Normal Approximation Versus Exact Sample Size

Why Compare Two Sample-Size Methods?

Sample-size planning for average bioequivalence (ABE) asks how many evaluable subjects are needed for a prespecified probability of concluding that the test and reference products are equivalent. A normal approximation is quick and often useful for an initial estimate. An exact calculation accounts for estimating the variability from a finite sample and uses the relevant \(t\) distribution. These methods can give different answers, particularly when the candidate sample is small or the expected treatment difference is close to an equivalence limit.

The comparison is meaningful only when both calculations use the same equivalence limits, expected log treatment difference, variability, significance level and target power. This tutorial uses a balanced two-period crossover and the usual log-scale ABE limits of 0.80 and 1.25. The sample size \(n\) denotes evaluable subjects, not the number to recruit before allowing for withdrawals.

Definition: Let \(D_i\) be subject \(i\)'s test-minus-reference contrast on the log scale, with \(D_i\sim N(\theta,\sigma_D^2)\), independently across subjects. Here \(\theta\) is the population mean log treatment difference and \(\sigma_D\) is the standard deviation of the within-subject contrast. ABE is assessed by two one-sided tests (TOST): \(H_{0L}:\theta\le L\) versus \(H_{aL}:\theta>L\), and \(H_{0U}:\theta\ge U\) versus \(H_{aU}:\theta<U\), where \(L=\ln(0.80)\) and \(U=\ln(1.25)\). Equivalence is concluded only when both null hypotheses are rejected.

For the exact calculation below, assume the standard balanced two-period crossover analysis: the contrast estimate is \(\bar D\), its standard error is \(S_D/\sqrt n\), the residual degrees of freedom are \(df=n-2\), and the variance estimate is independent of \(\bar D\). Under the normal model, \(\bar D\sim N(\theta,\sigma_D^2/n)\) and \(dfS_D^2/\sigma_D^2\sim\chi^2_{df}\). The calculations also assume that the planning values for \(\theta\) and \(\sigma_D\) are appropriate. These are planning assumptions, not facts guaranteed by the future study.

At one-sided significance level \(\alpha\), both tests use the critical value \(t_{1-\alpha,df}\). The lower-bound test rejects when \((\bar D-L)/(S_D/\sqrt n)>t_{1-\alpha,df}\); the upper-bound test rejects when \((\bar D-U)/(S_D/\sqrt n)<-t_{1-\alpha,df}\). Equivalently, the confidence interval for \(\theta\) formed using the corresponding two-sided \(1-2\alpha\) level must lie wholly within \((L,U)\).

The Normal Approximation

The normal approximation treats \(\sigma_D\) as known and replaces the random estimated standard error by \(\sigma_D/\sqrt n\). It also uses the standard normal critical value \(z_{1-\alpha}\) rather than a \(t\) critical value. Under these approximations, the event of passing both tests is

L+z_{1-\alpha}\frac{\sigma_D}{\sqrt n} < \bar D < U-z_{1-\alpha}\frac{\sigma_D}{\sqrt n}.

Since \(\bar D\) is normal with mean \(\theta\) and standard deviation \(\sigma_D/\sqrt n\), the approximate power is

\pi_N(n;\theta) = \Phi\!\left(\frac{\sqrt n(U-\theta)}{\sigma_D}-z_{1-\alpha}\right) - \Phi\!\left(\frac{\sqrt n(L-\theta)}{\sigma_D}+z_{1-\alpha}\right).

Here \(\Phi\) is the standard normal cumulative distribution function. This expression gives the probability of passing both tests under the planning value \(\theta\); it is not simply the power of one of the two tests. When \(\theta\) is the midpoint of symmetric log limits, \(L=-U\), the two rejection boundaries contribute equally. If the expected value is not at that midpoint, the nearer boundary generally constrains power more strongly.

