What a Two-Period Schedule Audit Checks
In a two-period crossover schedule, each participant is assigned an order of treatments. In the TR sequence, the participant receives test in Period 1 and reference in Period 2; in the RT sequence, the participant receives reference in Period 1 and test in Period 2. A schedule audit turns those sequence assignments into treatment counts for each period.
The audit has two specific questions. First, how many participants are assigned to each treatment in each period, and are those counts balanced? Second, does each treatment occur at least once in each period? These checks describe the proposed schedule; they do not by themselves establish that the design is suitable for a particular study or that the observed study data will follow the planned assignments.
The word “complete” matters: this count mapping assumes each participant has one assignment to a sequence and the planned sequence specifies a treatment in both periods. When auditing a planned schedule, count assignments. When auditing completed records, count the participants who actually contributed a treatment observation in each period, and state that this is an observed-completion audit rather than the original planned allocation.
From Sequence Counts to Period Counts
A compact way to organize the audit is to make a two-by-two table. Rows represent periods and columns represent treatments. The row total is the number scheduled in that period. For a complete two-period schedule, both row totals equal the total number assigned, \(n=n_{TR}+n_{RT}\).
| Period | Test count | Reference count | Period total |
|---|---|---|---|
| 1 | \(n_{TR}\) | \(n_{RT}\) | \(n_{TR}+n_{RT}\) |
| 2 | \(n_{RT}\) | \(n_{TR}\) | \(n_{TR}+n_{RT}\) |
For each period, define the signed difference as the test count minus the reference count. A positive value means more participants are assigned to test in that period; a negative value means more are assigned to reference. The absolute difference gives the size of the imbalance without its direction.
Thus, if the sequence counts are unequal, the direction of the imbalance reverses between periods: the sequence with more participants contributes more test assignments in one period and more reference assignments in the other. Period counts are equal within each period exactly when \(n_{TR}=n_{RT}\). This is a count-based property, not a statistical test.
Worked Audits
Worked Example: Equal Allocation to TR and RT
Hypothetical schedule. A proposed schedule assigns 24 participants: 12 to TR and 12 to RT. Audit the period counts, balance, and treatment presence.
Period 1. The 12 TR participants receive test, and the 12 RT participants receive reference. The counts are \(T=12\), \(R=12\), and the period total is \(12+12=24\). The signed difference is \(d_1=12-12=0\), so the period is balanced.
Period 2. The 12 RT participants receive test, and the 12 TR participants receive reference. The counts are \(T=12\), \(R=12\), and the period total is again \(24\). The signed difference is \(d_2=12-12=0\), so this period is also balanced.
Conclusion. Both treatments occur in both periods, and the treatment counts are equal within each period. This conclusion follows from the planned assignments; it does not guarantee that all 24 participants will complete both periods.
Worked Example: Both Treatments Present but Counts Unequal
Hypothetical schedule. A proposed schedule assigns 30 participants: 19 to TR and 11 to RT. Determine whether both treatments occur in each period and quantify the imbalance.
Period 1. The sequence mapping gives \(T=19\) and \(R=11\). The total is \(19+11=30\), and the signed difference is \(d_1=19-11=8\). There are eight more test than reference assignments in this period.
Period 2. The mapping reverses: \(T=11\) and \(R=19\). The total is \(11+19=30\), and \(d_2=11-19=-8\). There are eight more reference than test assignments in this period.
Conclusion. Both sequence counts are positive, so both treatments occur in both periods. The schedule is not treatment-balanced within either period: the absolute difference is \(8\) in each. It would be inaccurate to call the schedule balanced merely because every participant is assigned to receive both treatments across the two periods.
Worked Example: One Sequence Missing
Hypothetical schedule. A proposed schedule assigns all 16 participants to TR and none to RT. Audit the two periods.
Here \(n_{TR}=16\) and \(n_{RT}=0\). In Period 1, \(T=16\), \(R=0\), so \(d_1=16-0=16\). In Period 2, \(T=0\), \(R=16\), so \(d_2=0-16=-16\). Both period totals are 16.
Conclusion. The counts are not balanced, and the requirement that both treatments occur in both periods fails: Period 1 has no reference assignments, while Period 2 has no test assignments. Although each participant is scheduled to receive both treatments over the full study, that does not mean both treatments occur in each period.
Worked Example: Planned Assignments Versus Completed Records
Hypothetical schedule and records. A plan assigns 10 participants to TR and 10 to RT. By the end of the study, two TR participants have no Period 2 observation, and one RT participant has no Period 1 observation. Show why an audit should identify which counts it reports.
Planned counts. The original schedule has \(n_{TR}=10\) and \(n_{RT}=10\). It therefore plans \(T=10\), \(R=10\) in each period, with a signed difference of zero in each.
Available observations. In Period 1, all 10 TR participants contribute test observations, while 9 of the 10 RT participants contribute reference observations. The observed counts are \(T=10\), \(R=9\), so \(d_1=10-9=1\). In Period 2, 8 TR participants contribute reference observations, while all 10 RT participants contribute test observations. The observed counts are \(T=10\), \(R=8\), so \(d_2=10-8=2\).
