Page 400 - Six Sigma Advanced Tools for Black Belts and Master Black Belts

OTE/SPH
OTE/SPH
3:9
August 31, 2006
JWBK119-25
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Backward CUSUM 385
The likelihood ratio under the hypotheses H 0 and H j is
f
n i=1 μ=0 (y i )
λ n ( j) = j−1
f
i=1 f μ=0 (y i ) n i= j μ=μ 1 (y i )
n i= j μ=0 (y i )
f
= n
f
i= j μ=μ 1 (y i )
2

n (2π) −1/2 σ −1 exp y /2σ 2
i= j i
= n (25.4)
2
(2π) −1/2 σ −1 exp (y i − μ 1 ) /2σ 2
i= j
n n
1 2
2
= exp − y − (y i − μ 1 ) .
i
2σ 2
i= j i= j
The rejection region, using the Neyman--Pearson lemma, is given by λ n ( j) < k. Taking
the natural logarithm and simplifying (25.4), we therefore have the following rejection
region for testing the null hypothesis (25.2) against the alternative hypothesis (25.3):

n 2
i= j y i −σ ln k μ 1
> + . (25.5)
n − j + 1 μ 1 (n − j + 1) 2
The left-hand side of this inequality is simply the average of the latest n − j + 1 ob-
servations. We denote this average as ¯y j . Under H 0 ,¯y j is normally distributed with
2
mean zero and variance σ /(n − j + 1). Since the quantity on the right-hand side of
(25.5) is a constant (say, k ), it follows that ¯y j is the test statistic that yields the most

powerful test and the rejection region is ¯y j > k . That is, the rejection region, RR,is

given by

RR = ¯y j > k .

The precise value of k is determined by fixing α and noting that

α = P ( ¯y j in RR when μ = μ 0 = 0) .
Note that the form taken by the rejection region does not depend upon the particular
value assigned to μ 1 . That is, any value of μ greater than μ 0 = 0 would lead to exactly
the same rejection region. This gives rise to a UMP test for testing (25.2) against

μ = 0, for y 1 , y 2 ,..., y j−1 ,
H j =
μ>μ 1 , for y j , y j+1 ,..., y n .
The test is just a simple z-test with rejection region
n

i= j y i
√ ≥ z α . (25.6)
σ n − j + 1
It is well known that the value of k can be determined by fixing

n

i= j y i
α = P > k |μ = 0 .

n − j + 1