Normal Distribution
nothing is called 95% until it has been counted
The collection

Every essay — page 14

Essays 313 to 336 of 613, in the same order.

Choosing what the rule reads

A balancing rule cannot tell one basis from another with the same span, so choosing what to hand it is choosing a subspace, and everything about it is exact geometry: what a rule removes of a shape is that shape's squared multiple correlation on the span. The maximin basis over a list of shapes is then a finite problem with an exact answer — and over a whole subspace the guarantee is exactly zero rather than small, which is why the list cannot be avoided. Randomising which functions the rule reads is worth twice the best fixed choice.

The same number of assignments, held less tightly. For each basis size the box admits at a tolerance of 0.4, the Mahalanobis rule is given exactly the same number of admissible assignments — the ones with the smallest Mahalanobis distance — so the two differ only in the shape of the region. Each bar is the worst named function's remaining imbalance variance as a share of a coin's, with the mean over the named functions beside it. At one function the regions are the same interval: 0.0523 both ways. At 3 functions and 314 assignments, the box leaves 0.0771 in its worst function and the ellipse 0.1545; on average 0.0633 against 0.0927. The ellipse spends its assignments on a direction no function in the basis names.

What the ellipse spends its assignments on

Rerandomising on Mahalanobis distance never runs out of assignments, where a box of per-function tolerances does at four functions. But at the same number of admissible assignments it holds the worst named function half as tightly, because it spends its constraint on a combination of the basis nobody named.

4 figures · Allocation, part 6 Three regions at the box's own count. Across 24 sets of 16 units, on the sets where a box at a tolerance of 0.4 has any admissible assignment, the share of its own variance left in the worst of the k named functions at the box's count. the ellipse: 0.070, 0.134, 0.196, 0.194, 0.182. a sphere on the standardised functions: 0.062, 0.076, 0.091, 0.086, 0.080. the box: 0.062, 0.072, 0.081, 0.073, 0.069, at 2, 3, 4, 5, 6 functions, on 24, 20, 9, 4, 4 of the sets. The box's mean admissible count is 1045, 354, 115, 32, 13.

A box with a count

The ellipse never runs out of assignments and holds the named functions loosely; the box holds them tightly and can come up empty. Dropping the ellipse's thin directions was meant to give both, and it does not: no cutoff comes within a third of the box. But the box's emptiness was never a property of the box. It belongs to a tolerance fixed in advance. Set the tolerance by a count instead, and at that count the region is the box, on every set of units, with nothing left to run out.

3 figures · Allocation, part 7

The block size as a schedule

The blinded rule's block size is a dial between an interval's degrees of freedom, a stopping rule's, and the overshoot. Letting it change during the run keeps the coverage exact — the argument never needed the sizes to be equal — and then runs into an identity: the two kinds of degree of freedom add to N − 1 on every run, so a schedule cannot make more of both. What it can do is spend each where it is worth most, which is worth a few per cent and is not the free lunch the dial looked like.

One experiment, with the blocks getting smaller as the target comes into range. A single run at a requirement of 0.25, with the block sizes 5, 5, 11, 25, 11, 8, 3, 2 and a total of 70 observations in 8 blocks. The rule stops when the observations in hand reach z²σ̂²/d², with σ̂² pooled from the within-block contrasts — an estimate that moves as the run goes on, so the target moves too. Early blocks are large because the target is far away and cannot be overshot; late ones are small because a block is the granularity of the answer. The interval afterwards is built from the 8 block means and from nothing the rule looked at, and it has 7 degrees of freedom against the rule's 62.

A block size that changes

The blinded rule's exactness never needed the blocks to be the same size. Letting the size be chosen from the contrasts as the run goes on leaves the coverage exactly where it was — and runs straight into an identity that says what a schedule can and cannot buy.

6 figures · Stopping, part 8 No block size is best at both things the procedure claims. Two claims and one dial. The honest interval's half-width falls as the blocks get smaller, because the interval's degrees of freedom are the number of blocks: 0.2602 at blocks of two against 0.2933 at blocks of sixteen. The fixed-width claim — that the mean is within 0.25 of the truth — gets more reliable as they get larger, because the sample size is less variable: 92.40% against 95.00%. Both are computed from the same runs, and the second is reproduced to within a tenth of a point by E[2Φ(d√N/σ) − 1], which needs the sample-size distribution and nothing else. The schedules sit at the bottom left: as narrow as the smallest fixed block and as few observations, with the rule's spread estimate on half as many degrees of freedom again.

