Chapter 4 Inference About Process Quality

Report
Chapter 4 Inference About Process Quality
• Motivation
• Estimation
– point estimation
– interval estimation
• Hypothesis Testing
– Definition
– Testing on means
• known and Unknown variance
– Testing on Variance
The need of “Statistical Inference”
• In statistical quality control, the probability distribution is
used to model some quality characteristic (which is
related to process parameters).
• The parameters of a probability distribution are
2
unknown. N (,  ) ??
– Estimation of Process Parameters
– Point Estimation / Interval Estimation
• The parameters of a process can be time varying, how
do we identify a process change?
– Hypothesis Testing
Observations in a sample are used to
draw conclusions about the population
3
Random Samples
• Random Sample:
– Sampling from an infinite population or finite
population with replacement: A sample is selected so
that the observations are independently and
identically distributed.
– Sampling n samples from a finite population of N
N
items without replacement if each of the  n  possible
samples has an equal probability of being chosen
Random Sample = Independently and Identically Distributed
(i.i.d)
4
Terminology and Definition
• Statistic:
– Any function of the sample data that does not contain unknown
parameters.
• Estimate: a particular numerical value of an estimator, computed
from sample data. (An estimate is a particular statistic)
– Point estimate: a statistic that produces a single numerical
value as the estimate of the unknown parameter
– Interval estimate: a random interval (or called confidence
interval) in which the true value of the parameter falls with some
level of probability.
• Sampling distribution:
– The probability distribution of a statistic.
5
Point Estimation
Methods: [1]
• Method of Moment (MOM)
• Maximum Likelihood Estimation (MLE)
Distribution
Normal
Parameters

Estimator
ˆ  x
2
ˆ 2  S 2
ˆ  S / c4 (n>10) or ˆ  R / d 2 (n≤10); c4

Binomial
p
Poisson

and d2 are given in Appendix Table VI
1 n
pˆ   xi  x , {xi} are either 1 or 0,
n i 1
corresponding to “success” and “failure”
of the ith Bernoulli trial, respectively.
1 n
ˆ
   xi  x
n i 1
[1] “Statistical Inference”, George Casella and Roger L. Berger, 2nd edition
6
Interval Estimation
• Estimate the interval between two statistics that include the true
value of the parameter with some probability
Real mean of
– Example: Pr{ L   U}=1-
(0    1)
the population
– The interval L   U is called a 100(1- )% confidence interval (C.I.)
for the unknown mean 
– Two side C.I. (L is lower confidence limit, U is upper confidence limit)
– Single side C.I.:
• lower 100(1- )% C. I.: L  , Pr{ L  }=1-
• upper 100(1- )% C. I.:   U, Pr{   U}=1-
• Analysis procedures:
–
–
–
–
–
/2
/2
get the samples
L
U

compute the statistic
determine the statistic reference distribution
select confidence level
find the lower and/or upper confidence limits based on the reference
distribution
7
x

x1
x2
Q: how to
determine the
width?
U
L
Interval Estimation
If x is a random variable with unknown mean  and known
variance 2, what is the estimation interval for mean ?
n
– Select a statistic x  ( xi ) / n
i 1
2
– The approximate distribution of x is N (,  / n) regardless of the
distribution of x due to the central limit theorem.
– Given confidence level , then
• 100(1-)% two-side confidence interval on  is:
x  Z / 2


   x  Z / 2
n
n
where
Pr{z  Z / 2 }   / 2
• 100(1-)% upper confidence interval on  is:
  x  Z

