Post on 16-Jan-2016
1
SMUEMIS 7364
NTUTO-570-N
Inferences About Process QualityUpdated: 2/3/04
Statistical Quality ControlDr. Jerrell T. Stracener, SAE Fellow
2
Inferences about Process Quality
• Sampling & Sampling Distributions
• Inferences Based on Single Random Sample
• Inferences Based on Two Random Samples
• Inferences Based on More than Two Random Samples
3
Sampling & Sampling Distributions
4
Population vs. Sample
• Populationthe total of all possible values (measurement, counts, etc.) of a particular characteristic for aspecific group of objects.
• Samplea part of a population selected according tosome rule or plan.
Why sample?
5
Sampling
Characteristics that distinguish one type of sample from another:
• the manner in which the sample was obtained
• the purpose for which the sample was obtained
6
Simple Random Sample
The sample X1, X2, ... ,Xn is a random sample if X1, X2, ... , Xn are independent identically distributed random variables.
Remark: Each value in the population has an equal and independent chance of being included in the sample.
7
Generating Random Samplesusing Monte Carlo Simulation
8
Generating Random Numbers
f(y)
F(y)y
y
1.00.80.60.40.2 0
ri
yi
9
Generating Random Numbers
Generating values of a random variable using theprobability integral transformation to generate arandom value y from a given probability densityfunction f(y):
1. Generate a random value rU from a uniformdistribution over (0, 1).
2. Set rU = F(y)
3. Solve the resulting expression for y.
10
Generating Random Numbers with Excel
From the Tools menu, look for Data Analysis.
11
Generating Random Numbers with Excel
If it is not there, you must install it.
12
Generating Random Numbers with Excel
Once you select Data Analysis, the following window will appear. Scroll down to “Random Number Generation” and select it, then press “OK”
13
Generating Random Numbers with Excel
Choose which distribution you would like. Use uniform for an exponential or weibull distribution or normal for a normal or lognormal distribution
14
Generating Random Numbers with Excel
Uniform Distribution, U(0, 1). Select “Uniform” under the “Distribution” menu.Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.
15
Generating Random Numbers with Excel
Normal Distribution, N(, ). Select “Normal” under the “Distribution” menu.Type in “1” for number of variables and 10 for number of random numbers. Enter the values for the mean () and standard deviation () then press OK. 10 random numbers of uniform distribution will now appear on a new chart.
16
Generating Random Values from an ExponentialDistribution E() with ExcelFirst generate n random variables, r1, r2, …, rn, from U(0, 1). Select “Uniform” under the “Distribution” menu.Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.
17
Generating Random Values from an ExponentialDistribution E() with ExcelSelect a that you would like to use, we will use = 5.
Type in the equation xi=-ln(1 - ri), with filling in as 5, and ri as cell A1 (=-5*LN(1-A1)). Now with that cell selected, place the cursor over the bottom right hand corner of the cell. A cross will appear, drag this cross down to B10. This will transfer that equation to the cells below. Now we have n random values from the exponential distribution with parameter =5 in cells B1 - B10.
18
Generating Random Values from an WeibullDistribution W(,) with ExcelFirst generate n random variables, r1, r2, …, rn, from U(0, 1). Select “Uniform” under the “Distribution” menu.Type in “1” for number of variables and 10 for number of random numbers. Then press OK. 10 random numbers of uniform distribution will now appear on a new chart.
19
Generating Random Values from an WeibullDistribution W(,) with Excel Select a and that you would like to use, we will use = 100, = 20.
Type in the equation xi = [-ln(1 - ri)]1/, with filling in as 100, as 20, and ri as cell A1 (=100*(-LN(1-A1))^(1/20)). Now transfer that equation to the cells below. Now we have n random variables from the Weibull distribution with parameters =100 and =20 in cells B1 - B10.
20
Generating Random Values from an LognormalDistribution LN(, ) with ExcelFirst generate n random variables, r1, r2, …, rn, from N(, ). Select “Normal” under the “Distribution” menu.Type in “1” for number of variables and 10 for number of random numbers. Enter 0 for the mean and 1 for standard deviation then press OK. 10 random numbers of uniform distribution will now appear on a new chart.
21
Generating Random Values from an LognormalDistribution LN(, ) with ExcelSelect a and that you would like to use, we will use = 2, = 1.
Type in the equation , with filling in as 2, as 1, and ri as cell A1 (=EXP(2+A1*1)). Now transfer that equation to the cells below. Now we have an Lognormal distribution in cells B1 - B10.
iri ex
22
Flow Chart of Monte Carlo Simulation method
Input 1: Statistical distribution for each component variable.