Two degrees of freedom, one total

The block size is a dial, and the two things a fixed-width procedure claims move in opposite directions along it. Divide the width by the square root of the sample size and one of them turns out to depend on the number of blocks and on nothing else.

6 figures · Width, part 4 The overshoot is the last block size and nothing else. A run stops at a multiple of its own block sizes and cannot land between them, so it ends past its own target by about half a block. Fixed sizes overshoot by 1.5, 2.2, 3.1, 4.7, 8.5 observations as the size goes 2, 3, 5, 8, 16. Every schedule here ends in blocks of two and every one of them lands where blocks of two land — 1.62, 1.32, 1.37 against 1.48 — while having spent most of the run inside blocks four and eight times larger. That is the one thing on this page a schedule genuinely takes from both ends.

What a schedule actually buys

Big blocks early and small blocks late is the right instinct and it does not take both ends of the trade, because there are not two ends to take. What it does take is the overshoot — about four per cent of the observations — and a steadier stopping point.

6 figures · Nuisance, part 3 What a schedule is allowed to read, and what happens when it reads more. The construction allows the block sizes to be anything at all as long as they are functions of the within-block contrasts, which are independent of every block mean. A schedule that shrinks the block whenever the between-block spread is running above what the contrasts say is a direct attempt to hold down the quantity the interval will be built from, and it succeeds: the estimate lands at 0.8373σ² against the honest 0.9831, and the coverage goes with it. Growing the block on the same spread instead pushes the other way and over-covers — which is not a repair, it is the same violation with the sign reversed, and the level is no longer a property of the procedure at all.

A schedule that reads the mean

The block sizes may be anything at all provided they are functions of the contrasts. Two natural schedules break that, in opposite directions — and the most natural mistake of the three is not a schedule at all but a stopping rule, at 86.87% coverage and fewer observations.

9 figures · Stopping, part 9 How much of the forbidden spread a stopping rule may read. The rule stops when the observations in hand reach z²ŝ²/d², with ŝ² a blend: a share 1 − κ of the pooled within-block estimate it may read, and a share κ of the between-block estimate the interval is built from, which it may not. At κ = 0 it is the honest rule; at κ = 1 it stops on the reported spread. Each curve is the coverage lost against κ = 0 on the same 6,000 runs, with two paired standard errors, at three requirements. At a half-width of 0.35 the loss is 0.87% at κ = 0.2 and peaks at 3.97%. At 0.25 it is 0.20% at 0.2, 0.98% at 0.5 and 5.50% at 1. At 0.15 nothing measurable is lost until κ = 0.9, and 3.73% at 1. The cost of reading is not proportional to how much is read; it is concentrated where the rule leans on the forbidden spread, and a longer experiment can lean further before paying.

How much of the forbidden spread a rule may read

A stopping rule that reads the spread its interval will report breaks the interval. Blend that spread into what the rule reads, a share at a time, and the cost is not proportional to the share — at a tight requirement a rule can read four fifths of it before anything is lost, and the last fifth costs nearly four points.

4 figures · Stopping, part 10 Coverage lost against the honest rule. Coverage lost in points against the rule that reads only within-block spreads, on the same observations, blocks of 2, 6000 runs, at half-widths 0.5, 0.35, 0.25, 0.15. the reported spread in full: 0.32 ± 0.27, 2.53 ± 0.37, 5.33 ± 0.41, 3.75 ± 0.35. a fixed half: 0.72 ± 0.26, 2.43 ± 0.29, 0.98 ± 0.26, -0.25 ± 0.20. κ = (b − 1)/(N − 1): 0.37 ± 0.21, 1.18 ± 0.25, 0.58 ± 0.23, -0.18 ± 0.19. the same, phased in as b/(b + 20): 0.22 ± 0.10, 0.35 ± 0.16, 0.23 ± 0.18, -0.10 ± 0.19. the same, read in only after twenty blocks: 0.00 ± 0.00, -0.02 ± 0.04, 0.17 ± 0.19, -0.17 ± 0.19.

A rule that sets its own κ

A stopping rule may read the spread inside its blocks and may not read the spread between them, and the dial between the two costs nothing for a long stretch and then a lot. The share an efficiency argument recommends is each estimate's share of the degrees of freedom, and on blocks of two at a half-width of 0.15 that loses nothing and steadies the experiment by 30%. At 0.35 it loses 1.18 points. Deferring the forbidden reading until twenty blocks costs nothing measurable at any requirement, and it keeps most of the steadiness where steadiness can be had.

