Introduction to Statistical Quality Control, 4th Edition Chapter 3 Inferences About Process Quality

Introduction to Statistical Quality Control, 4th Edition

Chapter 3

Inferences About Process Quality


3-1. Statistics and Sampling Distributions

• Statistical methods are used to make decisions about a process– Is the process out of control? – Is the process average you were given the true

value?– What is the true process variability?


3-1. Statistics and Sampling Distributions

• Statistics are quantities calculated from a random sample taken from a population of interest.

• The probability distribution of a statistic is called a sampling distribution.


3-1.1 Sampling from a Normal Distribution

• Let X represent measurements taken from a normal distribution. X

• Select a sample of size n, at random, and calculate the sample mean,

• Then

~ ( , )N 2

x

x

n,N~

2



• Probability Example– The life of an automotive battery is normally

distributed with mean 900 days and standard deviation 35 days. What is the probability that a random sample of 25 batteries will have an average life of more than 1000 days?



• Chi-square (2) Distribution– If x1, x2, …, xn are normally and independently

distributed random variables with mean zero and variance one, then the random variable

is distributed as chi-square with n degrees of freedom

2n

22

21 x...xxy



• Chi-square (2) Distribution– Furthermore, the sampling distribution of

is chi-square with n – 1 degrees of freedom when sampling from a normal population.

2

2

2

n

1i

2i S)1n()xx(

y



– Chi-square (2) Distribution for various degrees of freedom.



• t-distribution– If x is a standard normal random variable and if

y is a chi-square random variable with k degrees of freedom, then

is distributed as t with k degrees of freedom.ky

xt


3-1.1 Sampling from a Normal Distribution• F-distribution

– If w and y are two independent chi-square random variables with u and v degrees of freedom, respectively, then

is distributed as F with u numerator degrees of freedom and v denominator degrees of freedom.

v/y

u/wF


3-1.2 Sampling from a Bernoulli Distribution• A random variable, x, with probability

function

is called a Bernoulli random variable.• The sumsum of a sample from a Bernoulli

process has a binomial distribution with parameters n and p.

0xq)p1(

1xp)x(p


3-1.2 Sampling from a Bernoulli Distribution

• x1, x2, …, xn taken from a Bernoulli process

• The sample mean is a discrete random variable given by

• The mean and variance of are

n

1iix

n

1x

x

n

)p1(p

p

2x

x


3-1.3 Sampling from a Poisson Distribution• Consider a random sample of size n, x1, x2, …, xn,

taken from a Poisson process with parameter

• The sum, x = x1 + x2 + … + xn is also Poisson with parameter n.

• The sample mean is a discrete random variable given by

• The mean and variance of are

n

1iix

n

1x

xn

, 2xx


3-2. Point Estimation of Process Parameters• Parameters are values representing the

population. Ex) The population mean and variance, respectively.

• Parameters in reality are often unknown and must be estimated.

• Statistics are estimates of parameters. Ex) The sample mean and sample

variance, respectively.

2,

2S,x


3-2. Point Estimation of Process Parameters

Two properties of good point estimators

1. The point estimator should be unbiased.

2. The point estimator should have minimum variance.


3-3. Statistical Inference for a Single Sample

Two categories of statistical inference:

1. Parameter Estimation

2. Hypothesis Testing



• A statistical hypothesis is a statement about the values of the parameters of a probability distribution.

01

00

:H

:H


3-3. Statistical Inference for a Single Sample• Steps in Hypothesis Testing

– Identify the parameter of interest

– State the null hypothesis, H0 and alternative hypotheses, H1.

– Choose a significance level

– State the appropriate test statistic

– State the rejection region

– Compare the value of test statistic to the rejection region. Can the null hypothesis be rejected?


3-3. Statistical Inference for a Single Sample• Example: An automobile manufacturer claims a

particular automobile can average 35 mpg (highway).

– Suppose we are interested in testing this claim. We will sample 25 of these particular autos and under identical conditions calculate the average mpg for this sample.