For a centered expectation, set \(d=U-\theta=\theta-L\). The formula reduces to \(2\Phi(\sqrt n d/\sigma_D-z_{1-\alpha})-1\). Solving this expression for a target power \(1-\beta\) gives the familiar continuous approximation:

n_N = \left[ \frac{(z_{1-\alpha}+z_{1-\beta/2})\sigma_D}{d} \right]^2.

This is a starting value, not the final design by itself. Round upward to an allowed integer and evaluate the power at that integer using the full two-boundary expression. For a balanced two-sequence crossover, the candidate may also need to be rounded to an even number.

Exact Power and Its Justification

In the exact calculation, the standard error is random and the critical value depends on \(df\). Let \(V=dfS_D^2/\sigma_D^2\), so \(V\sim\chi^2_{df}\). Conditional on \(V=v\), the standardized sample mean \(Z=\sqrt n(\bar D-\theta)/\sigma_D\) is standard normal. Both TOST conditions hold precisely when

\frac{\sqrt n(L-\theta)}{\sigma_D} +t_{1-\alpha,df}\sqrt{\frac{v}{df}} < Z < \frac{\sqrt n(U-\theta)}{\sigma_D} -t_{1-\alpha,df}\sqrt{\frac{v}{df}}.

The conditional probability is zero if the lower endpoint is not below the upper endpoint. Otherwise it is the difference between the normal CDF at the upper endpoint and the normal CDF at the lower endpoint. Averaging this conditional probability over the chi-square distribution of \(V\) yields exact power under the stated model:

\pi_E(n;\theta) = \int_0^\infty \left[ \Phi\!\left( \frac{\sqrt n(U-\theta)}{\sigma_D} -t_{1-\alpha,df}\sqrt{\frac{v}{df}} \right) - \Phi\!\left( \frac{\sqrt n(L-\theta)}{\sigma_D} +t_{1-\alpha,df}\sqrt{\frac{v}{df}} \right) \right]_+ f_{\chi^2_{df}}(v)\,dv,

where \([x]_+=\max(x,0)\) and \(f_{\chi^2_{df}}\) is the chi-square density. This is the joint probability of rejecting both null hypotheses; it preserves the dependence between the two tests because they share the same \(\bar D\) and \(S_D\). The integral is normally evaluated numerically. In a noncentral-\(t\) implementation, the two one-sided statistics have noncentralities \(\sqrt n(\theta-L)/\sigma_D\) and \(\sqrt n(\theta-U)/\sigma_D\), and their joint pass probability must still be calculated rather than treating the two tests as independent.

The justification is direct: conditional on the estimated variance, the only remaining random quantity in the pass event is the normal sample mean. Integrating its conditional probability over the distribution of the estimated variance includes the uncertainty omitted by the normal approximation. As \(df\) grows, the estimated variance is less variable and the \(t\) critical value approaches the normal critical value; that explains why the methods tend to converge as sample size increases. It does not guarantee that they are interchangeable at a particular small \(n\).

Worked Example: A Normal-Approximation Candidate

Suppose a hypothetical balanced crossover is planned at \(\alpha=0.05\) and target power 0.80. Let \(\sigma_D=0.18\) log units and let the expected treatment ratio be 1.03, so \(\theta=\ln(1.03)=0.02956\). The log limits are \(L=-0.22314\) and \(U=0.22314\). The distance to the nearer limit is \(U-\theta=0.19358\), reported as 0.1936; the distance to the other limit is \(\theta-L=0.25270\), reported as 0.2527.

Using \(z_{0.95}=1.6449\) and \(z_{0.90}=1.2816\) would correspond to a different power convention: for a centered treatment expectation, the target-power quantile in the two-sided power equation is \(z_{(1+0.80)/2}=z_{0.90}=1.2816\). Here, because the expectation is off-center, use the full two-boundary normal power expression and solve it numerically. For a rough one-boundary starting estimate, allocate the target one-sided power to the nearer boundary, giving \(z_{0.80}=0.8416\); this does not ensure the target joint power, so the full two-boundary power must be checked. The resulting continuous value is

\sqrt{n_N} \approx \frac{(1.6449+0.8416)(0.18)}{0.1936} = 2.312, \qquad n_N\approx (2.312)^2=5.345.