3 figures · Stopping, part 11

Scoring a search without spending data

A hold-out is expensive: every row spent scoring is a row not spent fitting. There are two standard ways of not paying — an information criterion, which predicts the hold-out from the in-sample numbers, and a resampling, which builds a reference distribution from one sample. Neither avoids what it replaces. The criterion's penalty *is* the displacement, computed rather than counted, and it beats a rolling hold-out at every split there is; both it and the resampling are undone by the same defect, which is rows that repeat each other, because both are counting independent things and there are fewer of those than there are rows.

What each rule gives up against an oracle that is arithmetic. Expected squared error of the candidate each rule selects, minus the expected squared error of the best candidate in the table, over 500 draws of 120 rows. Both quantities are closed forms — σ_S²(1 + q/(n − q − 1)) — so the only Monte Carlo here is over which candidate got picked. The hold-out spends half its rows measuring what the criterion computes, and pays 1.8 times as much for it. Schwarz's criterion is worst because it is answering a different question: which candidate contains the truth, rather than which one forecasts best.

A criterion is a prediction of the hold-out

A rolling hold-out spends half the sample measuring what a criterion computes from all of it. Against an oracle that is arithmetic rather than an estimate, the criterion gives up 0.01701 and the hold-out 0.03200 — and the number the hold-out reports for its own winner is optimistic by more than either.

3 figures · Order selection, part 3 The crossing is in the dependence, not in the split. Regret of each rule as the design and the errors are made persistent at the same coefficient, scored on fresh rows because the closed form assumes exactly what is being taken away. An optimism theorem counts rows; when the rows repeat each other there are fewer of them than there are rows, the penalty is too small for the fit it is correcting, and the criterion starts buying coefficients it should not — its average winner grows from 3.31 coefficients to 3.90. The hold-out never used the theorem and overtakes at ρ ≈ 0.81. Schwarz's criterion, worst of the three on independent rows, is best on repeating ones — its heavier penalty is right for the wrong reason.

Where the two searches cross

The obvious dial between a criterion and a hold-out is how much of the sample to hold out, and moving it never changes the answer. The dial that does is one nobody chooses — how much each row repeats the one before it — and the two rules change places at about 0.81.

3 figures · Forecast, part 8 Each repair is for its own defect, and one is for both. The 95% point of the statistic's own distribution in each world, against the mean 95% point of five reference distributions built from one sample. Where the error variance is a function of the design, the two resamplings that detach a residual from its row fall short and the two multipliers that keep it there do not; where the rows repeat each other it is the other way round. With both defects at once the blocked multiplier — drawn once per run of 5 rows, so the residual never moves and its neighbours share a sign — is the closest of the five, at 2.999 against a truth of 3.803. It is still short by 0.804, and that shortfall is the next figure.

Two defects and one resampling

Four resamplings, each the repair for one defect and wrong about the other. Put both defects in the same world and the statistic's 5% point is 3.8028, where the best of the four reaches 2.8326 — until a multiplier that stays on its own row and shares a sign with its neighbours reaches 2.9988.

4 figures · Bootstrap, part 7 A bias against a variance, with the answer in between. How wrong one sample's reference distribution is, split into the two things it is wrong by. Sharing the multiplier over more rows keeps more of the dependence and closes the bias from 1.688 to 0.835; every row it is shared over also removes an independent sign from the 101 the sample started with, and the spread of the resulting quantile rises from 1.307 to 2.172. The distance a practitioner with one sample is actually exposed to is the two together, and it is smallest at ℓ = 5.

How long a block a multiplier shares

Sharing a sign over more rows keeps more of the dependence and leaves fewer independent signs to build a distribution from. The bias falls from 1.6885 to 0.8479 and the spread rises from 1.3073 to 2.1716, and the rejection rate walks straight through its nominal level on the way from 11.3% to 1.3%.

