Download Sampling Distributions Binomial Distribution

274
Chap te rEi g h t
Sampling Distributions
In most Six Sigma projects involving enumerative statistics, we deal with samples, not
populations. We now consider the estimation of certain characteristics or parameters of
the distribution from the data.
The empirical distribution assigns the probability lin to each X i in the sample, thus
the mean of this distribution is
(8.54)
The symbol X is called "Xbar." Since the empirical distribution is determined by
a sample, X is simply called the sample mean.
The sample variance is given by
(8.55)
This equation for S2 is commonly referred to as the unbiased sample variance. The
sample standard deviation is given by
(8.56)
Another sampling statistic of special interest in Six Sigma is the standard deviation
of the sample average, also referred to as the standard error of the mean or simply the
standard error. This statistic is given by
(8.57)
As can be seen, the standard error of the mean is inversely proportional to the
square root of the sample size. That is, the larger the sample size, the smaller the standard deviation of the sample average. This relationship is shown in Fig. 8.34. It can be
seen that averages of n = 4 have a distribution half as variable as the population from
which the samples are drawn.
Binomial Distribution
Assume that a process is producing some proportion of nonconforming units, which
we will call p. If we are basing p on a sample we find p by dividing the number of nonconforming units in the sample by the number of items sampled. The equation that
will tell us the probability of getting x defectives in a sample of n units is shown by
Eq. (8.58) .
(8.58)
This equation is known as the binomial probability distribution. In addition to being
useful as the exact distribution of nonconforming units for processes in continuous
production, it is also an excellent approximation to the cumbersome hypergeometric
probability distribution when the sample size is less than 10% of the lot size.
Pro c es s Behay i 0 r Ch art s
Distribution of
Population
distribution - - - - I
FIGURE
8.34 Effect of sample size on the standard error.
Example of Applying the Binomial Probability Distribution
A process is producing glass bottles on a continuous basis. Past history shows that 1%
of the bottles have one or more flaws. If we draw a sample of 10 units from the process,
what is the probability that there will be 0 nonconforming bottles?
Using the above information, n = 10, P= .01, and x = O. Substituting these values into
Eq. (8.58) gives us
p(O) = qOO.01 °(1- 0.01)10-0 = 1 x 1 X 0.99 10
= 0.904 = 90.4%
Another way of interpreting the above example is that a sampling plan "inspect 10
units, accept the process if no nonconformances are found" has a 90.4% probability of
accepting a process that is averaging 1% nonconforming units.
Example of Binomial Probability Calculations Using Microsoft Excel"
Microsoft Excel has a built-in capability to analyze binomial probabilities. To solve the
above problem using Excel, enter the sample size, p value, and x value as shown in Fig. 8.35.
Note the formula result near the bottom of the screen.
Poisson Distribution
Another situation encountered often in quality control is that we are not just concerned
with units that don't conform to requirements, instead we are concerned with the number of nonconformances themselves. For example, let's say we are trying to control the
quality of a computer. A complete audit of the finished computer would almost certainly reveal some nonconformances, even though these nonconformances might be of
minor importance (for example, a decal on the back panel might not be perfectly
straight). If we tried to use the hypergeometric or binomial probability distributions to
evaluate sampling plans for this situation, we would find they didn't work because our
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Chap te rEi g h t
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Example of finding binomial probability using Microsoft Excel.
lot or process would be composed of 100% nonconforming units. Obviously, we are
interested not in the units per se, but in the non-conformances themselves. In other cases,
it isn't even possible to count sample units per se. For example, the number of accidents
must be counted as occurrences. The correct probability distribution for evaluating
counts of non-conformances is the Poisson distribution. The pdf is given in Eq. (8.59).
p(x) = 11 x:;~
(8.59)
In Eq. (8.59), 11 is the average number of nonconformances per unit, x is the number
of nonconformances in the sample, and e is the constant approximately equal to
2.7182818. P(x) gives the probability of exactly x occurrences in the sample.
Example of Applying the Poisson Distribution
A production line is producing guided missiles. When each missile is completed, an
audit is conducted by an Air Force representative and every nonconformance to requirements is noted. Even though any major nonconformance is cause for rejection, the prime
contractor wants to control minor nonconformances as well. Such minor problems as
blurred stencils, small burrs, etc., are recorded during the audit. Past history shows that
on the average each missile has 3 minor nonconformances. What is the probability that
the next missile will have 0 nonconformances?