Input 2: Relationshipbetween componentvariables and systemperformance
Select a random value from each of these distributions
Calculate the value of system performance for a system composed of components with the values obtained in the previous step.
Output: Summarize and plot resultingvalues of system performance. Thisprovides an approximation of the distribution of system performance.
Repeatmanytimes
23
Distribution of Sample Mean
24
Sampling Distribution of X with known
If X1, X2, ... ,Xn is a random sample of size n from a normal distribution with mean andknown standard deviation ,
and if ,
then
n
σμ,N~X
0,1N~
n
σμX
Z
and
n
1iiX
n
1X
25
Central Limit Theorem
If X is the mean of a random sample of size n, X1, X2, …, Xn, from a population with mean and finite standard deviation , then if n the limiting distribution of
n
XZ
is the standard normal distribution.
26
Central Limit Theorem
Remark: The Central Limit Theorem provides the basis for approximating the distribution of X witha normal distribution with mean and standard deviation
The approximation gets better as n gets larger.
n
27
Sampling Distribution of X with Unknown
Let X1, X2, ..., Xn be independent random variables that have normal distribution with mean and unknown standard deviation . Let
and
n
1iiX
n
1X
Then the random variable
n
1i
2
i2 XX
1n
1S
n
SμX
T
has a t-distribution with = n - 1 degrees of freedom.
28
Distribution of Sample Standard Deviation
29
Sampling Distributions of S2
If S2 is the variance of a random sample of size n taken from a normal population having the variance 2, then the statistic
n
i
i XX
12
2
2
22 s 1n
has a chi-squared distribution with = n - 1 degrees of freedom.
30
Inferences Based on a Single Random Sample
31
Estimation - Binomial Distribution
Estimation of a Proportion, p
• X1, X2, …, Xn is a random sample of size n fromB(n, p)
• Point estimate of p:
where fs = # of successes
n
fP s^
32
Estimation - Binomial Distribution
• Approximate (1 - ) ·100% confidence intervalfor p:
where and
where ,
and is the value of the standard normal random variable Z such that
ppp^
'L ppp
^'U
n
qpZp
^^
2/
22/
zZP
2
Z
'U
'L p,p
33
Estimation of the Mean - Normal Distribution
• X1, X2, …, Xn is a random sample of size n from N(, ), where both & are unknown.
• Point Estimate of
• (1 - ) 100% Confidence Interval for the mean
where ,
and
n
1ii
^
XXn
1μ
UL μ,μ
ΔμXμL ΔμXμU
n
stΔμ
1n,2
α
34
Estimation of the Mean - Infinite Population- Type Unknown
• X1, X2, …, Xn is a random sample of size n
• Point Estimate of
• An approximate(1 - ) 100% Confidence Interval for the mean
based on the Central Limit Theorem
where
and
n
1ii
^
XXn
1μ
UL μ,μ
ΔμXμU ΔμXμL
n
stμ
1n,2
α
35
Estimation of Means - Finite Populations
• X1, X2, ... , Xn is a random sample of size n from a population of size N with unknown parameters and
• Point Estimate of :
• An approximate (1 - ) · 100% Confidence Interval for is, where
X^
'U
'L ,
^
'L x
^'U xand ,
where ,1N
nN
n
stΔμ
1n,2
α
36
Estimation of Means - Finite Populations
where
• is the value of T ~ tdf for which
• is the finite population correction factor
1n,2
t
2
tTP1n,
2
1N
nN
2n
1ii
2 TT1n
1S
37
Estimation of Lognormal Distribution
• Random sample of size n, X1, X2, ... , Xn from LN (, )
• Let Yi = ln Xi for i = 1, 2, ..., n
• Treat Y1, Y2, ... , Yn as a random sample from N(, )
• Estimate and using the Normal DistributionMethods
38
Estimation of Weibull Distribution
• Random sample of size n, T1, T2, …, Tn, from W(, ), where both & are unknown.
• Point estimates
• is the solution of g() = 0
where
•
^
β
^^β
1n
1i
βi
^
Tn
1θ
n
1iin
1i
βi
n
1ii
βi
lnTn
1
β
1
T
lnTTβg
39
Estimation of Standard Deviation - Normal Distribution
• Point Estimate of
• (1 - ) · 100% Confidence Interval for is, where
and
n
ii XX
n 1
2^ 1
UL ,
21,2/
)1(
nL x
ns
2
1,2/1
)1(
nU x
ns
n
ns
1
40
Testing Hypotheses
There are two possible decision errors associated with testing a statistical hypothesis:
A Type I error is made when a true hypothesis is rejected.A Type II error is made when a false hypothesis is accepted.