n
• 100(1-)% lower confidence interval on  is:
x  Z


n
9
Example: The strength of a disposable plastic beverage
container is being investigated. The strengths are normally
distributed, with a known standard deviation of 15 psi. A sample
of 20 plastic containers has a mean strength of 246 psi.
Compute a 95% confidence interval for the process mean.
10
( z)  1   / 2
two size C.I. (90%)
( z)  1  
one side
C.I. (95%)
11
Example: A chemical process converts lead to gold. However, the
production varies due to the powers of the alchemist. It is known that
the process is normally distributed, with a standard deviation of 2.5 g.
How many samples must be taken to be 90% certain that an estimate
of the mean process is within 1.5 g of the true but unknown mean yield?
12
Interval Estimation of the Binomial Distribution
Parameter with A Larger Sample Size
• From the central limit theorem: p^ =x/n~ Normal (p, p(1-p) /n )
Example 5-1: A random sample of 200 printed circuit boards
contains 18 defective or nonconforming units. Estimate the process
fraction nonconforming. Construct a 90% two-sided confidence
interval on the true fraction nonconforming in the production
process.
13
Hypothesis Testing
•
•
•
•
Statistical hypothesis:
– a statement about the values of the parameters of a probability distribution
Hypothesis testing:
– Making a hypothesis concerning what we believe to be true and then use
sampled data to test it.
Two Hypotheses (Two Competing Propositions)
– Null Hypothesis H0: will be rejected if the sample data do not support it.
– Alternative Hypothesis H1: a hypothesis different from the null hypothesis
H 0 :   1.5
H 0 :   1.5
H 0 :   1.5
H1 :   1.5
H1 :   1.5
H1 :   1.5
Conclusion
– By Comparing the Test Statistic with Critical Value, determine whether
reject or NOT reject the null hypothesis.
14
Hypothesis Testing Procedures
1) State the null and alternative hypothesis, and
define the test statistic.
2) Specify the significance level .
3) Find the distribution of the test statistic and
the rejection region of H0.
4) Collect data and calculate the test statistic.
5) Compare the test statistic with the rejection
region.
6) Assess the risk.
15
Inference on the MEAN of a Normal
Population – Variance Known
Assuming Known Population Variance

x–
~ N(0,1)
/ n
x  Z / 2


   x  Z / 2
n
n


x
–


0
Reject H0:  0 if 
 > Z(/2)
/
n



x–
0
Reject H0: < 0 if
< -Z()
/ n

x – 0
Reject H0: > 0 if
> Z()
/ n
Critical value
Test Statistic: function of data and hypothetic value
16
Example: The response time of a distributed computer system is an
important quality characteristic. The system manager wants to know
whether the mean response time to a specific type of command
exceeds 75 millisec. From past experience, he knows that the
standard deviation of response time is 8 millisec. If the command is
executed 25 times and the response time for each trial is recorded.
The sample average response time is 79.25 millisec.
• Formulate an appropriate hypothesis and test the hypothesis.
• Find out the (lower/upper?) bound of the 95% C.I.
17
Inference on the MEAN of a Normal
Population – Variance Unknown
Assuming Unknown Population Variance

x–
~ t
(n-1)
s/ n
x  t / 2
s
s
   x  t / 2
n
n


x – 0
Reject H :   if 
 > t(/2, n-1)
0
0
s/
n



x–
0
Reject H0: < 0 if
<- t(n-1)
s/ n

x – 0
Reject H0: > 0 if
> t(n-1)
s/ n
H0 :    0
H1
18
19
20
Example: The mean time it takes a crew to restart an aluminum
rolling mill after a failure is of interest. The crew was observed
over 25 occasions, and the results were mean = 26.42 minutes
and variance S2 =12.28 minutes. If repair time is normally
distributed,
• Find a 95% confidence interval on the true but unknown mean
repair time.
• Test the hypothesis that the true mean repair time is 30 minutes.
21
Example 6-1: The life of a battery used in a cardiac pacemaker is
assumed to be normally distributed. A random sample of 10
batteries is subjected to an accelerated life test by running them
continuously at an elevated temperature until failure, and the
following lives are obtained.
25.5 h
26.1 h
26.8
23.2
24.2
28.4
25.0
27.8
27.3
25.7
• Construct a 90% two-sided confidence interval on mean life in the
accelerated test.
• Test the hypothesis, with =0.1 that the mean battery life is 26.5 h.
22
23
The Use of P-Values in Hypothesis Testing
1.
Traditional hypothesis testing:
– Given  to determine whether the null hypothesis was
rejected
– Disadvantage:
• No information on how close to/far away from the rejection region in a
probability sense
• predefined  may not reflect different decision maker’s risk assessments
2. P-Value
approach
– P-Value: the smallest level of significance that would lead to
rejection of the null hypothesis
– if the predefined >P= min, reject the null hypothesis
p _ value P(observingtest statisticsor extremevalue| H0 is true)
Underlying idea: “if H0 is really true, is it possible for test
statistic to be such big/small?”
0
x1
x2
Z0 
x  0