– Before actually collecting the data, we decide that if we get a sample average less than 33 mpg or more than 37 mpg, we will reject the makers claim. (Critical Values)


3-3. Statistical Inference for a Single Sample• Example (continued)

– H0:

H1:

• From the sample of 25 cars, the average mpg was found to be 31.5. What is your conclusion?

3535

33 35 37

x

RejectReject

Do not reject

Rejection Regions


3-3. Statistical Inference for a Single SampleChoice of Critical Values• How are the critical values chosen?• Wouldn’t it be easier to decide “how much room

for error you will allow” instead of finding the exact critical values for every problem you encounter?

OR• Wouldn’t be easier to set the size of the rejection

region, rather than setting the critical values for every problem?


3-3. Statistical Inference for a Single SampleSignificance Level• The level of significance, determines the size

of the rejection region.• The level of significance is a probability. It is

also known as the probability of a “Type I error” (want this to be small)

• Type I error - rejecting the null hypothesis when it is true.

• How small? Usually want 10.0


3-3. Statistical Inference for a Single SampleTypes of Error• Type I error - rejecting the null hypothesis when

it is true.• Pr(Type I error) = . Sometimes called the

producer’s risk.• Type II error - not rejecting the null hypothesis

when it is false.• Pr(Type II error) = . Sometimes called the

consumer’s risk.


3-3. Statistical Inference for a Single SampleAn Engine Explodes

H0: An automobile engine explodes when started.

H1: An automobile engine does not explode when started.

Which error would you take action to avoid? Whose risk is higher, the producer’s or the consumer’s?



Power of a Test

• The Power of a test of hypothesis is given by 1 -

• That is, 1 - is the probability of correctly rejecting the null hypothesis, or the probability of rejecting the null hypothesis when the alternative is true.


3-3.1 Inference on the Mean of a Population, Variance KnownHypothesis Testing

• Hypotheses: H0: H1:

• Test Statistic:

• Significance Level, • Rejection Region:

• If Z0 falls into either of the two regions above, reject H0

o o

n/

xZ 0

0

2/02/o ZZorZZ


3-3.1 Inference on the Mean of a Population, Variance KnownExample 3-1• Hypotheses: H0: H1:

• Test Statistic:

• Significance Level, = 0.05• Rejection Region: • Since 3.50 > 1.645, reject H0 and conclude that the

lot mean pressure strength exceeds 175 psi.

175 175

50.325/10

175182Z0

645.1ZZ0


3-3.1 Inference on the Mean of a Population, Variance KnownConfidence Intervals• A general 100(1- )% two-sided confidence

interval on the true population mean, is

• 100(1- )% One-sided confidence intervals are:

Upper Lower

)1(]UL[P

)1(]L[P)1(]U[P


3-3.1 Inference on the Mean of a Population, Variance Known

Confidence Interval on the Mean with Variance Known

• Two-Sided:

• See the text for one-sided confidence intervals.

)1(]n

Zxn

Zx[P22


3-3.1 Inference on the Mean of a Population, Variance KnownExample 3-2• Reconsider Example 3-1. Suppose a 95% two-sided

confidence interval is specified. Using Equation (3-28) we compute

• Our estimate of the mean bursting strength is 182 psi 3.92 psi with 95% confidence

92.18508.17825

1096.1182

25

1096.1182

nzx

nzx 2/2/


3-3.2 The Use of P-Values in Hypothesis Testing

• If it is not enough to know if your test statistic, Z0 falls into a rejection region, then a measure of just how significant your test statistic is can be computed - P-value.

• P-values are probabilities associated with the test statistic, Z0.



Definition• The P-value is the smallest level of significance

that would lead to rejection of the null hypothesis H0.



