 The proportion for group 2, which is the standard, reference, or control group.
OR0 The non-inferiority odds ratio, [P1/(1-P1)] / [P2/(1-P2)], assuming H0.
OR1 The non-inferiority odds ratio, [P1/(1-P1)] / [P2/(1-P2)], assuming H1.
Alpha The probability of rejecting a true null hypothesis.
Summary Statements
─────────────────────────────────────────────────────────────────────────
A parallel, two-group design will be used to test whether the Group 1 (treatment) proportion (P1) is non-inferior to
the Group 2 (reference) proportion (P2), with a non-inferiority odds ratio of 0.8 (H0: OR ≤ 0.8 versus H1: OR > 0.8).
The comparison will be made using a one-sided, two-sample Score test (Farrington & Manning) with a Type I error
rate (α) of 0.05. The reference group proportion is assumed to be 0.625. To detect an odds ratio (O1 / O2) of 1 (or
P1 of 0.625) with sample sizes of 50 for Group 1 (treatment) and 50 for Group 2 (reference), the power is 0.13427.
─────────────────────────────────────────────────────────────────────────
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Dropout-Inflated Sample Size
─────────────────────────────────────────────────────────────────────────
Dropout-Inflated Expected
Enrollment Number of
Sample Size Sample Size Dropouts
────────────── ────────────── ──────────────
Dropout Rate N1 N2 N N1' N2' N' D1 D2 D
────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
20% 50 50 100 63 63 126 13 13 26
20% 100 100 200 125 125 250 25 25 50
20% 150 150 300 188 188 376 38 38 76
20% 200 200 400 250 250 500 50 50 100
20% 250 250 500 313 313 626 63 63 126
20% 300 300 600 375 375 750 75 75 150
20% 350 350 700 438 438 876 88 88 176
20% 400 400 800 500 500 1000 100 100 200
20% 450 450 900 563 563 1126 113 113 226
20% 500 500 1000 625 625 1250 125 125 250
─────────────────────────────────────────────────────────────────────────
Dropout Rate The percentage of subjects (or items) that are expected to be lost at random during the course of the study
and for whom no response data will be collected (i.e., will be treated as "missing"). Abbreviated as DR.
N1, N2, and N The evaluable sample sizes at which power is computed (as entered by the user). If N1 and N2 subjects
are evaluated out of the N1' and N2' subjects that are enrolled in the study, the design will achieve the
stated power.
N1', N2', and N' The number of subjects that should be enrolled in the study in order to obtain N1, N2, and N evaluable
subjects, based on the assumed dropout rate. N1' and N2' are calculated by inflating N1 and N2 using the
formulas N1' = N1 / (1 - DR) and N2' = N2 / (1 - DR), with N1' and N2' always rounded up. (See Julious,
S.A. (2010) pages 52-53, or Chow, S.C., Shao, J., Wang, H., and Lokhnygina, Y. (2018) pages 32-33.)
D1, D2, and D The expected number of dropouts. D1 = N1' - N1, D2 = N2' - N2, and D = D1 + D2.
Dropout Summary Statements
─────────────────────────────────────────────────────────────────────────
Anticipating a 20% dropout rate, 63 subjects should be enrolled in Group 1, and 63 in Group 2, to obtain final group
sample sizes of 50 and 50, respectively.
─────────────────────────────────────────────────────────────────────────
References
─────────────────────────────────────────────────────────────────────────
Chow, S.C., Shao, J., and Wang, H. 2008. Sample Size Calculations in Clinical Research, Second Edition.
Chapman & Hall/CRC. Boca Raton, Florida.
Farrington, C. P. and Manning, G. 1990. 'Test Statistics and Sample Size Formulae for Comparative Binomial
Trials with Null Hypothesis of Non-Zero Risk Difference or Non-Unity Relative Risk.' Statistics in Medicine, Vol. 9,
pages 1447-1454.
Fleiss, J. L., Levin, B., Paik, M.C. 2003. Statistical Methods for Rates and Proportions. Third Edition. John Wiley &
Sons. New York.
Gart, John J. and Nam, Jun-mo. 1988. 'Approximate Interval Estimation of the Ratio in Binomial Parameters: A
Review and Corrections for Skewness.' Biometrics, Volume 44, Issue 2, 323-338.
Gart, John J. and Nam, Jun-mo. 1990. 'Approximate Interval Estimation of the Difference in Binomial Parameters:
Correction for Skewness and Extension to Multiple Tables.' Biometrics, Volume 46, Issue 3, 637-643.
