506 Optimal Tests of Hypothesessayx 1 ,x 2 ,x 3 ,.... These are usually plotted; and if they are between the LCL and
UCL, we say that the process isin control. If one falls outside the limits, this
would suggest that the meanμhas shifted, and the process would be investigated.
It was recognized by some that there could be a shift in the mean, say fromμto
μ+(σ/√
n); and it would still be difficult to detect that shift with a single sample
mean, for now the probability of a singlexexceeding UCL is only about 0.023. This
means that we would need about 1/ 0. 023 ≈43 samples, each of sizen, on the average
before detecting such a shift. This seems too long; so statisticians recognized that
they should be cumulating experience as the sequenceX 1 ,X 2 ,X 3 ,...is observed
in order to help them detect the shift sooner. It is the practice to compute the
standardized variableZ=(X−μ)/(σ/
√
n); thus, we state the problem in these
terms and provide the solution given by a sequential probability ratio test.
HereZisN(θ,1), and we wish to testH 0 :θ= 0 againstH 1 :θ= 1 using the
sequence of iid random variablesZ 1 ,Z 2 ,...,Zm,....Weusemrather thann,as
the latter is the size of the samples taken periodically. We have
L(0,m)
L(1,m)=exp[
−∑
z^2 i/ 2]exp [−∑
(zi−1)^2 /2]=exp[
−∑mi=1(zi− 0 .5)]
.Thus
k 0 <exp[
−∑mi=1(zi− 0 .5)]
<k 1canbewrittenash=−logk 0 >∑mi=1(zi− 0 .5)>−logk 1 =−h.It is true that−logk 0 =logk 1 whenαa =βa.Often,h=−logk 0 is taken
to be about 4 or 5, suggesting thatαa=βais small, like 0.01. As
∑
(zi− 0 .5)
is cumulating the sum ofzi− 0. 5 ,i=1, 2 , 3 ,..., these procedures are often called
CUSUMS. If the CUSUM =
∑
(zi− 0 .5) exceedsh, we would investigate the process,
as it seems that the mean has shifted upward. If this shift is toθ=1,thetheory
associated with these procedures shows that we need only eight or nine samples on
the average, rather than 43, to detect this shift. For more information about these
methods, the reader is referred to one of the many books on quality improvement
through statistical methods. What we would like to emphasize here is that through
sequential methods (not only the sequential probability ratio test), we should take
advantage of all past experience that we can gather in making inferences.
EXERCISES
8.4.1. LetXbeN(0,θ) and, in the notation of this section, letθ′=4,θ′′=9,
αa=0.05, andβa=0.10. Show that the sequential probability ratio test can be
based upon the statistic
∑n
1 X
2
i. Determinec^0 (n)andc^1 (n).