Example• Reconsider Example 3-1. The test statistic was

calculated to be Z0 = 3.50 for a right-tailed hypothesis test. The P-value for this problem is thenP = 1 - (3.50) = 0.00023

• Thus, H0: = 175 would be rejected at any level of significance P = 0.00023


3-3.3 Inference on the Mean of a Population, Variance Unknown

Hypothesis Testing

• Hypotheses: H0: H1:• Test Statistic:

• Significance Level, • Rejection Region: • Reject H0 if

o o

n/s

xt 0

0

1n,2/0 tt 1n,2/0 tt


3-3.3 Inference on the Mean of a Population, Variance Unknown

Confidence Interval on the Mean with Variance Unknown

• Two-Sided:

• See the text for the one-sided confidence intervals.

)1(n

stx

n

stxP 1n,2/1n,2/


3-3.3 Inference on the Mean of a Population, Variance UnknownComputer Output

Table 3-2. Minitab Output for Example 3-3 Welcome to Minitab, press F1 for help. One-Sample T: Strength Test of mu = 50 vs mu not = 50 Variable N Mean StDev SE Mean Strength 16 49.864 1.661 0.415 Variable 95.0% CI T P Strength (48.979, 50.750) -0.33 0.749


3-3.4 Inference on the Variance of a Normal Distribution

Hypothesis Testing

• Hypotheses: H0: H1:

• Test Statistic:

• Significance Level,

• Rejection Region:

20

2 20

2

20

220

S)1n(

2

1n,2

1

20

2

1n,2

20 or


3-3.4 Inference on the Variance of a Normal Distribution

Confidence Interval on the Variance

• Two-Sided:


1s)1n(s)1n(

P2

1n,2/1

22

21n,2/

2


3-3.5 Inference on a Population ProportionHypothesis Testing

• Hypotheses: H0: p = p0 H1: p p0

• Test Statistic:


0

00

0

0

00

0

0

npx)p1(np

np)5.0x(

npx)p1(np

np)5.0x(

Z

2/0 ZZ


3-3.5 Inference on a Population Proportion

Confidence Interval on the Population Proportion

• Two-Sided:


1

n

)p̂1(p̂Zp̂p

n

)p̂1(p̂Zp̂P 2/2/


3-3.6 The Probability of Type II Error

Calculation of P(Type II Error)

• Assume the test of interest is H0: H1:

• P(Type II Error) is found to be

• The Power of the test is then 1 -

o o

nZ

nZ

22



Operating Characteristic (OC) Curves

• Operating Characteristic (OC) curve is a graph representing the relationship between , , and n.

• OC curves are useful in determining how large a sample is required to detect a specified difference with a particular probability.



Operating Characteristic (OC) Curves


3-3.7 Probability Plotting

• Probability plotting is a graphical method for determining whether sample data conform to a hypothesized distribution based on a subjective visual examination of the data.

• Probability plotting uses special graph paper known as probability paper. Probability paper is available for the normal, lognormal, and Weibull distributions among others.

• Can also use the computer.


3-3.7 Probability Plotting

Example 3-8

j x(j) (j – 0.5)/10 1 1176 0.05 2 1183 0.15 3 1185 0.25 4 1190 0.35 5 1191 0.45 6 1192 0.55 7 1201 0.65 8 1205 0.75 9 1214 0.85 10 1220 0.95

1150 1160 1170 1180 1190 1200 1210 1220 1230 1240

1

5

10

20

3040

506070

80

90

95

99

Data

Per

cent

Normal Probability Plot for Life

ML Estimates

Mean:

StDev:

1195.7

13.3120


3-4. Statistical Inference for Two Samples

• Previous section presented hypothesis testing and confidence intervals for a single population parameter.

• Results are extended to the case of two independent populations

• Statistical inference on the difference in population means, 21


3-4.1 Inference For a Difference in Means, Variances Known

Assumptions1. X11, X12, …, X1n1 is a random sample from population

1.

