Multiattribute acceptance sampling plans - Library(ISI Kolkata ...
Multiattribute acceptance sampling plans - Library(ISI Kolkata ...
Multiattribute acceptance sampling plans - Library(ISI Kolkata ...
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where a 0 and n 0 are the parameters of the moment equivalent <strong>plans</strong> obtained from (2.4.5)<br />
for ρ i = p i /(p 1 + p 2 + ... + p r ) for i = 1, 2, ..., r,<br />
then n 0 < n.<br />
....(2.4.6)<br />
Proof<br />
For a fixed s ≥ 2 let (2.4.6) hold for all r, 2 ≤ r ≤ s and we take r = s.<br />
Consider a C kind plan with sample size n and <strong>acceptance</strong> numbers c 1 , c 2 , ..., c s .<br />
Now,<br />
P (p [s] ) = G(c 1 , np 1 )G(c 2 , np 2 )...G(c s , np s ) = G(a 0 , n 0 p (s) ),<br />
where p (s) = p 1 +p 2 +...+p s and a 0 and n 0 have been obtained from (2.4.5) for ρ i = p i /p (s) i =<br />
1, 2, ...s.<br />
Adding one more attribute to the study system,<br />
P (p [s+1] ) = G(c 1 , np 1 )G(c 2 , np 2 )....G(c s , np s )G(c s+1 , np s+1 )<br />
= G(a 0 , n 0 p (s) )G(c s+1 , np s+1 )<br />
= G(a 0 , n(n 0 /n)p (s) )G(c s+1 , np s+1 )<br />
= G(a 0 , n.p ′ (s) )G(c s+1, np s+1 ), where p ′ (s) = (n 0/n)p (s) < 1.<br />
Note that n 0 /n < 1 by supposition and therefore p ′ (s) < p (s) < 1.<br />
We may now construct a SSP with parameter (a 1 , n 1 ) for the C kind plan with number<br />
of attributes, r =2 using (2.4.5) such that,<br />
G(a 1 , n 1 (p s+1 + p ′ (s)) = G(a 0 , n.p ′ (s))G(c s+1 , np s+1 ).<br />
Further let this be equal to G(a 1 , n 2 .(p s+1 + p (s) )).<br />
But p ′ (s) < p (s) and, therefore, n 2 < n 1 < n.<br />
This implies that if (2.4.6) holds for r = 2 then it holds for all r ≥ 2, by mathmatical<br />
induction.<br />
We now prove that (2.4.6) holds for r = 2.<br />
To prove that n 0 < n we must show that<br />
E(m 2 ) − E 2 (m) > E(m) or E(m 2 ) − E(m) > E 2 (m).<br />
Defining ρ = m 1 /m = p 1 /p,<br />
E(m) =<br />
c 1 ∑<br />
x 1 =0<br />
c 2 ∑+x 1<br />
x=x 1<br />
b(x 1 , x, ρ) =<br />
and b(x 1 , x, ρ) = ( x<br />
x 1<br />
)<br />
ρ<br />
x 1<br />
(1 − ρ) x−x 1<br />
,we get<br />
c 1 ∑<br />
x∑<br />
x=0 x 1 =0<br />
b(x 1 , x, ρ)+<br />
c 2 ∑+c 1<br />
c 1 ∑<br />
x=c 1 +1 x 1 =0<br />
b(x 1 , x, ρ) = (c 1 +1)+<br />
c 2 ∑<br />
i=1<br />
B(c 1 , c 1 +i, ρ).<br />
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