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The Combination Test Principle · Tutorial 188 of 799

Worked Example: Combining Two Stage P-Values

Learn to compute and interpret inverse normal and Fisher combination tests for two independent stages, and see why their calibrated decisions can differ.

Advanced 11 min read

What You'll Learn

  • Identify the assumptions needed for exact two-stage inverse normal and Fisher combination tests.
  • Convert one-sided stagewise p-values to normal scores and calculate an inverse normal statistic.
  • Calculate Fisher’s statistic and its chi-square reference p-value.
  • Compare decisions from the two tests at a prespecified significance level.
  • Recognize dependence, invalid stagewise p-values, and post hoc method selection as threats to valid inference.
  • Interpret different combination-test results without treating either as a pooled-data analysis.

Why Combine Stagewise P-Values?

A two-stage trial produces evidence at two points in time. A combination test uses each stage’s p-value to form a prespecified final test, rather than treating the two p-values as if they could simply be averaged or compared separately with the significance level. The choice of combination function matters: inverse normal and Fisher tests can give different final p-values from the same stagewise evidence, while each can control the type I error rate under its own stated conditions.

This worked tutorial uses two stages, independent stagewise p-values, and combination rules specified before the results are observed. The examples use one-sided p-values in a common direction, so smaller p-values consistently represent evidence against the null in that direction. This is an important condition for interpreting the inverse normal result as a directional test.

Definition: Let \(p_1\) and \(p_2\) be the stagewise p-values from stages 1 and 2. A combination test applies a prespecified function to these p-values and compares the resulting statistic with its null reference distribution, or equivalently reports a calibrated overall p-value. A rejection rule at significance level \(\alpha\) rejects \(H_0\) when the overall p-value is at most \(\alpha\).

Rules and Conditions for the Two Procedures

For the equal-weight inverse normal test, transform each one-sided stagewise p-value into a normal score and average the scores with weights \(1/\sqrt{2}\). Here \(\Phi\) is the standard normal cumulative distribution function and \(\Phi^{-1}\) is its quantile function.

$$ Z_i=\Phi^{-1}(1-p_i), \qquad Z_{\mathrm{IN}}=\frac{Z_1+Z_2}{\sqrt{2}}, \qquad p_{\mathrm{IN}}=1-\Phi(Z_{\mathrm{IN}}). $$

For Fisher’s two-stage test, multiply the stagewise p-values inside a logarithmic statistic. With independent, continuous uniform p-values under the null, the reference distribution has four degrees of freedom.

$$ X_F=-2\ln(p_1p_2), \qquad X_F\sim\chi^2_4 \text{ under } H_0, \qquad p_F=P(\chi^2_4\ge X_F). $$

The assumptions matter. For these reference distributions to be exact, each stagewise p-value must be valid under the null, and the two p-values must be independent under the null. In the examples, they are treated as continuous and uniform under \(H_0\). Stage independence is a design and analysis property to establish, not something that can be verified merely by looking at the two observed p-values. The equal inverse normal weights and Fisher rule must also be chosen in advance, not selected after comparing their results.

The reference distributions have short justifications. If \(p_i\) is uniform on \((0,1)\) under \(H_0\), then \(1-p_i\) is also uniform, and applying \(\Phi^{-1}\) gives a standard normal score. Independence of the two scores then makes their sum divided by \(\sqrt{2}\) standard normal. For Fisher’s rule, \(-2\ln(p_i)\) has a chi-square distribution with 2 degrees of freedom under a uniform null p-value; independence makes the sum of the two such terms chi-square with 4 degrees of freedom. These facts explain why each procedure has a calibrated null reference distribution under the stated assumptions.

Worked Example: Both Procedures Reject

Suppose two independent stagewise one-sided p-values are \(p_1=0.040\) and \(p_2=0.010\). Use the equal-weight inverse normal test and Fisher’s test, each at \(\alpha=0.025\). Assume the p-values are valid and independent under \(H_0\), and that both tests and the one-sided direction were prespecified.