4 figures · Reference, part 5 A penalty that counts independent rows. Regret of five selection rules over fifteen candidates as the design and the errors are made persistent at the same coefficient, 300 draws at each of 7 settings, scored on fresh rows. Akaike's criterion gives up 0.0193, 0.0196, 0.0294, 0.0428, 0.0666, 0.1075, 0.1281; charging each coefficient its own variance inflation, with the persistences read off the sample, gives up 0.0188, 0.0231, 0.0339, 0.0375, 0.0496, 0.0656, 0.0652, and with them given 0.0193, 0.0236, 0.0348, 0.0374, 0.0426, 0.0469, 0.0318. The hold-out gives up 0.0358, 0.0365, 0.0393, 0.0453, 0.0594, 0.0913, 0.1087 and Schwarz's criterion 0.0355, 0.0295, 0.0331, 0.0397, 0.0575, 0.0849, 0.1047.

A penalty that counts independent rows

Akaike's penalty charges a fit two error variances a coefficient because each coefficient is fitted from n independent rows. When the design and the errors both persist, a row is worth less than a row, and the honest charge is twice the coefficient's variance inflation. On a hundred and twenty rows at persistence 0.7 that charge is 34.7 against an optimism counted at 34.5, where Akaike's charges 10. Charged it, the criterion's regret at 0.95 falls from 0.128 to 0.032 with the persistences given and to 0.065 when they are read off the sample — and at 0.5, where the charge is right and Akaike's is half of it, Akaike's still selects better.

4 figures · Forecast, part 9 Three charges and two counts. The optimism of an in-sample fit of the widest candidate on 120 rows, in units of its own error variance, over 1500 draws, by persistence. Counted on the 40 rows straight after the sample: 11.1, 21.5, 35.6, 51.5, 65.5, 87.8, 119.1. Counted on rows 600 further on: 9.3, 18.3, 33.0, 50.6, 67.0, 96.1, 159.0. Charged by the long-series inflation: 10.0, 19.3, 34.7, 54.4, 74.3, 114.2, 234.1. Charged by the same inflation summed over the sample's rows: 10.0, 19.2, 34.2, 53.1, 71.8, 108.2, 208.8. Charged by twice the trace of the hat matrix against the errors' correlation on the sample's own design: 10.0, 18.7, 32.0, 47.5, 61.5, 85.2, 132.2.

A charge the sample can afford

The inflation charge for dependent rows runs to twice the counted optimism at persistence 0.95, and the reason offered was that a series of n rows cannot be optimistic about movements slower than n rows. Summing the inflation over n rows does almost nothing: 234 becomes 209, against 119 counted. Evaluating the same variance on the sample's own design gives 132, and moving the fresh rows away from the sample raises the count to 159. So the charge can be made nearly right, and when it is, the regret doubles from 0.032 to 0.061. A selection wants a much larger charge than the optimism, and past about fifty per predictor it is no longer selecting.

3 figures · Forecast, part 10

When the set is too large to walk

Two constructions that were each measured by enumerating everything, at the size where enumerating everything stops being possible. A balancing dictionary over two covariates is an outer product, every inner product in it is still closed form, and a rule holding every main effect of both removes exactly nothing of any interaction. A trial of two hundred units has assignments that cannot be walked — and the acceptance rate turns out not to depend on the size of the trial at all, so the exhaustion a sixteen-unit count runs into is a fact about sixteen units.

Two covariates make the dictionary an outer product. Four functions of each covariate, and everything a balancing rule may be handed. The margins are the 8 main effects and the block between them is the 16 interactions, which are 66.7% of the dictionary. Every inner product in it is closed form — ⟨f₁g₁, f₂g₂⟩ = ⟨f₁,f₂⟩⟨g₁,g₂⟩ when the covariates are independent — so nothing about the geometry gets harder. What gets harder is the counting: choosing k of 24 is C(24, k), which is 10,626 at four and 735,471 at eight.

A dictionary that is a product

Two covariates make what a balancing rule may read an outer product — eight main effects and sixteen interactions — and every inner product in it is still closed form. What a rule holding all eight main effects removes of a pure interaction is not small. It is zero.

5 figures · Blocking, part 6 One curve is a binomial coefficient and the other is a line. The number of subsets a maximin over this dictionary would have to score, against the number the exchange algorithm actually scores. At three functions the walk is 2,024 subsets and is the honest answer; at eight it is 735,471 and the exchange algorithm has looked at 421. The warrant for the second curve is the four sizes where both exist and agree, which is a weak warrant — it says the algorithm has not yet been wrong, not that it cannot be — and it is the only one available past the point the first curve leaves the page.

Where the enumeration stops

A maximin over an eight-function dictionary is a walk over seventy subsets. Over twenty-four it is 735,471 at eight functions, and the exchange algorithm that replaces the walk scores 421. What licenses the second curve is four sizes where both exist and agree, which is a weaker warrant than it looks.

