1. X-Chart: Control charts without subgroups but with single observations
6th Discussion Section(March, 2, 2010) Jihye Lee
So far, we’ve learned about control chart by using a subgroup.
Xbar, R and s chart are the control chart for subgroup.
We calculated average of number of subgroups under the Xbar chart.
However, there are constant numbers which can’t measure by using subgroups. (e.g. time, pressure and temperature). So, to monitor these individual or single observations, we are using different control chart. We call this as control chart for using individual observation or measurements.
There is X-chart will be the control chart by using single observations.
X-Chart: Control charts without subgroups
but with single observations
cf. Xbar Chart-control charts with subgroups
p (R book)

2. Now, we learn about detecting a mean shift
Now, we learn about detecting a mean shift. What I mean about mean shift. After you shift your mean from the your original mean, you will see difference between X-bar and X chart when you detect for shift mean.
Control limits for X- bar chart: µ ±3σx-bar(this is sub notation, I hope you know that)
Control limits for X chart: µ ±3σ(because we are dealing with single observation, so, we don’t need to x-bar here)
With this understanding of control limits, I would like to shift my mean to aσ.
And with this a shift in the process mean of aσ, I want to know the probability of
single observation plotting above the UCL and subgroup plotting above their UCL.
In other words, I’m asking you the probability that the next producing units will be defect.
Let’s see what I meant detecting a mean shift.
After mean shift of aσ, probability will be changed as well.
z(probability- standardization) = x- µ/σ
In the X-chart for detecting a mean shift(aσ), it will be changed as follows,
µ+3σ-(µ+aσ)/σ= 3-a
In the Xbar-chart for detecting a mean shift(aσ), it’ll be changed as follows,
µ ±3σx-bar-(µ+aσ)/σ= 3-a√n (because of the fact that σx-bar=σ/√n)

3. Let’s review by using a numerical example.
Imagine we have a computer production w/ a certain number of defects. Now the mean of the defect computers shifts
(rises to 2σ). How many computers would we produce before we detect the mean shift(signal)?
Formula(p.133):
z= µ+3σ-(µ+aσ)/σ=3-a(with a=change of mean=2)
We are looking for P(Z>3-a)=P(Z>3-2)=P(Z>1)
Look on p.52: statement 4: the area for z=a &subtract from .5
Look up the value for z=1(p.534-ztable)=.34134
= is the probability that we get a signal producing the next computer
1/p gives us the number of computers we expect to produce before we detect the mean shift
1/.15866=6.303=>we expect a signal at the 7th computer produced(detection of the mean shift)

4. Imagine we would have a subgroups instead of individual observation
Imagine we would have a subgroups instead of individual observation. Would we get a signal earlier or later?
Problem)
We now produce groups of 3 computers(subgroups)
Formula(p.134):
z= 3-a*√n (with n=subgroup size=3)
We are looking for P(Z>3-2*√3)=Z=-.465(rounded)
Look on p.52: statement 5: the area for z=-a &add it to .5
Look up the value for z=.465(p.534-ztable)= (find the memo)
= is the probability that we get a signal producing the next subgroup
1/p gives us the number of subgroups we expect to produce before we detect the mean shift(get a signal)
1/.67903=1.473=>we expect a signal in the 2nd subgroup produced (detection of the mean shift)

5. We get an earlier signal when we use subgroups than
single observation because the probability to detect the
mean shift in a group of computers is more likely than to
detect the mean shift while producing a single computer
Note!!
If we can choose between a X-Chart and Xbar Chart,
Which one we should use?
You should choose X-bar chart because X-bar chart is more sensitive and you see earlier signal(expect producing of a defect computer in the next) to detect in the mean shift.
Go back to the previous chart which mention about probability of an
individual observation plotting above an UCL and subgroup plotting
an above an UCL. Compare different between X-bar and X chart probabilities! You definitely know what I meant here by seeing the probabilities from the previous slides.

6. Constructing a X-Chart
Sample to construct the X-chart: 13 6 8 11 14
Control limits for X-chart: µ ^ ± 3σ^(pg. 134)
There are 2 ways estimate your sigma(σ^ - again, ^ stands for estimating )
σ^= MR bar/d2 and s/c4 (we know what is “s”(s.d) and d2 and c4(from tableE).
So, what is MR bar? MR is moving range. As noticed from the name, you’re moving range by subtract 2nd value from 1st, 3rd value from 2nd value, 4th value from 3rd value and so on and skip negative signs(-).
Okay, let’s find the control limits, before that we need to estimating σ^
by calculate MR bar/d2. Let’s do MR bar first.
Here, 13-6=7, 6-8=-2(->2), 8-11=-3(3) and 11-14=-3(3).
So, after you calculating of your MR and you average of them, that will be MR bar, thus, MR bar would be 3.75.
And, now we need to know d2. Here even though we have 5 observation, our moving average size was 2(I made pair with 2 observation, so MR should be 2, different from number of size) because we are dealing with moving average by pairing two numbers. So, find d2 with n=2. It’ll be 1.128

7. Constructing a X-Chart
Sample to construct the X-char: (continued from previous example) 13 6 8 11 14
Knowing that µ ^ ±3 σ^ and σ^ MR bar/d2 and s/c4
MR bar is 3.75.
d2 with n=2. It’ll be
Plug these numbers into the formula = 3.75/1.128=
And µ^ will be the average of the variables( /5=10.4)
Now, you’re ready to plug the numbers into the above formula.
Control limits for X chart = µ ^ ±3 σ^
=10.4 ± 3*3.3245
UCL =
LCL = 0.427
You can draw chart with these UCL and LCL, which shows all of the observation is in the inside of control limits.

8. Find UCL and LCL for Xchart!
Another example,
MR bar=18.63
d2 =1.128 (Table page 540 with n=2)
µ^ = 24.65
Find UCL and LCL for Xchart! 7 2 3 4 27 20 25 32 23 41 47 43

9. p and np-Chart
p Charts and np Charts can be used to measure the
Percentage(p chart) or number(np chart) of
nonconforming units(binomial distributed process-only
2 possible outcomes, conforming and nonconforming unit)
in a process with X=number of nonconforming units,
n=size of the sample and p^=X/n

10. p-Chart
The p-Chart shows how many percent of the units are
nonconforming. It is often used when sample sizes vary.
Control limits(p.155)=p±3√p*(1-p)/n (3* because of the 3σ limit)
X=18
N=36
P^=.5
CL=.5±3 √.5*.5/36=.5±.25=UCL=.75 and LCL=.25

11. Control limits(p.153): np±3√np*(1-p) (3 because of the 3σ limit)
np-Chart
Control limits(p.153): np±3√np*(1-p) (3 because of the 3σ limit)
36*.5± 3 √36*.5(1-.5)=18 ±3*3
UCL=27
LCL=9
With the formula for the binomial distribution we can prove whether
the distribution follows normal distribution. If the distribution follows
a normal distribution, we should get a result close to the
3σ limit area of

12. 6th Discussion Section(March, 2, 2010) Jihye Lee
Regression-based limits: The Regression-based calculation of the CL of np-Charts and p-Charts(again, n=36,p=.5) is a slightly modified 3σ methods and produces therefore slightly different limits: Control limits(pg 164) UCL= np+2.983*√np = LCL= np *√np =