1 SMU EMIS 7364 NTU TO-570-N Inferences About Process Quality Updated: 2/3/04 Statistical Quality...

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SMUEMIS 7364

NTUTO-570-N

Inferences About Process QualityUpdated: 2/3/04

Statistical Quality ControlDr. Jerrell T. Stracener, SAE Fellow

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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

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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

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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


If it is not there, you must install it.

12


Once you select Data Analysis, the following window will appear. Scroll down to “Random Number Generation” and select it, then press “OK”

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Choose which distribution you would like. Use uniform for an exponential or weibull distribution or normal for a normal or lognormal distribution

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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.

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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.

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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.

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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.

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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.

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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.

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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.

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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

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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

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Distribution of Sample Mean

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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

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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.

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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

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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.

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Distribution of Sample Standard Deviation

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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.

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Inferences Based on a Single Random Sample

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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^

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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

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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

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Estimation of the Mean - Infinite Population- Type Unknown

• X1, X2, …, Xn is a random sample of size n


• 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

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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

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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

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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

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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

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Estimation of Standard Deviation - Normal Distribution


• (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

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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)

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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.

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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.

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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

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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|)

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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|)

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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 ,

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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,

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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 ,

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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

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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 ,

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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

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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 ,

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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

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Inferences Based onTwo Random Samples

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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

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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

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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

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• 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

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where = degrees of freedom

11 2

2

2

22

1

2

1

21

2

2

22

1

21

n

ns

n

ns

ns

ns

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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

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• (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

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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

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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

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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'

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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

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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

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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

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Reject Ho if

or

or

depending on the alternative hypothesis.

),(FF 211 vv

)(FF 21α ,vv

),(FFor ),(FF 212/212/1 vvvv

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Inferences Based onMore than Two Random Samples

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Normal Distribution - Estimation of

X1, X2, …, Xn is a random sample of size n from N(, ), where both & are unknown.


• (1 - )·100% Confidence Interval for is ,

where

and

n

1ii

^

XXn

1μ

UL μ,μ

ΔμXμL

ΔμXμU

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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

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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

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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

1 SMU EMIS 7364 NTU TO-570-N Inferences About Process Quality Updated: 2/3/04 Statistical Quality...

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Transcript of 1 SMU EMIS 7364 NTU TO-570-N Inferences About Process Quality Updated: 2/3/04 Statistical Quality...