Conclusion. The plan is balanced, but the available observations are not balanced in either period. Report these as separate facts: the schedule’s planned allocation is \(10\) per treatment per period, whereas the observed counts are \(10\) versus \(9\) in Period 1 and \(10\) versus \(8\) in Period 2. Missing observations do not change the original schedule, but they do change the count audit of available data.
Why the Count Identities Hold
The formulas follow by listing what each sequence contributes, rather than by applying an inferential result. In Period 1, every TR assignment contributes one test assignment and every RT assignment contributes one reference assignment. In Period 2, those roles reverse. Therefore the period totals are each \(n_{TR}+n_{RT}\), while the signed differences are opposites.
A further check is that the schedule assigns each participant one test period and one reference period. Summing across periods, the total number of planned test assignments is \(n_{TR}+n_{RT}=n\), and the same is true for reference:
This equality is a useful arithmetic check on a complete TR/RT schedule. It does not imply equal test and reference counts within each period. For that, the sequence counts themselves must be equal.
A Practical Audit and Its Limits
A reliable audit starts from sequence assignments, not from an assumption that the schedule is balanced. Count the TR and RT assignments, translate them into period-by-treatment counts, calculate the within-period differences, and state whether either sequence is absent. If the schedule is supplied as a participant-level list, counting the sequence labels is usually the quickest route; the treatment-by-period table is the clearest way to present the result.
For equal period treatment counts, the required condition is \(n_{TR}=n_{RT}\). If the counts differ, report the difference rather than hiding it behind a general label such as “approximately balanced.” If a design has a prespecified unequal allocation target, compare the counts with that target and identify the departure; do not silently treat unequal counts as equal. The treatment-presence check is simpler: verify that both sequence counts are greater than zero.
An audit can be checked in several ways. Confirm that every period total equals the number assigned to the schedule; confirm that the test and reference totals across both periods each equal that number; and confirm that the Period 2 counts reverse the Period 1 sequence contributions. If a table violates any of these identities, revisit the sequence labels, period headings, or count arithmetic. A common clerical error is to read RT as test followed by reference, which is TR; writing out each sequence in words before tallying prevents that reversal.
A sequence-count table and a period-by-treatment table answer closely related but not identical presentation needs. The sequence-count table shows how participants are allocated to orders. The period-by-treatment table makes period-specific balance and missing treatment presence immediately visible. A participant-level list is useful for finding duplicated IDs, incorrect labels, or missing assignments, but it is not a substitute for reporting the aggregate counts. Using both a list check and a count table can catch transcription errors.
The audit is deliberately limited. Equal counts do not prove that assignments were randomized as intended, that participants completed the schedule, or that the design meets every scientific or operational requirement. Unequal counts do not, by themselves, establish that the study is invalid. This check also does not assess the order of treatment transitions, carryover concerns, or the suitability of a design for a study question. It answers only the count and treatment-presence questions stated at the outset.
Common Mistakes
- Checking only overall treatment totals. A complete TR/RT schedule gives each participant one assignment to each treatment overall, even when the counts differ greatly within each period. Calculate counts separately for Period 1 and Period 2.
- Equating treatment presence with balance. If both sequence counts are positive, both treatments occur in both periods; the counts can still be unequal. Report the presence check and the balance check separately.
- Calling the two periods separate participant groups. In the planned crossover schedule, the same participants are assigned across both periods. Each period total is \(n_{TR}+n_{RT}\), not half the sample.
- Mixing planned assignments with observed records. Withdrawals or missing period observations can alter observed counts without changing the schedule as originally planned. Label which set of counts is being audited.
- Inferring broader design quality from a count table. A balanced table is a useful allocation property, not proof of randomization quality or overall design adequacy.
Check Your Understanding
For each question, use the sequence-to-period mapping and show the relevant count check.
- A schedule assigns 18 participants to TR and 18 to RT. What are the treatment counts in each period, and are both periods balanced?
- A schedule assigns 14 participants to TR and 10 to RT. Find the signed difference in each period. Does each treatment occur in each period?
- A schedule assigns 20 participants to RT and none to TR. Which treatment is absent from each period?
- Explain why equal total numbers of test and reference assignments across the two periods do not establish period balance.
- A plan has 8 TR and 8 RT assignments, but observed Period 1 records include 8 test and 7 reference observations. State the planned Period 1 counts and the observed Period 1 imbalance.
Solutions. (1) Period 1 has \(T=18\), \(R=18\); Period 2 also has \(T=18\), \(R=18\). Both are balanced, and both treatments occur in both periods. (2) Period 1 has \(T=14\), \(R=10\), so \(d_1=14-10=4\). Period 2 has \(T=10\), \(R=14\), so \(d_2=10-14=-4\). Both treatments occur in both periods because both sequence counts are positive, but neither period is balanced. (3) Period 1 has \(T=0\), \(R=20\), so test is absent. Period 2 has \(T=20\), \(R=0\), so reference is absent. (4) Each participant contributes one test and one reference assignment overall, so the total assignments are equal across periods combined regardless of how many participants are in each sequence. Period balance requires equal test and reference counts separately within each period, which occurs only when the sequence counts are equal. (5) The plan has Period 1 counts \(T=8\), \(R=8\), balanced. The observed records have \(T=8\), \(R=7\), so the observed signed difference is \(8-7=1\), with one more test than reference observation.