2. X21, X22, …, X2n2 is a random sample from population 2.

3. The two populations represented by X1 and X2 are independent

4. Both populations are normal, or if they are not normal, the conditions of the central limit theorem apply



• Point estimator for is

where

21 21 XX

212121 XEXEXXE

2

22

1

21

2121 nnXVXVXXV



Hypothesis Tests for a Difference in Means, Variances Known

Null Hypothesis:

Test Statistic:

0210 :H

2

22

1

21

0210

nn

XXZ



Hypothesis Tests for a Difference in Means, Variances Known

Alternative Hypotheses Rejection Criterion

0211 :H 2/02/0 zzorzz

0211 :H zz0

0211 :H zz0



Confidence Interval on a Difference in Means, Variances Known

100(1 - )% confidence interval on the difference in means is given by

2

22

1

21

2/21212

22

1

21

2/21 nnzxx

nnzxx


3-4.2 Inference For a Difference in Means, Variances Unknown

Hypothesis Tests for a Difference in Means,

Case I:

• Point estimator for is

where

222

21

21 21 XX

21

2

2

2

1

2

21 n

1

n

1

nnXXV




Case I:

The pooled estimate of , denoted by is defined by

222

21

2pS2

2nn

S1nS1nS

21

222

2112

p




Case I:

Null Hypothesis:

Test Statistic:

0210 :H

21p

0210

n1

n1

S

XXt

222

21



Hypothesis Tests for a Difference in Means, Variances Unknown

Alternative Hypotheses Rejection Criterion

0211 :H

2nn,2/0

2nn,2/0

21

21

tt

ortt

0211 :H

0211 :H

2nn,2/0 21tt

2nn,2/0 21tt




Case II:

Null Hypothesis:

Test Statistic:

0210 :H

2

22

1

21

0210

nS

nS

XXt

22

21




Case II:• The degrees of freedom for are given by

2

1nnS

1nnS

nS

nS

2

2

222

1

2

121

2

2

22

1

21

22

21

0t



Confidence Intervals on a Difference in Means, Case I:


21p2nn,2/2121

21p2nn,2/21 n

1

n

1stxx

n

1

n

1stxx

2121

222

21



Confidence Intervals on a Difference in Means, Case II:


2

22

1

21

,2/21212

22

1

21

,2/21 n

s

n

stxx

n

s

n

stxx

22

21


3-4.2 Paired Data

• Observations in an experiment are often paired to prevent extraneous factors from inflating the estimate of the variance.

• Difference is obtained on each pair of observations, dj = x1j – x2j, where j = 1, 2, …, n.

• Test the hypothesis that the mean of the difference, d, is zero.


3-4.2 Paired Data

• The differences, dj, represent the “new” set of data with the summary statistics:

1n

ddS

dn

1d

n

1j

2

j2d

n

1jj


3-4.2 Paired Data

Hypothesis Testing

• Hypotheses: H0: d = 0 H1: d 0

• Test Statistic:


• Rejection Region: |t0| t/2,n-1

nS

dt

d

0


3-4.3 Inferences on the Variances of Two Normal Distributions

Hypothesis Testing• Consider testing the hypothesis that the variances of two

independent normal distributions are equal.

• Assume random samples of sizes n1 and n2 are taken from populations 1 and 2, respectively

22

211

22

210

:H

:H



Hypothesis Testing

• Hypotheses:

• Test Statistic:


• Rejection Region:

22

21

0 S

SF

22

211

22

210 :H:H

1n,1n),2/1(01n,1n,2/0 2121FFFF



Alternative Test Rejection Hypothesis Statistic Region

22

211 :H

22

211 :H 1n,1n,0 21

FF

1n,1n,0 12FF

21

22

0 S

SF

22

21

0 S

SF



Confidence Intervals on Ratio of the Variances of Two Normal Distributions

100(1 - )% two-sided confidence interval on the ratio of variances is given by

1n,1n,2/22

21

22

21

1n,1n),2/1(22

21

1212F

S

SF

S

S


3-4.4 Inference on Two Population

ProportionsLarge-Sample Hypothesis Testing

• Hypotheses: H0: p1 = p2 H1: p1 p2

• Test Statistic:


2

22

1

11

21210

n)p1(p

n)p1(p

)pp(P̂P̂Z

2/0 ZZ



Proportions

Alternative Hypothesis Rejection Region

211 pp:H zz0

211 pp:H zz0



ProportionsConfidence Interval on the Difference in Two Population

Proportions

• Two-Sided:


2

22

1

112/21

212

22

1

112/21

n

)p1(p

n

)p1(pZP̂P̂

ppn

)p1(p

n

)p1(pZP̂P̂


3-5. What If We Have More Than Two Populations?