True SituationDecision H0 true H0 falseAccept H0 correct Type II errorReject H0 Type I error correct(Accept H1)
41
Testing Hypotheses
The decision risks are measured in terms of probability.
= P(Type I error)= P(reject H0|H0 is true)= Producers risk
= P(Type II error) = P(accept H0|H1 is true)= Consumers risk
Remark: 100% · is commonly referred to as the significance level of a test.
Note: For fixed n, increases as decreases, and vice versa, as n increases, both and decrease.
42
Power Function
Before applying a test procedure, i.e., a decision rule, we need to analyze its discriminating power,i.e., how good the test is. A function called the power function enables us to make this analysis.
Power Function = P(rejecting H0|true parameter value)
OC Function = P(accepting H0|true parameter value)= 1 - Power Function
where OC is Operating Characteristic.
43
Power Function
A plot of the power function vs the test parametervalue is called the power curve and 1 - power curveis the OC curve.
1
0
PR()
ideal power curve
H0 H1
44
Power Function
The power function of a statistical test of hypothesis is the probability of rejecting H0 as a function of the true value of the parameter being tested, say , i.e.,
PF() = PR()
= P(reject H0|)
= P(test statistic falls in CA|)
45
Operating Characteristic Function
The operating characteristic function of a statisticaltest of hypothesis is the probability of acceptingH0 as a function of the true value of the parameterbeing tested, say , i.e.,
OC() = PA()
= P(accept H0|)
= P(test statistic falls in CR|)
46
Tests of Proportions
Let X1, X2, . . ., Xn be a random sample of size nfrom B(n, p).
Case 1: small sample sizes
To test the Null HypothesisH0: p = p0, a specified value, against the
appropriate Alternative Hypothesis
1. HA: p < p0 ,or
2. HA: p > p0 ,or
3. HA: p p0 ,
47
Tests of Proportions
at the 100 · % Level of Significance, calculatethe value of the test statistic using X ~ B(n, p = p0).
Find the number of successes and compute the appropriate P-Value, depending upon the alternativehypothesis and reject H0 if P , where
1. P = P(X x|p = p0) ,or
2. P = P(X x|p = p0) ,or
3. P = 2P(X x|p = p0) if x < np0, or
P = 2P(X x|p = p0) if x > np0,
48
Tests of Proportions
Case 2: large sample sizes with p not extremelyclose to 0 or 1.
To test the Null HypothesisH0: p = p0, a specified value, against the
appropriate Alternative Hypothesis
1. HA: p < p0 ,or
2. HA: p > p0 ,or
3. HA: p p0 ,
49
Tests of Proportions
Calculate the value of the test statistic
and reject H0 if
1. ,or
2. ,or
3. or ,
depending on the alternative hypothesis.
00
0
qnp
npxZ
zz
zz
2
αzz 2
αzz
50
Test of Means
Let X1, …, Xn, be a random sample of size n, from a normal distribution with mean and standard deviation , both unknown.
To test the Null HypothesisH0: = 0 , a given or specified value
against the appropriate Alternative Hypothesis
1. HA: < 0 ,or
2. HA: > 0 ,or
3. HA: 0 ,
51
Test of Means
at the 100 % level of significance. Calculate the value of the test statistic
Reject H0 if
1. t < -t, n-1 ,
2. t > t, n-1 ,
3. t < -t/2, n-1 , or if t > t/2, n-1 ,
depending on the Alternative Hypothesis.
n
sX
t 0
52
Test of Variances
Let X1, …, Xn, be a random sample of size n, from a normal distribution with mean and standard deviation , both unknown.
To test the Null HypothesisH0: 2 = 2
0, a specified valueagainst the appropriate Alternative Hypothesis
1. HA: 2 < 20 ,
or2. HA: 2 > 2
0 ,or
3. HA: 2 20 ,
53
Test of Variances
at the 100 % level of significance. Calculate the value of the test statistic
Reject H0 if
1. 2 < 21-, n-1 ,
2. 2 > 2, n-1 ,
3. 2 < 21-/2, n-1 , or if 2 > 2
/2, n-1 ,
depending on the Alternative Hypothesis.