Inverse normal calculation. The stage scores are \(Z_1=\Phi^{-1}(0.960)\approx1.7507\) and \(Z_2=\Phi^{-1}(0.990)\approx2.3263\). Therefore,

$$ Z_{\mathrm{IN}} =\frac{1.7507+2.3263}{\sqrt{2}} \approx 2.8829, \qquad p_{\mathrm{IN}}=1-\Phi(2.8829)\approx0.00197. $$

Since \(0.00197<0.025\), the inverse normal test rejects \(H_0\). Under this test’s specified one-sided combination rule, the combined evidence is statistically significant at the 2.5% level.

Fisher calculation. The product is \(p_1p_2=(0.040)(0.010)=0.000400\). Thus,

$$ X_F=-2\ln(0.000400)\approx15.6481. $$

For 4 degrees of freedom, the upper-tail probability is \(e^{-X_F/2}(1+X_F/2)\). Here \(e^{-X_F/2}=0.000400\), so

$$ p_F=0.000400(1+7.82405)\approx0.003530. $$

Since \(0.003530<0.025\), Fisher’s test also rejects \(H_0\). The agreement in this example does not mean the procedures are interchangeable; each has a different combination function.

Different Combination Rules Can Give Different Decisions

Both tests are calibrated to their own null distributions. They do not, however, give the same numerical summary of stagewise evidence. The inverse normal test combines transformed p-values, while Fisher’s test is driven by the logarithm of their product. It is therefore possible for them to fall on opposite sides of the same significance threshold.

Worked Example: Opposite Decisions at a 5% Level

Suppose \(p_1=p_2=0.100\), with independent, valid one-sided p-values and prespecified equal inverse normal weights. Compare the tests at \(\alpha=0.05\).

For inverse normal, \(\Phi^{-1}(0.900)\approx1.2816\) at each stage. Substitution gives

$$ Z_{\mathrm{IN}} =\frac{1.2816+1.2816}{\sqrt{2}} \approx1.8124, \qquad p_{\mathrm{IN}}=1-\Phi(1.8124)\approx0.0350. $$

Because \(0.0350<0.05\), the inverse normal test rejects \(H_0\) at the 5% level.

For Fisher’s test, \(p_1p_2=(0.100)(0.100)=0.0100\), so

$$ X_F=-2\ln(0.0100)\approx9.2103, \qquad p_F=e^{-9.2103/2}(1+9.2103/2)\approx0.05605. $$

Since \(0.05605>0.05\), Fisher’s test does not reject \(H_0\). Neither result is an arithmetic error: the tests apply different prespecified functions, and their calibrated overall p-values can differ. The method should not be chosen after seeing which one rejects.

Worked Example: Decisions Near a Stricter Threshold

Now let \(p_1=0.150\) and \(p_2=0.0300\), again assuming independent, valid stagewise p-values. Compare the same two procedures at \(\alpha=0.025\).

The inverse normal scores are \(\Phi^{-1}(0.850)\approx1.0364\) and \(\Phi^{-1}(0.970)\approx1.8808\). Hence,

$$ Z_{\mathrm{IN}} =\frac{1.0364+1.8808}{\sqrt{2}} \approx2.0628, \qquad p_{\mathrm{IN}}=1-\Phi(2.0628)\approx0.0196. $$

Because \(0.0196<0.025\), the inverse normal test rejects \(H_0\) at the 2.5% level.

For Fisher’s test, \(p_1p_2=(0.150)(0.0300)=0.00450\). Therefore,

$$ X_F=-2\ln(0.00450)\approx10.8074\approx10.81, \qquad p_F=e^{-10.8074/2}(1+10.8074/2)\approx0.02882. $$

Fisher’s p-value is greater than \(0.025\), so Fisher’s test does not reject \(H_0\). The exact same stagewise p-values have led to different decisions because the tests combine evidence differently. This comparison is meaningful only if the tests were specified and calibrated under the conditions described above.

How to Check a Calculation and Interpret Its Result

A careful calculation proceeds from the design assumptions to the final decision, rather than from a desired conclusion back to a chosen test. Keep enough precision in the stagewise p-values and intermediate calculations; round the reported results only after computing the final statistic and p-value. Small rounding differences can matter when a p-value is near \(\alpha\).

