Why Sample Size Does Not Always Equal Information
An interim look is often described by the number of participants enrolled so far. That count is useful, but it is not automatically the fraction of the planned statistical information. The distinction matters because a spending function takes information fraction as its input. If the fraction is calculated from the wrong model, the spending target will be evaluated at the wrong point on the curve.
The simplest case is a fixed-variance model in which each participant contributes the same amount of information. Then the information fraction equals the accrued-sample fraction. Other designs—for example, a treatment comparison with changing allocation or unequal outcome variances—need a variance-based calculation. The design’s statistical model determines the conversion.
For a chosen cumulative spending function \(\alpha(t)\), the target cumulative spending by look \(k\) is \(\alpha(t_k)\). The target increment assigned between looks \(k-1\) and \(k\) is \(\alpha(t_k)-\alpha(t_{k-1})\), with \(\alpha(t_0)=0\). These are spending targets used when calibrating boundaries. The target at a look is not, by itself, the probability of crossing the boundary at that look; the sequential boundary calibration accounts for the chance of having crossed at earlier looks.
When Information Is Proportional to Sample Size
Consider independent observations \(X_1,\ldots,X_n\) from a normal population with mean \(\mu\) and known variance \(\sigma^2\). The sample mean has variance \(\sigma^2/n\). Its information for \(\mu\), the reciprocal of that variance, is \(I(n)=n/\sigma^2\). If the variance model is unchanged across looks, the common factor \(1/\sigma^2\) cancels in the ratio:
where \(n_k\) is the accrued sample size and \(N\) is the planned final sample size. This result requires that observations contribute the same information and that the analysis model and measurement precision remain as specified. Under those assumptions, sample size is a valid shortcut for information.
Worked Example: A One-Sample Mean With Fixed Variance
Suppose a hypothetical study plans \(N=300\) independent observations to estimate a mean under a normal, known-variance model. Each observation has the same variance, so information is proportional to sample size. The first two looks occur after 120 and 225 observations. Find the information fractions and evaluate a one-sided Hwang-Shih-DeCani spending target with total \(\alpha=0.025\) and shape parameter \(\gamma=-2\).
The model assumptions make the sample-size ratio appropriate. The fractions are $$ t_1=\frac{120}{300}=0.4000,\qquad t_2=\frac{225}{300}=0.7500,\qquad t_3=1. $$
For \(\gamma\ne0\), the Hwang-Shih-DeCani cumulative spending function is $$ \alpha(t)=\alpha\frac{1-\exp(-\gamma t)}{1-\exp(-\gamma)}. $$ With \(\gamma=-2\), this becomes \(\alpha(t)=0.025\{\exp(2t)-1\}/\{\exp(2)-1\}\). At the first look, $$ \alpha(0.4)=0.025\frac{\exp(0.8)-1}{\exp(2)-1} =0.025\frac{1.225541}{6.389056} \approx0.0047955. $$
At the second look, $$ \alpha(0.75)=0.025\frac{\exp(1.5)-1}{\exp(2)-1} =0.025\frac{3.481689}{6.389056} \approx0.0136236. $$ The target increment from the first to the second look, using the unrounded targets, is \(0.0136236-0.0047955\approx0.0088281\). At the final look, \(\alpha(1)=0.025\), so the remaining increment is approximately \(0.025-0.0136236=0.0113764\). These are cumulative and incremental spending targets, not look-specific crossing probabilities.
The cancellation in the one-sample calculation is the key justification for the shortcut: when information has the form \(I(n)=c n\) for a constant \(c\), its fraction is \(I(n_k)/I(N)=n_k/N\). The constant may reflect a common variance or another fixed feature of the model. If the per-participant contribution changes, the cancellation no longer gives a simple sample-size ratio.
Two-Arm Comparisons: Allocation Affects Information
For two independent treatment groups with normally distributed outcomes and known variances \(\sigma_T^2\) and \(\sigma_C^2\), the estimated treatment difference has variance $$ \operatorname{Var}(\bar{X}_T-\bar{X}_C)=\frac{\sigma_T^2}{n_T}+\frac{\sigma_C^2}{n_C}. $$
Its information for the difference is the reciprocal of this variance. With equal variances and a fixed allocation ratio, information is proportional to total sample size. But when the allocation changes, the same total number of participants can yield a different variance and therefore a different information fraction. With unequal variances, the allocation and the group-specific sample sizes both matter.
Worked Example: A Change in Allocation During Accrual
In a hypothetical equal-variance trial, the planned final sample is 360 participants: 180 assigned to treatment and 180 to control. At an interim look, 180 participants have accrued, but the allocation so far is 120 treatment and 60 control. Assume independent normal outcomes with the same known variance \(\sigma^2\) in both groups, and use the treatment-difference information model. Calculate the information fraction and compare it with the raw sample-size fraction.
For equal variances, $$ I(n_T,n_C)=\frac{1}{\sigma^2(1/n_T+1/n_C)} =\frac{n_Tn_C}{\sigma^2(n_T+n_C)}. $$ At the interim look, \(I_k=120(60)/\{180\sigma^2\}=40/\sigma^2\). At the planned final look, \(I_K=180(180)/\{360\sigma^2\}=90/\sigma^2\). Therefore, $$ t_k=\frac{I_k}{I_K}=\frac{40}{90}=\frac{4}{9}\approx0.4444. $$
The raw sample-size fraction is \(180/360=0.5000\), which overstates the information fraction here because the interim allocation is less balanced than the planned final allocation. If the one-sided spending function is linear, \(\alpha(t)=0.025t\), the target at this look is \(0.025(4/9)\approx0.01111\), not the \(0.01250\) obtained by incorrectly substituting the raw sample-size fraction. The assumptions used in the calculation are that the group outcomes are independent, the variances are equal and stable, and the stated allocation counts describe the analyzed participants.
