13.2Control Charts for Average Values: TheX-Control Chart 549
things do not change again) until the chart will indicate that the process is now out of
control? To answer this, note that a subgroup average will be within the control limits if
− 3 <√
nX−μ
σ< 3or, equivalently, if
− 3 −a√
n
σ<√
nX−μ
σ−a√
n
σ< 3 −a√
n
σor
− 3 −a√
n
σ<√
nX−μ−a
σ< 3 −a√
n
σHence, sinceX is normal with meanμ+aand varianceσ^2 /n— and so
√
n(X −
μ−a)/σhas a standard normal distribution — the probability that it will fall within
the control limits is
P{
− 3 −a√
n
σ<Z< 3 −a√
n
σ}
=(
3 −a√
n
σ)
−(
− 3 −a√
n
σ)≈(
3 −a√
n
σ)and so the probability that it falls outside is approximately 1− (3−a
√
n/σ). For
instance, if the subgroup size isn=4, then an increase in the mean value of 1 standard
deviation — that is, a =σ— will result in the subgroup average falling outside of the
control limits with probability 1− (1)=.1587. Because each subgroup average will
independently fall outside the control limits with probability 1− (3−a
√
n/σ), it follows
that the number of subgroups that will be needed to detect this shift has a geometric
distribution with mean{ 1 − (3−a
√
n/σ)}−^1. (In the case mentioned before with
n=4, the number of subgroups one would have to chart to detect a change in the mean
of 1 standard deviation has a geometric distribution with mean 6.3.)
13.2.1 Case of Unknownμandσ
If one is just starting up a control chart and does not have reliable historical data, thenμ
andσwould not be known and would have to be estimated. To do so, we employkof the
subgroups wherekshould be chosen so thatk≥20 andnk≥100. IfXi,i=1,...,kis
the average of theith subgroup, then it is natural to estimateμbyXthe average of these
subgroup averages. That is,
X=X 1 +···+Xk
k