ExampleInvestigating the effect of one factor (with several levels) on some response. See Table 3-5

Hardwood ObservationsConcentration 1 2 3 4 5 6 Totals Avg

5% 7 8 15 11 9 10 60 10 10 12 17 13 18 19 15 94 15.67 15 14 18 19 17 16 18 102 17 20 19 25 22 23 18 20 127 21.17

Overall 383 15.96



Analysis of Variance

• Always a good practice to compare the levels of the factor using graphical methods such as boxplots.

• Comparative boxplots show the variability of the observations within a factor level and the variability between factor levels.



Figure 3-14 (a)

5 10 15 20

5

15

25

Hardwood Concentration (%)

Ten

sile

stren

gth

(psi

)



• The observations yij can be modeled by

a = number of factor levels

n = number of replicates (# of observations

per treatment (factor) level.)

n,...,2,1j

a,...,2,1iY ijiij



• The hypotheses being tested are

H0 :

H1 : for at least one i

• Total variability can be measured by the “total corrected sum of squares”:

1 2 0 ... a

0i

2a

1i

n

1j..ijT )yy(SS



• The sum of squares identity is

• Notationally, this is often written as

SST = SSTreatments + SSE

2a

1i

n

1j.iij

a

1i

2...i

2a

1i

n

1j..ij )yy()yy(n)yy(



• The expected value of the treatment sum of squares is

• If the null hypothesis is true, then

a

1i

2i

2Treatments n)1a()SS(E

2Treatments

1a

SSE



• The error mean square

• If the null hypothesis is true, the ratio

has an F-distribution with a – 1 and a(n – 1)

degrees of freedom.

)1n(a

ssMS E

E

E

Treatments

E

Treatments0 MS

MS

)]1n(a/[SS

)1a/(SSF



The following formulas can be used to calculate the sums of squares.

• Total Sum of Squares (SST):

• Sum of Squares for the Treatments (SSTreatment):

• Sum of Squares for error (SSE):

SSE = SST -SSTreatment

a

1i

n

1j

2..2

ijT an

yySS

a

1i

2..

2.i

Treatment an

y

n

ySS



• Analysis of Variance Table 3-7

Source of Variation

Sum of Squares

Degrees of Freedom

Mean Square

F0

Treatments SSTreatments a - 1 MSTreatments

E

Treatments

MS

MS

Error SSE a(n – 1) MSE Total SST an - 1



• Analysis of Variance Table 3-8

Analysis of Variance Source DF SS MS F P Factor 3 382.79 127.60 19.61 0.000 Error 20 130.17 6.51 Total 23 512.96 Individual 95% CIs For Mean Based on Pooled StDev Level N Mean StDev -----+---------+---------+---------+- 5 6 10.000 2.828 (---*---) 10 6 15.667 2.805 (---*----) 15 6 17.000 1.789 (---*---) 20 6 21.167 2.639 (---*----) -----+---------+---------+---------+- Pooled StDev = 2.551 10.0 15.0 20.0 25.0



Residual Analysis

• Assumptions: model errors are normally and independently distributed with equal variance.

• Check the assumptions by looking at residual plots.


3-5. What If We Have More Than Two Populations?• Residual Analysis• Plot of residuals versus factor levels

5 10 15 20

-4

-3

-2

-1

0

1

2

3

4

5

Percent Hardwood

Res

idua

l


3-5. What If We Have More Than Two Populations?• Residual Analysis• Normal probability plot of residuals

-4 -3 -2 -1 0 1 2 3 4 5

.001

.01

.05

.20

.50

.80

.95

.99

.999

Pro

babi

lity

Residuals

Normal Probability Plot

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Introduction to Statistical Quality Control, 4th Edition Chapter 3 Inferences About Process Quality