Julious, S. A. and Campbell, M. J. 2012. 'Tutorial in biostatistics: sample sizes for parallel group clinical trials with
binary data.' Statistics in Medicine, 31:2904-2936.
Lachin, John M. 2000. Biostatistical Methods. John Wiley & Sons. New York.
Machin, D., Campbell, M., Fayers, P., and Pinol, A. 1997. Sample Size Tables for Clinical Studies, 2nd Edition.
Blackwell Science. Malden, Mass.
Miettinen, O.S. and Nurminen, M. 1985. 'Comparative analysis of two rates.' Statistics in Medicine 4: 213-226.
─────────────────────────────────────────────────────────────────────────
This report shows the values of each of the parameters, one scenario per row.
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Plots Section
Plots
─────────────────────────────────────────────────────────────────────────
The values from the table are displayed in the above chart. These charts give us a quick look at the sample
size that will be required for various sample sizes.
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Example 2 – Finding the Sample Size
Continuing with the scenario given in Example 1, the researchers want to determine the sample size
necessary to achieve a power of 0.80.
Setup
If the procedure window is not already open, use the PASS Home window to open it. The parameters for this
example are listed below and are stored in the Example 2 settings file. To load these settings to the
procedure window, click Open Example Settings File in the Help Center or File menu.
Design Tab
_____________ _______________________________________
Solve For ....................................................... Sample Size
Power Calculation Method ............................. Normal Approximation
Higher Proportions Are .................................. Better (H1: OR > OR0)
Test Type ....................................................... Likelihood Score (Farr. & Mann.)
Power............................................................. 0.8
Alpha.............................................................. 0.05
Group Allocation ............................................ Equal (N1 = N2)
OR0 (Non-Inferiority Odds Ratio) ................... 0.80
OR1 (Actual Odds Ratio) ............................... 1.0
P2 (Group 2 Proportion) ................................. 0.625
Output
Click the Calculate button to perform the calculations and generate the following output.
Numeric Results
───────────────────────────────────────────────────────────────────────────────────────────────
Solve For: Sample Size
Groups: 1 = Treatment, 2 = Reference
Test Statistic: Farrington & Manning Likelihood Score Test
Hypotheses: H0: OR ≤ OR0 vs. H1: OR > OR0
───────────────────────────────────────────────────────────────────────────────────────────────
Proportions Odds Ratio
Power Sample Size ─────────────────────────── ──────────────────
───────────── ──────────────── Non-Inferiority Actual Reference Non-Inferiority Actual
Target Actual* N1 N2 N P1.0 P1.1 P2 OR0 OR1 Alpha
──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────
0.8 0.80003 1057 1057 2114 0.5714 0.625 0.625 0.8 1 0.05
───────────────────────────────────────────────────────────────────────────────────────────────
* Power was computed using the normal approximation method.
The required sample size is 1057 per group.
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Example 3 – Comparing the Power of the Two Test
Statistics
Continuing with Example 2, the researchers want to determine which of the two possible test statistics to
adopt by using the comparative reports and charts that PASS produces. They decide to compare the powers
from binomial enumeration and actual alphas for various sample sizes between 1000 and 1200.
Setup
If the procedure window is not already open, use the PASS Home window to open it. The parameters for this
example are listed below and are stored in the Example 3 settings file. To load these settings to the
procedure window, click Open Example Settings File in the Help Center or File menu.
Design Tab
_____________ _______________________________________
Solve For ................................................................... Power
Power Calculation Method ......................................... Binomial Enumeration
Maximum N1 or N2 for Binomial Enumeration ........... 5000
Zero Count Adjustment Method ................................. Add to zero cells only
Zero Count Adjustment Value .................................... 0.0001
Higher Proportions Are .............................................. Better (H1: OR > OR0)
Test Type ................................................................... Likelihood Score (Farr. & Mann.)
Alpha.......................................................................... 0.05
Group Allocation ........................................................ Equal (N1 = N2)
Sample Size Per Group ............................................. 1000 1100 1200
OR0 (Non-Inferiority Odds Ratio) ............................... 0.80
OR1 (Actual Odds Ratio) ........................................... 1.0
P2 (Group 2 Proportion) ............................................. 0.625
Reports Tab
_____________ _______________________________________
Show Comparative Reports ....................................... Checked
Comparative Plots Tab
_____________ _______________________________________
Show Comparative Plots ............................................ Checked
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Output
Click the Calculate button to perform the calculations and generate the following output.
