Sous l'hypothèse nulle que les distributions sont les mêmes et que les deux échantillons sont obtenus de manière aléatoire et indépendamment de la distribution commune, nous pouvons déterminer les tailles de tous les tests 5×5 (déterministes) qui peuvent être effectués en comparant une valeur de lettre à une autre. Certains de ces tests semblent avoir un pouvoir raisonnable pour détecter les différences de distribution.
Une analyse
La définition originale du résumé à 5 lettres de tout lot ordonné de nombres x1≤x2≤⋯≤xn est la suivante [Tukey EDA 1977]:
Pour tout nombre m=(i+(i+1))/2 dans {(1+2)/2,(2+3)/2,…,(n−1+n)/2} définir xm=(xi+xi+1)/2.
Soit i¯=n+1−i .
Soit m=(n+1)/2 et h=(⌊m⌋+1)/2.
Le résumé à 5 lettres est l'ensemble {X−=x1,H−=xh,M=xm,H+=xh¯,X+=xn}. Ses éléments sont connus sous les noms de charnière minimale, inférieure, médiane, supérieure et maximale, respectivement.
Par exemple, dans le lot de données (−3,1,1,2,3,5,5,5,7,13,21) on peut calculer que n=12 , m=13/2 et h=7/2 , d'où
X−H−MH+X+=−3,=x7/2=(x3+x4)/2=(1+2)/2=3/2,=x13/2=(x6+x7)/2=(5+5)/2=5,=x7/2¯¯¯¯¯¯¯¯=x19/2=(x9+x10)/2=(5+7)/2=6,=x12=21.
Les charnières sont proches (mais généralement pas exactement les mêmes que) des quartiles. Si des quartiles sont utilisés, notez qu'en général, ils seront des moyennes arithmétiques pondérées de deux des statistiques d'ordre et se situeront ainsi dans l'un des intervalles [xi,xi+1] où i peut être déterminé à partir de n et l'algorithme utilisé pour calculer les quartiles. En général, lorsque q est dans un intervalle [i,i+1] , j'écrirai librement xq pour faire référence à une telle moyenne pondérée de xi et .xi+1
Avec deux lots de données et ( y j , j = 1 , … , m ) , il existe deux résumés distincts de cinq lettres. Nous pouvons tester l'hypothèse nulle selon laquelle les deux sont des échantillons aléatoires iid d'une distribution commune F en comparant l'une des x- lettres x q à l'une des y- lettres y r . Par exemple, nous pourrions comparer la charnière supérieure de x(xi,i=1,…,n)(yj,j=1,…,m),Fxxqyyrx to the lower hinge of y in order to see whether x is significantly less than y. This leads to a definite question: how to compute this chance,
PrF(xq<yr).
For fractional q and r this is not possible without knowing F. However, because xq≤x⌈q⌉ and y⌊r⌋≤yr, then a fortiori
PrF(xq<yr)≤PrF(x⌈q⌉<y⌊r⌋).
Nous pouvons ainsi obtenir des limites supérieures universelles (indépendantes de ) sur les probabilités souhaitées en calculant la probabilité de droite, qui compare les statistiques d'ordre individuel. La question générale qui se pose à nous estF
What is the chance that the qth highest of n values will be less than the rth highest of m values drawn iid from a common distribution?
Even this does not have a universal answer unless we rule out the possibility that probability is too heavily concentrated on individual values: in other words, we need to assume that ties are not possible. This means F must be a continuous distribution. Although this is an assumption, it is a weak one and it is non-parametric.
Solution
The distribution F plays no role in the calculation, because upon re-expressing all values by means of the probability transform F, we obtain new batches
X(F)=F(x1)≤F(x2)≤⋯≤F(xn)
and
Y(F)=F(y1)≤F(y2)≤⋯≤F(ym).
Moreover, this re-expression is monotonic and increasing: it preserves order and in so doing preserves the event xq<yr. Because F is continuous, these new batches are drawn from a Uniform[0,1] distribution. Under this distribution--and dropping the now superfluous "F" from the notation--we easily find that xq has a Beta(q,n+1−q) = Beta(q,q¯) distribution:
Pr(xq≤x)=n!(n−q)!(q−1)!∫x0tq−1(1−t)n−qdt.