Thus the upward-rounded initial candidate is 6 evaluable subjects. The quotient is \(\sqrt n\), not \(n\): retaining this distinction avoids reporting 2.312 as a subject count. Check the full normal power at \(n=6\), where \(\sqrt6=2.4495\):

\pi_N(6;\theta) = \Phi\!\left(\frac{\sqrt6(0.1936)}{0.18}-1.6449\right) - \Phi\!\left(-\frac{\sqrt6(0.2527)}{0.18}+1.6449\right) \approx \Phi(0.990)-\Phi(-1.794) \approx 0.802.

The arithmetic indicates approximate power just above 0.80 for six subjects under the known-variance normal calculation. The numerical integration for exact power is a separate check; the normal result alone does not establish that six subjects meet the target under the finite-sample \(t\) procedure.

Comparing Candidate Sample Sizes

For each integer candidate \(n\), exact sample-size planning evaluates \(\pi_E(n;\theta)\) and selects the smallest allowed \(n\) for which that power is at least the target. In a balanced design, evaluate only allowed allocations. A candidate that has approximate power above target can fail the exact criterion because it uses a more variable estimated standard error and a larger \(t\) critical value at small degrees of freedom.

Worked Example: Exact Power Near a Small Candidate

Continue the hypothetical inputs above, with \(n=6\), \(df=4\), \(t_{0.95,4}=2.132\), \(\theta=0.02956\), and \(\sigma_D=0.18\). The exact power integral uses the standardized distances \(\sqrt6(U-\theta)/\sigma_D\approx2.634\) and \(\sqrt6(L-\theta)/\sigma_D\approx-3.439\). Substitution gives

\pi_E(6;\theta) = \int_0^\infty \left[ \Phi\!\left(2.634-2.132\sqrt{\frac v4}\right) - \Phi\!\left(-3.439+2.132\sqrt{\frac v4}\right) \right]_+ f_{\chi^2_4}(v)\,dv.

Numerical quadrature of this expression gives power of approximately 0.60. The exact result is below the 0.80 target even though the normal approximation at the same \(n\) was approximately 0.802. This large difference is practically important: the approximation would leave the design substantially underpowered if used as the final decision. Increasing \(n\) changes both \(df\) and the variance of the estimated standard error, so the exact calculation must be repeated at each allowed candidate; the difference is not corrected by merely replacing one critical value at \(n=6\).

A reproducible calculation should report the input limits, expected \(\theta\), \(\sigma_D\), \(\alpha\), degrees of freedom, and the power method. For example, a numerical search evaluates the integral at successive allowed sample sizes and stops at the first whose exact power reaches the specified target. The result is conditional on the assumed \(\theta\) and variability; it is not a guarantee of the achieved power if those planning inputs differ from the study values.

Worked Example: Centered Expectation and Sequence Balance

Consider another hypothetical plan with \(\theta=0\), \(\sigma_D=0.18\), \(\alpha=0.05\), and target power 0.80. The limits are symmetric, so \(d=0.22314\). The centered normal formula gives

n_N \approx \left[ \frac{(1.6449+1.2816)(0.18)}{0.22314} \right]^2 \approx (2.3607)^2 \approx 5.573.

Rounding upward gives six subjects, which is also compatible with equal allocation to two sequences. At \(n=6\), however, the exact calculation uses \(df=4\) and \(t_{0.95,4}=2.132\), not \(z_{0.95}=1.6449\). Substituting the symmetric values into the exact integral gives

\pi_E(6;0) = \int_0^\infty \left[ \Phi\!\left(3.037-2.132\sqrt{\frac v4}\right) - \Phi\!\left(-3.037+2.132\sqrt{\frac v4}\right) \right]_+ f_{\chi^2_4}(v)\,dv.

This integral can be evaluated directly by quadrature or a validated noncentral-\(t\) power routine. If it falls below 0.80, six is not the exact sample-size answer even though it is the rounded normal estimate; continue to the next allowed balanced size and select the first one meeting the exact target. This example illustrates why integer rounding, sequence constraints and exact power evaluation are distinct steps.