4 figures · Optimum, part 3 A rate that does not know how large the trial is. The share of equal splits admitted by a tolerance of 1 coin-spreads on 3 functions, at six trial sizes. The first two are exact — 12,870 and 184,756 splits, walked, averaged over eight draws of the units — and the rest are sampled. From a hundred units on, the rate sits on (2Φ(1) − 1)^3 = 0.3182, which contains no n at all. The two small trials are 29.2% and 27.7% short of it, so the sixteen-unit measurement understates the rate rather than bracketing it. Meanwhile the admissible count — the rate times C(n, n/2) — goes from 2^11.5 to 2^393.7: the exhaustion a small trial runs into is a fact about small trials.

A count that has to be estimated

At sixteen units the admissible assignments can be counted by walking all 12,870 of them. At four hundred there are about 2^393.70, and the share admitted is 0.31885 against a closed form of 0.31818 that has no trial size in it at all. The exhaustion a small trial runs into is a fact about small trials.

5 figures · Randomisation, part 6 The rate falls geometrically; the count does not fall at all. The share of equal splits of two hundred units that a tolerance of 1 coin-spreads admits, against the number of functions the tolerance is stated for, with the closed form (2Φ(1) − 1)^k drawn beside it. The rate falls by about two thirds with every constraint. The admissible count is that rate times C(200, 100), and it goes from 2^195 to 2^192 — it does not fall in any sense a trial cares about. What the falling rate costs is sampling: 9,878 draws to collect a thousand admissible ones at six constraints, against 1,465 at one.

What a reference distribution costs to sample

A randomisation test on a trial too large to enumerate has to sample its reference distribution, at 1/p attempts per draw and a p-value resolved to 1/(B + 1). Six constraints cost 9,878 attempts per thousand draws, and a thousand draws resolve p to 9.99·10⁻⁴ and not one digit finer.

3 figures · Reference, part 6 A basis drawn rather than chosen. The share of the worst of seven shapes a balancing rule removes, over two covariates and a dictionary of twenty-four functions, for the best fixed basis of each size (lower line) and for a basis drawn at random from the least favourable mixture (upper line), at one to four functions. The fixed basis guarantees 0.0000, 0.0000, 0.4244, 0.6366: nothing at one function or two, because two of the shapes are pure interactions and two of them are main effects of different covariates, and no one or two functions touch all of them. The drawn basis guarantees 0.2042, 0.4084, 0.6133, 0.7920 in expectation over the draw — about a fifth of a shape for every function up to three. The marks above the upper line are the best reply to nature's mixture, 0.2051, 0.4098, 0.6142, 0.7931: the game's value lies between the two.

A basis drawn rather than chosen

On two covariates no fixed basis of one or two functions protects every shape an outcome might have — its worst case is exactly zero. Draw the basis at random from the right mixture and the guarantee is a fifth of a shape for every function. The price is per trial — at three functions, two draws in five leave one shape wholly unprotected.

4 figures · Optimum, part 4 What a floor costs a drawn basis. The expected guarantee of a basis drawn from a mixture restricted to bases that each remove at least a stated share of every shape, solved at each distinct floor. 2 functions, of 276 bases: 40.8% at a floor of 0.0%; the best fixed basis floors at 0.0%. 3 functions, of 2024 bases: 61.3% at a floor of 0.0%, 50.0% at a floor of 31.8%, 50.0% at a floor of 34.7%, 49.7% at a floor of 40.5%, 42.4% at a floor of 42.4%; the best fixed basis floors at 42.4%. 4 functions, of 10626 bases: 79.2% at a floor of 0.0%, 77.1% at a floor of 31.8%, 77.1% at a floor of 34.7%, 77.1% at a floor of 35.1%, 77.1% at a floor of 35.6%, 77.1% at a floor of 37.0%, 77.1% at a floor of 38.3%, 77.1% at a floor of 40.5%, 76.5% at a floor of 41.0%, 76.5% at a floor of 42.4%, 76.5% at a floor of 42.8%, 76.5% at a floor of 42.9%, 76.5% at a floor of 43.5%, 76.5% at a floor of 44.7%, 76.5% at a floor of 45.3%, 76.5% at a floor of 46.0%, 76.5% at a floor of 46.8%, 76.5% at a floor of 47.2%, 76.5% at a floor of 50.0%, 76.5% at a floor of 56.0%, 76.5% at a floor of 63.7%; the best fixed basis floors at 63.7%.