Similarly the distribution of yr is Beta(r,m+1−r). By performing the double integration over the region xq<yr we can obtain the desired probability,
Pr(xq<yr)=Γ(m+1)Γ(n+1)Γ(q+r)3F~2(q,q−n,q+r; q+1,m+q+1; 1)Γ(r)Γ(n−q+1)
Because all values n,m,q,r are integral, all the Γ values are really just factorials: Γ(k)=(k−1)!=(k−1)(k−2)⋯(2)(1) for integral k≥0.
The little-known function 3F~2 is a regularized hypergeometric function. In this case it can be computed as a rather simple alternating sum of length n−q+1, normalized by some factorials:
Γ(q+1)Γ(m+q+1) 3F~2(q,q−n,q+r; q+1,m+q+1; 1)=∑i=0n−q(−1)i(n−qi)q(q+r)⋯(q+r+i−1)(q+i)(1+m+q)(2+m+q)⋯(i+m+q)=1−(n−q1)q(q+r)(1+q)(1+m+q)+(n−q2)q(q+r)(1+q+r)(2+q)(1+m+q)(2+m+q)−⋯.
This has reduced the calculation of the probability to nothing more complicated than addition, subtraction, multiplication, and division. The computational effort scales as O((n−q)2). By exploiting the symmetry
Pr(xq<yr)=1−Pr(yr<xq)
the new calculation scales as O((m−r)2), allowing us to pick the easier of the two sums if we wish. This will rarely be necessary, though, because 5-letter summaries tend to be used only for small batches, rarely exceeding n,m≈300.
Application
Suppose the two batches have sizes n=8 and m=12. The relevant order statistics for x and y are 1,3,5,7,8 and 1,3,6,9,12, respectively. Here is a table of the chance that xq<yr with q indexing the rows and r indexing the columns:
q\r 1 3 6 9 12
1 0.4 0.807 0.9762 0.9987 1.
3 0.0491 0.2962 0.7404 0.9601 0.9993
5 0.0036 0.0521 0.325 0.7492 0.9856
7 0.0001 0.0032 0.0542 0.3065 0.8526
8 0. 0.0004 0.0102 0.1022 0.6
A simulation of 10,000 iid sample pairs from a standard Normal distribution gave results close to these.
To construct a one-sided test at size α, such as α=5%, to determine whether the x batch is significantly less than the y batch, look for values in this table close to or just under α. Good choices are at (q,r)=(3,1), where the chance is 0.0491, at (5,3) with a chance of 0.0521, and at (7,6) with a chance of 0.0542. Which one to use depends on your thoughts about the alternative hypothesis. For instance, the (3,1) test compares the lower hinge of x to the smallest value of y and finds a significant difference when that lower hinge is the smaller one. This test is sensitive to an extreme value of y; if there is some concern about outlying data, this might be a risky test to choose. On the other hand the test (7,6) compares the upper hinge of x to the median of y. This one is very robust to outlying values in the y batch and moderately robust to outliers in x. However, it compares middle values of x to middle values of y. Although this is probably a good comparison to make, it will not detect differences in the distributions that occur only in either tail.
Being able to compute these critical values analytically helps in selecting a test. Once one (or several) tests are identified, their power to detect changes is probably best evaluated through simulation. The power will depend heavily on how the distributions differ. To get a sense of whether these tests have any power at all, I conducted the (5,3) test with the yj drawn iid from a Normal(1,1) distribution: that is, its median was shifted by one standard deviation. In a simulation the test was significant 54.4% of the time: that is appreciable power for datasets this small.
Much more can be said, but all of it is routine stuff about conducting two-sided tests, how to assess effects sizes, and so on. The principal point has been demonstrated: given the 5-letter summaries (and sizes) of two batches of data, it is possible to construct reasonably powerful non-parametric tests to detect differences in their underlying populations and in many cases we might even have several choices of test to select from. The theory developed here has a broader application to comparing two populations by means of a appropriately selected order statistics from their samples (not just those approximating the letter summaries).
These results have other useful applications. For instance, a boxplot is a graphical depiction of a 5-letter summary. Thus, along with knowledge of the sample size shown by a boxplot, we have available a number of simple tests (based on comparing parts of one box and whisker to another one) to assess the significance of visually apparent differences in those plots.
I don't see how there could be such a test, at least without some assumptions.
You can have two different distributions that have the same 5 number summary:
Here is a trivial example, where I change only 2 numbers, but clearly more numbers could be changed
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