1
Confirm the inputs.
Identify the two stagewise p-values, the one-sided direction, the prespecified combination function, and the decision threshold \(\alpha\). Check that the p-values are valid under the null and that stage independence is justified.
2
Calculate each procedure separately.
For inverse normal, transform each p-value with \(\Phi^{-1}(1-p_i)\), combine using the stated weights, and find the standard normal upper-tail probability. For Fisher, calculate \(-2\ln(p_1p_2)\) and its chi-square upper-tail probability with 4 degrees of freedom.
3
Compare with the prespecified level.
For each procedure, reject \(H_0\) if its calibrated overall p-value is at most \(\alpha\); otherwise, do not reject. State the rule and the conclusion in the trial context.

These procedures answer a testing question: whether the stagewise evidence, combined by a specified rule, is sufficiently inconsistent with \(H_0\). An overall p-value is not the probability that \(H_0\) is true, and a non-rejection does not establish that there is no treatment effect. Neither combination statistic is, by itself, a pooled estimate of treatment effect or a confidence interval.

Diagnostics, Alternatives, and Limitations

The central diagnostic is whether the assumptions and the prespecified analysis match the trial actually conducted. Independence can be compromised if stages reuse participants or observations, share data in a way that induces dependence, or use stagewise tests whose inputs overlap. A p-value that is not valid under the null also invalidates the stated reference calculation. In either case, simply applying the standard normal or chi-square reference distribution does not repair the problem.

The inverse normal formula here requires a consistent one-sided direction: a small stagewise p-value must correspond to evidence in the same favorable direction at both stages. If stagewise p-values are two-sided, or if effects point in different directions, this calculation cannot be interpreted as a simple directional combination without an explicitly specified rule for handling direction. Fisher’s statistic is based on p-value magnitude and does not encode the sign of an effect; it can combine small p-values even when effects point in opposite directions.

The main alternative in this tutorial is not “combine” versus “do not combine,” but one prespecified combination function versus another. Inverse normal expresses evidence on a normal-score scale and, in this example, uses equal fixed weights. Fisher’s rule aggregates through the product of p-values. Other valid combination rules may be appropriate for other design goals, but they require their own prespecified calibration. A pooled-data test is a different analysis and should not be assumed equivalent to either combination test.

A common mistake is to report whichever method gives the smaller p-value and describe it as the trial’s final test. That procedure makes the test depend on the observed data and can inflate the type I error rate unless the selection is itself accounted for in the design and calibration. The correction is to define the primary combination rule and its rejection threshold in advance. If alternative combination results are also reported, distinguish them clearly from the prespecified primary decision.

Key takeaway: For two independent, valid stagewise p-values, calculate the prespecified inverse normal or Fisher statistic, use its matching null reference distribution, and compare the resulting overall p-value with the prespecified \(\alpha\). Different calibrated combination tests can yield different results; their assumptions and rules must not be chosen after examining the outcomes.

Check Your Understanding

Use the stated assumptions in each question. The solutions follow the questions so the calculations and decisions can be checked directly.

  1. For independent, valid stagewise p-values \(p_1=0.250\) and \(p_2=0.040\), calculate the equal-weight inverse normal statistic and overall p-value. What is the decision at \(\alpha=0.05\)?
  2. For those same p-values, calculate Fisher’s statistic and overall p-value, then give the decision at \(\alpha=0.05\).
  3. Why is the standard Fisher chi-square reference distribution not justified merely because two observed stagewise p-values are available?
  4. Suppose both combination tests were prespecified but give different decisions. Is it valid to select the rejecting test after seeing the results as the primary analysis? Explain.

Solutions. For question 1, \(\Phi^{-1}(0.750)\approx0.6745\) and \(\Phi^{-1}(0.960)\approx1.7507\). Thus \(Z_{\mathrm{IN}}=(0.6745+1.7507)/\sqrt{2}\approx1.7159\), giving \(p_{\mathrm{IN}}\approx0.0431\). Since \(0.0431<0.05\), reject \(H_0\) under this test. For question 2, the product is \(0.250(0.040)=0.0100\), so \(X_F=-2\ln(0.0100)\approx9.2103\), and \(p_F=e^{-9.2103/2}(1+9.2103/2)\approx0.05605\). Since \(0.05605>0.05\), do not reject \(H_0\) under Fisher’s test. For question 3, the exact reference calculation requires valid null p-values and independence under the null; observing two p-values alone establishes neither condition. For question 4, no: choosing the primary test based on which one rejects makes the analysis data-dependent and can change its error rate unless that selection rule was prespecified and calibrated.

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