We have 11 = 3, x = O. Substituting these values into Eq. (8.59) gives us
0
3
p(0)=3 e- =lxO.05
O!
0.05=5%
1
In other words, 100% - 5% = 95% of the missiles will have at least one nonconformance.
Process Behavior Charts
The Poisson distribution, in addition to being the exact distribution for the number
of non-conformances, is also a good approximation to the binomial distribution in certain cases. To use the Poisson approximation, you simply let /.l = np in Eq. (8.59). Juran
(1988) recommends considering the Poisson approximation if the sample size is at least
16, the population size is at least 10 times the sample size, and the probability of occurrence p on each trial is less than 0.1. The major advantage of this approach is that it
allows you to use the tables of the Poisson distribution, such as in Appendix 7. Also, the
approach is useful for designing sampling plans.
Example of Poisson Probability Calculations Using Microsoft Excel
Microsoft Excel has a built-in capability to analyze Poisson probabilities. To solve the
above problem using Excel, enter the average and x values as shown in Fig. 8.36. Note
the formula result near the bottom of the screen.
Hypergeometric Distribution
Assume we have received a lot of 12 parts from a distributor. We need the parts badly
and are willing to accept the lot if it has fewer than 3 nonconforming parts. We decide
to inspect only 4 parts since we can't spare the time to check every part. Checking the
sample, we find 1 part that doesn't conform to the requirements. Should we reject the
remainder of the lot?
This situation involves sampling without replacement. We draw a unit from the lot,
inspect it, and draw another unit from the lot. Furthermore, the lot is quite small, the
sample is 25% of the entire lot. The formula needed to compute probabilities for this
POISSON
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= 0.0497870158
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- 13 l
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14
the Poisson probab llit:,t mass function, use F.il.LSE
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.J 6 ! - - - - - - - - - - - - - - - - - - - - - - - - -.. . . . . . .
1
FIGURE
8.36
Example of finding Poisson probability using Microsoft Excel.
I
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Chapter Eight
procedure is known as the hypergeometric probability distribution, and it is shown in
Eq. (8.60).
(8.60)
In Eq. (8.60), N is the lot size, m is the number of defectives in the lot, n is the sample
size, x is the number of defectives in the sample, and P(x) is the probability of getting
exactly x defectives in the sample. Note that the numerator term c::-~m gives the number of combinations of non-defectives while C;Z is the number of combinations of defectives. Thus the numerator gives the total number of arrangements of samples from lots
of size N with m defectives where the sample n contains exactly x defectives. The term
C~ the denominator is the total number of combinations of samples of size n from lots
of size N, regardless of the number of defectives. Thus, the probability is a ratio of the
likelihood of getting the result under the assumed conditions.
For our example, we must solve the above equation for x = 0 as well as x = I, since
we would also accept the lot if we had no defectives. The solution is shown as follows.
P(O)=
P(l)=
C 12 - 3C 3
q2
4-0
0
C12 -3C 3
4- 1
1
q2
126 x 1
=--=0255
495
.
84 x 3 252
=--=-=0509
495
495
.
P(l or less) = P(O) + P(l)
Adding the two probabilities tells uOOOOs the probability that our sampling plan
will accept lots of 12 with 3 nonconforming units. The plan of inspecting 4 parts and
accepting the lot if we have 0 or 1 nonconforming has a probability of 0.255 + 0.509 =
0.764, or 76.4%, of accepting this "bad" quality lot. This is the "consumer 's risk" for this
sampling plan. Such a high sampling risk would be unacceptable to most people.
Example of Hypergeometric Probability Calculations Using Microsoft Excel
Microsoft Excel has a built-in capability to analyze hypergeometric probabilities. To
solve the above problem using Excel, enter the population and sample values as shown
in Fig. 8.37. Note the formula result near the bottom of the screen (0.509) gives the probability for x = 1. To find the cumulative probability you need to sum the probabilities for
x = 0 and x = 1 etc.
Normal Distribution
The most common continuous distribution encountered in Six Sigma work is, by far, the
normal distribution. Sometimes the process itself produces an approximately normal
distribution, other times a normal distribution can be obtained by performing a mathematical transformation on the data or by using averages. The probability density function for the normal distribution is given by Eq. (8.61).
(8.61)