When the Approximation Is Useful—and When It Fails

The normal approximation is useful for rapid planning, checking calculations, and understanding how sample size scales with the distance to an equivalence boundary and the assumed variability. It can also be a reasonable final approximation when the candidate sample is large enough that the \(t\) critical value and variance-estimation uncertainty have little practical effect. The exact calculation is preferable for the final choice when sample sizes are small, the target is stringent, or a one- or two-subject difference changes the study's feasibility.

A practical comparison uses the same inputs and reports both the continuous approximation and the exact integer search. The absolute difference in subjects matters, but so does its design consequence: the number may need to accommodate balanced sequences, and the recruited total may exceed the evaluable target if attrition is anticipated. Those adjustments should not be mistaken for a difference in statistical power methods.

Diagnostics include checking that the implementation uses the intended log limits and the correct \(\sigma_D\), not a coefficient of variation or a standard deviation on the original scale; verifying \(df\) for the specified design and analysis; and confirming that exact power is nondecreasing over candidate sample sizes for fixed planning inputs. Compare one candidate against an independent numerical implementation or a hand-checked integral. A surprising gap can signal a sign error in the upper-bound test, use of the wrong tail probability, an incorrect degrees-of-freedom convention, or accidental addition of the two marginal test powers instead of calculating their joint event.

The method has limitations. Exact power is exact only under its distributional and design assumptions, including the normal contrast model and the variance-estimator distribution. It does not account automatically for misspecified variability, protocol deviations, missing profiles, multiplicity across endpoints, or uncertainty in planning inputs. Further, a single expected \(\theta\) gives a power calculation conditional on that value; it does not describe power averaged over a distribution of possible treatment effects.

Common Mistakes

Calling \(\sqrt n\) the sample size. In the normal formula, the ratio on the right estimates \(\sqrt n\). Square it, then round the subject count upward.
Using the one-boundary formula as the final power. It can provide an initial estimate, but the full two-boundary probability is needed when the expected difference is off-center.
Assuming exact and approximate power are identical. The normal calculation fixes the standard error and uses a normal critical value. Exact power includes variability in the estimated standard error and the finite-\(df\) \(t\) critical value.
Adding marginal powers. The two TOST decisions share the same estimate and standard error. Equivalence requires both decisions to pass, so their joint probability is needed.
Interpreting the planned power as guaranteed. Power is calculated under specified assumptions. It is not the probability that the study will succeed after its data are observed.

Key takeaway: Use the normal approximation to obtain and understand a candidate, but use the exact noncentral-\(t\) or equivalent joint-power calculation to verify the smallest allowed evaluable sample size when finite-sample accuracy matters. Reconcile differences by checking inputs, degrees of freedom, and whether both equivalence tests are included.

Check Your Understanding

Answer each question, then compare with the brief solutions.

Why does the exact calculation have a random standard error while the normal approximation does not?
For \(\alpha=0.05\), why must both one-sided tests reject before equivalence is concluded?
If a normal calculation gives \(\sqrt n=2.312\), what is the continuous sample-size estimate, and what is the upward-rounded candidate?
Name two inputs that must be identical when comparing normal and exact sample-size methods.
Give one reason an exact-power calculation can still be wrong for a real study.

Solutions. (1) Exact power uses the sample standard deviation \(S_D\), which varies from sample to sample; the approximation substitutes the planning value \(\sigma_D\). (2) Equivalence requires evidence against both the lower-limit and upper-limit null hypotheses; passing only one leaves the other equivalence condition unestablished. (3) \(n=(2.312)^2=5.345\) approximately, so the upward-rounded candidate is 6 subjects, before any sequence-allocation constraint. (4) Examples include the limits, \(\theta\), \(\sigma_D\), \(\alpha\), and target power. (5) The assumed variance or contrast distribution may not describe the study population, so exact computation under a misspecified model does not ensure correct achieved power.