318 Some Elementary Statistical InferencesWe also note that the Jacobian is equal to 1 and that the space of positive probability
density is{(w 1 ,w 2 ,...,wn):0<wi,i=1, 2 ,...,n, w 1 +···+wn< 1 }.Since the joint pdf ofZ 1 ,Z 2 ,...,Znisn!, 0 <z 1 <z 2 <···<zn<1, zero
elsewhere, the joint pdf of thencoverages is
k(w 1 ,...,wn)={
n!0<wi,i=1,...,n, w 1 +···wn< 1
0elsewhere.Because the pdfk(w 1 ,...,wn) is symmetric inw 1 ,w 2 ,...,wn, it is evident that the
distribution of every sum ofr, r < n, of these coveragesW 1 ,...,Wnis exactly the
same for each fixed value ofr. For instance, ifi<jandr=j−i, the distribution
ofZj−Zi=F(Yj)−F(Yi)=Wi+1+Wi+2+···+Wjis exactly the same as that
ofZj−i=F(Yj−i)=W 1 +W 2 +···+Wj−i. But we know that the pdf ofZj−iis
the beta pdf of the form
hj−i(v)={
Γ(n+1)
Γ(j−i)Γ(n−j+i+1)vj−i− (^1) (1−v)n−j+i 0 <v< 1
0elsewhere.
Consequently,F(Yj)−F(Yi)hasthispdfand
P[F(Yj)−F(Yi)≥p]=
∫ 1
p
hj−i(v)dv.
Example 4.10.1. LetY 1 <Y 2 <···<Y 6 be the order statistics of a random
sample of size 6 from a distribution of the continuous type. We want to use the
observed interval (y 1 ,y 6 ) as a tolerance interval for 80% of the distribution. Then
γ=P[F(Y 6 )−F(Y 1 )≥ 0 .8]
=1−
∫ 0. 8
0
30 v^4 (1−v)dv,
because the integrand is the pdf ofF(Y 6 )−F(Y 1 ). Accordingly,
γ=1−6(0.8)^5 +5(0.8)^6 =0. 34 ,
approximately. That is, the observed values ofY 1 andY 6 define a 34% tolerance
interval for 80% the probability for the distribution.
Remark 4.10.1.Tolerance intervals are extremely important and often they are
more desirable than confidence intervals. For illustration, consider a “fill” problem
in which a manufacturer says that each container has at least 12 ounces of the
product. LetXbe the amount in a container. The company would be pleased to
note that the interval (12. 1 , 12 .3), for instance, is a 95% tolerance interval for 99%
of the distribution ofX. This would be true in this case, because the FDA allows
a very small fraction of the containers to be less than 12 ounces.