20
22 1
s
n
54
Inferences Based onTwo Random Samples
55
Estimation - Binomial Populations
Estimation of the difference between two proportions
• Let X11, X12, …, , and X21, X22, …, ,be random samples from B(n1, p1) and B(n2, p2) respectively
• Point estimation of p1 - p2
2
^
1
^^
ppp
21 XX
2
2
1
1
n
f
n
f
11X n 22X n
56
Estimation - Binomial Populations
• Approximate (1 - ) · 100% confidence intervalfor
where
and
UL pp ,
21 ppp
2
^
2
^
2
1
^
1
^
1
2
^
n
qp
n
qpZppL
2
^
2
^
2
1
^
1
^
1
2
^
n
qp
n
qpZppU
57
Estimation of Difference Between Two Means - Normal Distribution
• Let X11, X12, …, , and X21, X22, …, be random samples from N(1, 1) and N(2, 2),respectively, where , , and are all unknown
• Point estimation of = 1 - 2
2
^
1
^^
μμμΔ
22X n11X n
21 XX
58
Estimation of Difference Between Two Means - Normal Distribution
• An approximate (1 - ) · 100% Confidence Interval for = 1 - 2
where
''
, UL
2
22
1
21
,2
^'
n
s
n
stL
2
22
1
21
,2
^'
n
s
n
stU
59
Estimation of Difference Between Two Means - Normal Distribution
where = degrees of freedom
11 2
2
2
22
1
2
1
21
2
2
22
1
21
n
ns
n
ns
ns
ns
60
Estimation of Ratio of Two Standard Deviations - Normal Distribution
• Let X11, X12, …, , and X21, X22, …, be random samples from n(1, 1) and n(2, 2),respectively
• Point estimation of
where
for i = 1, 2
22X n11X n
2
1
r
2
1^
s
sr
in
j
iiji XXn
s1
2
1
1
61
Estimation of Ratio of Two Standard Deviations - Normal Distribution
• (1 - ) · 100% Confidence Interval for
where
and
ULrr ,
21
2
α2
1σ υ,υ,F
1
s
sr
L
2
1
r
12
2
α2
1σ υ,υ,F
s
sr
U
62
Estimation of Ratio of Two Standard Deviations - Normal Distribution
where is the value of the F-Distribution with
and degrees of freedom for which
21
2
,, F
111 n 122 n
2,, 21
2
FFP
63
Test on Two Means
Let X11, X12, …, X1n1 be a random sample of size n1 from N(1,
1) and X21, X22, …, X2n2 be a random sample of size n2
from N(2, 2), where 1, 1, 2 and 2 are all unknown.
To test
H0: 1 - 2 = do, where do 0,
against the appropriate alternative hypothesis
64
Test on Two Means
1. H1: 1 - 2 < do, where do 0,
or2. H1: 1 - 2 > do, where do 0,
or3. H1: 1 - 2 do, where do 0,
at the 100% level of significance, calculate the value of the test statistic.
2
22
1
21
021
ns
ns
dXXt'
65
Test on Two Means
Reject Ho if
1. t' < t
or 2. t' > t
or 3. t' < t or t' > t depending on the
alternative hypothesis.
1
ns
1
ns
ns
ns
2
2
2
22
1
2
1
21
2
2
22
1
21
nn
66
Test on Two Variances
Let X11, X12, …, X1n1 be a random sample of size n1 from N(1,
1) and X21, X22, …, X2n2 be a random sample of size n2
from N(2, 2), where 1, 1, 2 and 2 are all unknown.
To test
H0:
against the appropriate alternative hypothesis
22
21 σσ
67
Test on Two Variances
1. H1:
or2. H1:
or3. H1:
at the 100% level of significance, calculate the value of the test statistic.
σσ 21
21
21
21 σσ
21
21 σσ
22
21
S
SF
68
Test on Two Variances
Reject Ho if
or
or
depending on the alternative hypothesis.
),(FF 211 vv
)(FF 21α ,vv
),(FFor ),(FF 212/212/1 vvvv
69
Inferences Based onMore than Two Random Samples
70
Normal Distribution - Estimation of
X1, X2, …, Xn is a random sample of size n from N(, ), where both & are unknown.
• Point Estimate of
• (1 - )·100% Confidence Interval for is ,
where
and
n
1ii
^
XXn
1μ
UL μ,μ
ΔμXμL
ΔμXμU
71
Normal Distribution - Estimation of
where is the value of the t-distribution with
parameter = n-1
which P(T> ) = /2
and may be obtained from the table t-distribution (Located in the resource section on the website).
n
stΔμ
1n,2
α
1n,2
αt
1n,2
αt
72
Estimation of Lognormal Distribution
• Random sample of size n, X1, X2, ... , Xn from LN (, )
• Let Yi = ln Xi for i = 1, 2, ..., n
• Treat Y1, Y2, ... , Yn as a random sample from N(, )
• Estimate and using the Normal DistributionMethods
73
Estimation of Weibull Distribution
• Random sample of size n, T1, T2, …, Tn, from W(, ), where both & are unknown.
• Point estimates
• is the solution of g() = 0
where
•
^
β
^^β
1n
1i
βi
^
Tn
1θ
n
1iin
1i
βi
n
1ii
βi
lnTn
1
β
1
T
lnTTβg