Power Comparison of Two Different Tests
───────────────────────────────────────────────────────────────────────────────────────────────
Hypotheses: H0: OR ≤ OR0 vs. H1: OR > OR0
───────────────────────────────────────────────────────────────────────────────────────────────
Power
Sample Size ────────────
──────────────── Target F.M. M.N.
N1 N2 N P2 OR0 OR1 Alpha Score Score
──────────────────────────────────────────────────────────────────────────────────
1000 1000 2000 0.625 0.8 1 0.05 0.7790 0.7790
1100 1100 2200 0.625 0.8 1 0.05 0.8129 0.8125
1200 1200 2400 0.625 0.8 1 0.05 0.8414 0.8411
───────────────────────────────────────────────────────────────────────────────────────────────
Note: Power was computed using binomial enumeration of all possible outcomes.
Actual Alpha Comparison of Two Different Tests
───────────────────────────────────────────────────────────────────────────────────────────────
Hypotheses: H0: OR ≤ OR0 vs. H1: OR > OR0
───────────────────────────────────────────────────────────────────────────────────────────────
Alpha
Sample Size ────────────────────
──────────────── F.M. M.N.
N1 N2 N P2 OR0 OR1 Target Score Score
──────────────────────────────────────────────────────────────────────────────────
1000 1000 2000 0.625 0.8 1 0.05 0.0499 0.0498
1100 1100 2200 0.625 0.8 1 0.05 0.0502 0.0501
1200 1200 2400 0.625 0.8 1 0.05 0.0500 0.0498
───────────────────────────────────────────────────────────────────────────────────────────────
Note: Actual alpha was computed using binomial enumeration of all possible outcomes.
Plots
─────────────────────────────────────────────────────────────────────────
The power is almost exactly the same for both tests, which is not surprising given the large sample size.
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Example 4 – Comparing Power Calculation Methods
Continuing with Example 3, let’s see how the results compare if we were to use approximate power
calculations instead of power calculations based on binomial enumeration.
Setup
If the procedure window is not already open, use the PASS Home window to open it. The parameters for this
example are listed below and are stored in the Example 4 settings file. To load these settings to the
procedure window, click Open Example Settings File in the Help Center or File menu.
Design Tab
_____________ _______________________________________
Solve For ....................................................... Power
Power Calculation Method ............................. Normal Approximation
Higher Proportions Are .................................. Better (H1: OR > OR0)
Test Type ....................................................... Likelihood Score (Farr. & Mann.)
Alpha.............................................................. 0.05
Group Allocation ............................................ Equal (N1 = N2)
Sample Size Per Group ................................. 1000 1100 1200
OR0 (Non-Inferiority Odds Ratio) ................... 0.80
OR1 (Actual Odds Ratio) ............................... 1.0
P2 (Group 2 Proportion) ................................. 0.625
Reports Tab
_____________ _______________________________________
Show Power Detail Report ............................. Checked
Click the Calculate button to perform the calculations and generate the following output.
Output
Power Detail Report
───────────────────────────────────────────────────────────────────────────────────────────────
Test Statistic: Farrington & Manning Likelihood Score Test
Hypotheses: H0: OR ≤ OR0 vs. H1: OR > OR0
───────────────────────────────────────────────────────────────────────────────────────────────
Normal Binomial
Sample Size Approximation Enumeration
──────────────── ──────────── ─────────────
N1 N2 N P2 OR0 OR1 Power Alpha Power Alpha
─────────────────────────────────────────────────────────────────────────────────────────────────────
1000 1000 2000 0.625 0.8 1 0.78044 0.05 0.77899 0.0499
1100 1100 2200 0.625 0.8 1 0.81377 0.05 0.81289 0.0502
1200 1200 2400 0.625 0.8 1 0.84250 0.05 0.84139 0.0500
───────────────────────────────────────────────────────────────────────────────────────────────
Notice that the approximate power values are close to the binomial enumeration values for all sample sizes.
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Example 5 – Validation
We could not find a validation example for a non-inferiority test for the odds ratio of two proportions. The
calculations are basically the same as those for a non-inferiority test of the ratio of two proportions, which
has been validated using Blackwelder (1993). We refer you to Example 5 of Chapter 211, “Non-Inferiority
Tests for the Ratio of Two Proportions,” for a validation example.