This example illustrates why knowing only the total accrued count can be insufficient. For equal variances, the information depends on \(n_Tn_C/(n_T+n_C)\), not simply on \(n_T+n_C\), unless allocation is fixed at the same proportion throughout.
Using the Specified Variance Model
The same principle applies when the treatment groups have unequal variances. In that case, the reciprocal variance of the treatment difference weights each group’s sample size according to its variance. A raw count fraction can be misleading even if the total planned and accrued sample sizes are known.
Worked Example: Unequal Variances and a Linear Spending Target
Consider a hypothetical trial with independent normal outcomes. The known variances are \(\sigma_T^2=9\) and \(\sigma_C^2=4\) response-units squared. The planned final sample has 120 treatment and 180 control participants. At an interim look, there are 50 treatment and 100 control participants. Calculate the information fraction and the target under a one-sided linear spending function with total \(\alpha=0.025\).
The treatment-difference variance at the final look is $$ \frac{9}{120}+\frac{4}{180}=0.07500+0.02222=0.09722 $$ response-units squared, so \(I_K=1/0.09722\approx10.286\) per response-unit squared. At the interim look, the variance is $$ \frac{9}{50}+\frac{4}{100}=0.1800+0.04000=0.2200, $$ so \(I_k=1/0.2200\approx4.545\) per response-unit squared. Thus $$ t_k=\frac{I_k}{I_K} =\frac{0.09722}{0.2200} \approx0.4419. $$
The total sample-size fraction is \(150/300=0.5000\), but the information fraction is approximately \(0.4419\). Under linear spending, \(\alpha(t)=0.025t\), the target is \(0.025(0.4419)\approx0.01105\). Using the raw count fraction instead would give \(0.01250\), a different target. This calculation assumes the stated variances are appropriate and stable, the group outcomes are independent, and the treatment effect is analyzed using the difference in group means.
Checks, Alternatives, and Limitations
Before substituting a fraction into a spending function, verify that it represents the design’s information model. In practice, check the following:
- Match the denominator to the design. Use the planned final information \(I_K\), not an arbitrary target count, unless that count implies the same information under the model.
- Use the analyzed sample sizes. Counts should reflect participants with usable outcomes for the planned analysis, not merely those enrolled, if missing outcomes reduce information.
- Check the information sequence. Planned information should increase across looks. A computed fraction outside \(0<t_k\le1\) may indicate a wrong denominator, inconsistent counts, or a departure from the planned information model.
- Revisit the model when precision changes. Changing allocation, variance assumptions, measurement precision, or analysis population can break the proportionality between sample size and information.
The main alternative to a raw sample-size fraction is direct calculation from the variance of the effect estimator: \(I_k=1/\operatorname{Var}(\hat{\theta}_k)\), using the model appropriate to the design. When the variance is available only through a model-based estimate, the resulting information fraction is only as reliable as that model and estimate. A planned sample-size fraction remains convenient when equal per-participant information and a fixed allocation justify it; it is not a universal substitute.
Information fractions also have limits. They summarize statistical precision for the effect parameter, not the amount of follow-up, the number of calendar days, or the quality of data. A fraction computed under an assumed variance model may differ from the information actually achieved. The method for handling irregular realized looks or recalibrating boundaries after timing changes is a separate design question.
A second interpretive error is to call \(\alpha(t_k)\) the probability of a false-positive result at look \(k\). It is the cumulative spending target through that look. Boundary calibration accounts for dependence across looks and prior opportunities to cross; the spending target alone does not describe the probability of crossing at a particular look.
Check Your Understanding
For each problem, identify the appropriate information fraction and explain what it implies for the spending target. Solutions follow the questions.
- A one-sample normal study with constant known variance plans 500 observations and looks after 150 and 350 observations. What are the two information fractions?
- A two-arm equal-variance study plans 100 participants per arm. At an interim look there are 60 treatment and 40 control participants. Calculate the information fraction and compare it with the total sample-size fraction.
- For a one-sided linear spending function with total \(\alpha=0.02\), what is the cumulative spending target at an information fraction of \(0.60\)?
- Explain why the fraction of enrolled participants may not equal the information fraction when some participants lack usable outcome measurements.
Solutions. (1) Constant variance and equal per-observation information give \(t_1=150/500=0.3000\) and \(t_2=350/500=0.7000\). (2) Interim information is \(60(40)/\{100\sigma^2\}=24/\sigma^2\); final information is \(100(100)/\{200\sigma^2\}=50/\sigma^2\). Hence \(t=24/50=0.4800\). The raw sample-size fraction is \(100/200=0.5000\), so it is not the correct information fraction for this imbalanced interim allocation. (3) The linear target is \(\alpha(t)=0.02t\), giving \(0.02(0.60)=0.01200\). (4) Enrollment counts participants whether or not their outcomes contribute to the analysis. Missing or unusable outcomes can reduce the analyzed sample and its precision, so information must be based on the analysis model and available data rather than enrollment alone.