n
H0 :   0
Two-sided
p_value
 Z / 2
Z / 2
Use of P-Value for the Normal Distribution
H0: =0, standard normal statistic Z0~N(0,1)
– P=2[1-(|Z0|)] with two-sided H1, i.e., H1: 0
– P=1-(Z0) for one-sided H1, H1: >0
Z0
– P=(Z0) for one-sided H1, H1: <0
x  0

/ n
A small p-value is evidence against the null hypothesis while a
large p-value means little or no evidence against the null
f(x)
hypothesis
(Z0)
1(Z0)
Z0<0
=0
x
Z0>0
If p-value is small, it is less likely that the test statistic is small. So, H0 is
NOT true.
Inference on the MEAN of a Normal
Population – Variance Known
Assuming Known Population Variance

x–
~ N(0,1)
/ n
x  Z / 2


   x  Z / 2
n
n


x
–


0
Reject H0:  0 if 
 > Z(/2)
/
n



x–
0
Reject H0: < 0 if
< -Z()
/ n

x–
0
Reject H : >  if
> Z()
0
0
/ n
H0 :    0
H1
p_value

x  0
2 1  (
/ n

(

)

x  0
)
/ n
1 (
x  0
)
/ n
27
Example: The response time of a distributed computer system is an
important quality characteristic. The system manager wants to know
whether the mean response time to a specific type of command
exceeds 75 millisec. From past experience, he knows that the
standard deviation of response time is 8 millisec. If the command is
executed 25 times and the response time for each trial is recorded.
The sample average response time is 79.25 millisec. Formulate an
appropriate hypothesis and test the hypothesis.
• Calculate the p-value of the true mean response time is as low as 75
millisec.
28
Inference on the MEAN of a Normal
Population – Variance Unknown
Assuming Unknown Population Variance

x–
~ t
(n-1)
s/ n
x  t / 2
s
s
   x  t / 2
n
n
p_value
CDF


x
–


0
Reject H0:  0 if 
 > t(/2, n-1)
s/
n



x–
0
Reject H0: < 0 if
<- t(n-1)
s/ n

x – 0
Reject H : >  if
> t(n-1)
0
0
s/ n
H0 :    0
H1

x  0
2 1  t( n 1) (
s/ n

t n 1 (

)

