IEOR 165 Discussion 3
February 20, 2015
Question 1. Suppose that when a process is in control each item will be defective with
probability .04. Suppose that your three sigma (z1−α/2 = 3) control chart calls for taking
daily samples of size 500. What is the probability that, if the probability of defective item
should suddenly shift to .08, your control chart would detect this shift on the next sample?
Question 2. The following data present the number of defective bearing and seal assemblies
in samples of size 100.
Does it appear that the process was in control throughout? If not, determine revised
control limits if possible.
1
Solution 1.
X is the number of defective items in a subgroup of n items
F = X/n is the fraction of the subgroup that is defective
X ∼ Binom(n, p) However if n is large we can approximate X by N (np, np(1 − p))
F ∼ N (E[X/n], V ar(X/n)) ∼ N (p, p(1 − p)/n))
p = 0.04 n = 500
r
r
p(1 − p)
0.04(0.96)
= 0.04 − 3
= 0.014
LCL = p − 3
n
500
r
r
p(1 − p)
0.04(0.96)
U CL = p + 3
= 0.04 + 3
= 0.066
n
500
Then for X, LCL = 500(0.014) = 6.85, U CL = 500(0.066) = 33.15.
If p0 = 0.08, X ∼ Binom(500, 0.08) since n is large X ∼ N (500(0.08), 500(0.08)(0.92)).
P (detect|p0 = 0.08) ≈ P (X ≥ 34) = 1 − P (X ≤ 33) = 0.86
Solution 2.
k = 20 n = 100 p is unknown
Estimate of p is given by
89
X1 + X2 + ... + Xn
=
= 0.0445
kn
20 · 100
r
r
F̄ (1 − F̄ )
0.0445(0.9555)
LCL = F̄ − 3
= 0.04 − 3
= −0.017 → 0
n
100
r
r
0.0445(0.96)
F̄ (1 − F̄ )
U CL = F̄ + 3
= 0.04 + 3
= 0.106
n
100
Then for X, LCL = 0 and U CL = 10.6. Thus no more than 10 defectives in a sample. All
samples satisfy this condition. The process was in control throughout. No need to revise the
control limits.
F̄ =
2