Two-shift entrance test, one raw cut-off, and a candidate's complaint
A university conducts a 100-mark entrance test for 60 M.A. seats in a subject. Because of hall capacity the test is held in two shifts. Shift 1 has 900 candidates with a mean of 54 and a standard deviation of 12. Shift 2 has 600 candidates with a mean of 45 and a standard deviation of 12. The examination committee has so far applied a single raw cut-off of 60 marks. A Shift 2 candidate who scored 58 has written to the Vice-Chancellor pointing out that her score would almost certainly have cleared the cut-off had she sat in Shift 1. The Registrar argues that shift-wise normalisation would be 'unfair to Shift 1 candidates who worked harder'. The Vice-Chancellor asks you, as a committee member, for a written note before the next meeting. You have the summary statistics above; the raw score lists are held by the controller of examinations.
1. Using the normal approximation, estimate how many candidates clear 60 in each shift, and what a strict raw-score merit list filling 60 seats would mean for Shift 2 candidates. Show your working.
Read the worked answer
Standardise with z = (60 - mean) / SD and read the upper tail of the normal distribution.
- Shift 1: z = (60 - 54)/12 = 0.50, upper tail 30.85 per cent, so 0.3085 x 900 = about 278 clear the cut-off.
- Shift 2: z = (60 - 45)/12 = 1.25, upper tail 10.56 per cent, so 0.1056 x 600 = about 63 clear the cut-off.
- Total qualifying: about 341 for 60 seats, so a second filter is unavoidable.
What a strict raw-score list does. If seats go to the top 60 raw scores, the implied threshold is around 72-73 marks: at 72.5, roughly 55 Shift 1 candidates and only about 7 Shift 2 candidates lie above it. So Shift 2 would take roughly 6-7 of the 60 seats, about 11 per cent, while constituting 40 per cent of the pool (600 of 1,500).
The petitioner. Her 58 in Shift 2 is z = 1.083, roughly the top 14 per cent of her shift; the same 58 in Shift 1 is z = 0.333, roughly the top 37 per cent. Her point about the comparison is arithmetically well founded.
Caveat for the note: the calculation assumes approximate within-shift normality and comparable candidate quality across shifts. Both should be checked against the actual distributions before any decision.
2. Compute the percentile-based normalisation alternative (seats allocated in proportion to shift size) and explain who gains and who loses. Does it answer the petitioner's grievance?
Read the worked answer
Method. Replace 'same raw mark' with 'same relative standing'. With 60 seats for 1,500 candidates, award 4 per cent of each shift: 36 seats from Shift 1 and 24 from Shift 2. The cut-off is the 96th percentile, z = 1.751.
- Shift 1 cut-off: 54 + 1.751 x 12 = about 75.0
- Shift 2 cut-off: 45 + 1.751 x 12 = about 66.0
Who gains, who loses. Shift 2 receives a 6-mark difficulty credit and its share of seats rises from roughly 11 per cent to 40 per cent, matching its share of the pool. Shift 1's effective requirement rises from 60 to 75: candidates in the 60-74 band who would have been admitted under the raw rule are displaced.
Does it answer the petitioner? Only partly, and the note must say so plainly. Her grievance about the comparison method is valid - 58 in Shift 2 is a far stronger relative performance than 58 in Shift 1. But under normalisation she would still not qualify, because her shift's cut-off would be about 66. She has identified a real defect in the procedure without being its beneficiary.
On the Registrar's objection: 'worked harder' is not measurable from these data. If shifts were assigned randomly or by roll number, the mean gap is a property of the paper, not of the candidates. If candidates chose shifts, or centres differ in quality, normalisation is still the lesser distortion but the allocation rule must be published in advance.
3. What should the Vice-Chancellor be advised to do about this year's admissions, and about the process for future years?
Read the worked answer
This year. Do not improvise a new rule after results are known. Changing the method post hoc disadvantages candidates who relied on the announced rule, and it is exactly the kind of retro-fitting that is hardest to defend if challenged. Advise instead that the committee: (a) confirm in writing what rule was actually published before the test; (b) obtain the raw distributions and check whether the normality assumption and the cross-shift comparability hold; (c) if the announced rule was raw cut-off, apply it, and record the decision and its reasoning in the minutes. If no rule was ever published, the committee has discretion and should exercise it transparently, with a written technical note explaining the chosen method.
Future years - the real fix is process, not statistics:
- Eliminate the two-shift problem where possible: one paper, one shift, or parallel shifts with randomised allocation of candidates.
- If two shifts are unavoidable, pre-announce equipercentile normalisation in the prospectus, with the formula and the seat-allocation rule, before candidates apply.
- Publish a technical note after results: shift sizes and distributions, the method used, the resulting cut-offs, and reliability of the paper.
- Have the controller of examinations, not the academic committee alone, certify the computation, and keep the audit trail.
Reasoning: a selection rule is defensible when it is announced before it is applied, applied mechanically, and explainable to the candidate who loses under it.