x  0
)
s/ n
1  tn1 (
x  0
)
s/ n
29
Example: The mean time it takes a crew to restart an aluminum
rolling mill after a failure is of interest. The crew was observed
over 25 occasions, and the results were mean = 26.42 minutes
and variance S2 =12.28 minutes. If repair time is normally
distributed, find a 95% confidence interval on the true but
unknown mean repair time.
• Test the H0: μ = 30 v.s. μ ≠ 30.
• Calculate the p-value of the hypothesis that μ = 30.
30
Example: The life of a battery used in a cardiac pacemaker is
assumed to be normally distributed. A random sample of 10
batteries is subjected to an accelerated life test by running them
continuously at an elevated temperature until failure, and the
following lives are obtained.
25.5 h
26.1 h
26.8
23.2
24.2
28.4
25.0
27.8
27.3
25.7
Construct a 90% two-sided confidence interval on mean life in the
accelerated test. Test the H0: μ = 26.5 v.s. μ ≠ 26.5.
• Calculate the p-value of the hypothesis that μ = 26.5.
31
Confidence Interval v.s.
Hypothesis Testing
• If the value of the parameter specified by the null
hypothesis is contained in the 100(1- )% interval,
then the null hypothesis cannot be rejected at the
 level.
• If the value specified by the null hypothesis is not
in the interval, then the null hypothesis will be
rejected at the  level
32
Understanding the result of Hypothesis Test
• When we reject the null hypothesis, it is a strong
conclusion: there is a strong evidence that the null
hypothesis is false.
• When we fail to reject the null hypothesis, it is a
weak conclusion: It does not mean that the null
hypothesis is correct. It only means we do not have
strong evidence to reject it.
33
Court System and Hypothesis Testing
Hypothesis testing in science is a lot like the criminal court system in the
United States. How do we decide guilt?
• Assume innocence until “proven” guilty.
• Evidence is presented at a trial.
• Proof has to be “beyond a reasonable doubt.”
A jury's possible decision:
• guilty
• not guilty
Note that a jury cannot declare somebody ``innocent,'' just ``not
guilty.'' This is an important point.
34
Interrelationships between statistical
inferences
Confidence
Interval
Statistical
Distribution
Hypothesis
Testing
p_Value
Interrelationships between statistical
inferences


   x  Z / 2
Assuming Known Population
n Variance
n
x  Z / 2
Assuming Known Population Variance
x – 
~ N(0,1)
/ n
x – 
~ N(0,1)
/ n
x –   0|)] with
P=2[1-(|Z
two-sided
0 H1, i.e.,
Reject H0:  0 if H1: 0 > Z(/2)
 / n 
x – 
x –  

0
Reject H0:  0 if 
 > Z(/2)
 / n 
x – 
0
Reject H0: < 0 if
< -Z()
/ n
x – 
0
Inference on the Difference in Means
of Two Populations – Variance Known
Assume Known Population Variances Observations in TWO samples are all i.i.d.
_
_
x -x
1
2
1

n
1

2
2
n
H 0 : 1   2
2
~ N ( 0,1)
_
_
x -x
1
2
 Z / 2
1
2
n
1

2
n
2
_
_
 1   2  x1 - x 2  Z  / 2
2
1
2
n
1

2
n
p_value
2
H1
Reject H0 : 1  2 if
_
x1 - x 2  Z
 /2
 12  22
n1
_
Reject H 0 : 1   2 if
_
_

n2
1
2



n1 n2
2
1
2
2
x -x

1
2
2
1

 Z
n1
Reject H 0 : 1  2 if
_
x1 - x 2



n1 n2
2
1
2
2
 Z
2
2
)]
n2
 ( x1 x 2 )
 12  22

_
_

n
_ 1 _
_
x -x
2[1   (
_
n2
_
1   ( x1 x 2 )
 12
n1

 22
n2
37
2
2
Example: A bakery has a line making Binkies, a big-selling junk food.
Another line has just been installed, and the plant manager wants to
know if the output of the new line is greater than that of the old line, as
promised by the bakery equipment firm. 12 days of data are selected at
random from line 1 and 10 days of data are selected at random from line
–
–
2, with x = 1124.3 cases and x = 1138.7. It is known that  2= 52
1
2
1
and  2 = 60. Test the appropriate hypotheses at  = 0.05, given that
2
the outputs are normally distributed.
38
Inference on the Difference in Means
of Two Populations – Variance Unknown
a) Assume Homogeneity ( 
_
  12   22 )
i.i.d.
_
x -x
1
Sp
2
2
1 1

n1 n2
_
_
~ t (n1  n2  2) ; x1 - x 2  t / 2,n  n  2 S p
1
2

_
1
n n
1
_
 1   2  x1 - x 2  t / 2,n1  n2  2 S p
2
1
H1
Reject H0 : 1  2 if
_
1
x1 - x 2
1 1