A floor for the draw

A basis drawn at random guarantees more on average than any fixed one, at the price of leaving some shape wholly untouched in a large share of trials. Requiring every basis in the draw to keep a floor looks like a smooth trade between the two. It is a staircase, and the first step does nearly all the work. No basis has a floor between zero and 31.83%, so there is nothing to give up until a floor of that size is asked for. And at four functions, seven bases tie at the fixed basis's own 63.66%: a mixture of them keeps that floor exactly and adds thirteen points to the guarantee.

3 figures · Optimum, part 5

A promise about two arms

The fixed-width interval whose coverage is exact was built for one mean, and almost nothing anybody runs an experiment for is one mean. The construction survives with the harmonic effective size in place of the block size, and four things do not: a unit of that size costs four observations rather than one, the second variance puts Neyman's allocation in reach of a blinded rule, the conservation identity comes up one degree of freedom per block short, and there is an exact estimator under either of two conditions and none under both.

The construction survives a difference of two weighted means. Coverage of δ̂ ± t√(S_D²/H) on b − 1 degrees of freedom, over 900 runs at a requirement of 0.3, where δ̂ is the block differences weighted by h_b = (1/m_A + 1/m_B)⁻¹ and H is their total. The theorem the one-mean field rests on goes through with h_b in place of the block size, and the reason is that the weights a weighted least squares decomposition needs are the inverse variances — which is exactly what h_b is. The stopping rule reads only within-arm within-block contrasts, so it is a function of nothing the interval reports, whatever it does with the block sizes. Each bar is within 2.9% of the level it claims.

A width promised for a difference

The exact fixed-width interval was built for one mean. Two arms make the target 42.7 units of effective size and each unit costs four observations, so the same promise about a difference costs 169.4 rather than 42.7 — and the theorem survives untouched with the harmonic size in place of the block size.

5 figures · Width, part 5 Three sets of weights, five designs, and no estimator that is exact everywhere. Coverage of the same interval under three weightings. h_b is the inverse variance when the arms share a variance or the allocation is constant; equal weights are right when every block has the same two counts; the estimated precision weights are right in the limit and exact nowhere, because the decomposition needs the weights to be the constants they are only estimating. In the corner — two variances, changing sizes, changing allocation — the two exact estimators are the ones that miss, at 98.45% and 95.65%, and the one with no theorem behind it is at 95.05%. That is the whole statement: there is an exact estimator under either condition, and none under both.

Which weights are the inverse variances

There is an exact estimator when the two arms share a variance and another when every block has the same two counts, and between them they cover every trial anybody designs on purpose. In the corner where neither holds, both cover 98.45% instead of 95%, and the only estimator at its level is the one with no theorem behind it.

3 figures · Nuisance, part 4 Two arms leave one degree of freedom per block unaccounted for. Each point is one run. The one-mean field's identity is (b − 1) + (N − b) = N − 1, and every schedule moves along that line rather than off it. Two arms give the rule N − 2b and the interval b − 1, which come to N − b − 1 — short of the N − 2 two arms leave by exactly one per block, since a block's arm counts absorb one degree of freedom each and only one of the two directions carries the difference. The hollow points add what the block sums are worth, b − 1 more, and land on the total. The missing degrees of freedom are not lost; they are in a place the interval has to be shown it may read.

The degrees of freedom in the sums

One arm partitions N − 1 exactly. Two arms give the rule N − 2b and the interval b − 1, which is short by one per block — and the missing ones are in the block sums, which are correlated with the differences at −0.79 and are usable anyway.

5 figures · Blocking, part 7 Free until the sums stop seeing what the differences see. Coverage with and without the block sums pooled into the interval's variance estimate. With one effect and one level they are free. With an effect that varies between blocks they are still free, because a block's sum picks that variation up exactly as its difference does. With a level that varies they make the interval 37% wider and conservative. And where the effect falls as the level rises — a ceiling, and not an exotic thing to suppose — the sums carry none of the between-block variation while the differences carry all of it, the pooled estimate is short, and the interval that uses it covers 88.75% on a width 20% narrower than the honest one.

What a two-arm rule may not pool

A spread computed "within the block" without the arm label carries a share of the effect, so the trial runs 173 observations at a null and 282 at an effect of 1.5. The stopping rule is reading the thing it exists to measure, and the phrase that produced it is one word long.

4 figures · Allocation, part 6

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