n1 n2
Sp
_
Reject H 0 : 1   2 if
1
2
1 1

n1 n2
_
x -x
Sp
 t (, n1  n2  2)
_
1
Reject H 0 : 1  2 if
 t ( / 2, n1  n2  2)
2[1  t( n1  n2  2) (
_
x -x
Sp
p_value
_
2
1 1

n1 n2
 t (, n1  n2  2)
t( n1  n2 2) (
x1  x2
)]
1 1
Sp

n1 n2
x1  x2
)
1 1
Sp

n1 n2
1  t( n1  n2 2) (
x1  x2
)
1 1
Sp

n1 n2
39

1
n n
Equal variance
(n  1)s 2  (n  1)s 2
1
2
2
1
2
S

p
where
n1  n2  2
H 0 : 1   2
1
2
Inference on the Difference in Means
of Two Populations – Variance Unknown
b) Assume Heterogeneity (  1
2
_
_
x1 - x 2
2
1
2
2
s
s

n1 n2
H 0 : 1   2
  22 )
2
~ t (v )
(s / n  s / n )
v

where
(s / n ) (s / n )
2
1
1
2
1
n1  1
H1
Reject H0 : 1  2 if
_
x -x
1
Reject H 0 : 1  2 if
2
2
1
s
s22

n1 n2
2
2
n2  1
 t ( / 2, v)
_
x -x
1
2
2
1
s
s2
 2
n1 n2
_

2
2
_
_
Reject H 0 : 1   2 if
2
2
1
2
2
2
 t (  , v )
_
x1 - x 2
s12 s22

n1 n2
 t (, v )
40
Example: Two quality-control technicians measured the surface
finish of a metal part, obtaining the data shown below. Assume that
the measurements are normally distributed.
Technician 1
Technician 2
1.45
1.54
1.37
1.41
1.21
1.56
1.54
1.37
1.48
1.20
1.29
1.31
1.34
1.27
1.35
Assuming that the variances are equal, construct a 95% confidence
interval on the mean difference in surface-finish measurements. Also,
test H0: μ1= μ2 v.s. H1: μ1≠ μ2 and compute the p-value.
41
42
t ,
Text Book
Page P696
43
Review
• P-value: a probability value
p _ value P(observingtest statisticsor extremevalue| H0 is true)
• With the same α value, C.I., Hypothesis testing and p-value
give the same inferential conclusion.
• Inferences on mean of two populations: what are the
statistics used? What are the reference distributions? How
to define the reject region?
Inference on the Variance of a Normal
Distribution
(n  1)s 2
~ 2(n – 1)
2


2
1 / 2, n 1
H0 :   0 H1
(n - 1) s2
2
Reject H :   if
>

(/2,n - 1)
0
o
 2
o
(n - 1) s2
Reject H : <  if
<
2
0
o

o
(n - 1) s2
Reject H : >  if
>
2
0
o

o
C.I.

(n  1) s 2
2
 2 / 2, n 1
(n - 1) s2
2
<

or
(1 - /2,n - 1)
 2
o
2(1-,n - 1)
2(,n - 1)
(n  1)S 2
(n  1)S 2
2
  2
,
2
 / 2,n1
1 / 2,n1
Pr{2n1  2 / 2,n1}   / 2
45
Example Construct a 90% two-sided confidence interval on the
variance of battery life. Convert this into a corresponding
confidence interval on the standard deviation of battery life.
25.5 h
26.1 h
26.8
23.2
24.2
28.4
25.0
27.8
27.3
25.7
46
Text P695
47
Inference on the Variances of
Two Normal Distributions
S12 / 12
~ Fn1 1, n2 1 With H :  2 =  2
2
2
0
1
2
S2 / 2
for H :  2   2
1
1
2
Reject H
if
0
s 2
1
2
s
2
>F
(/2,n –1,n –1) or
1
2
s 2
1
2
s
2
< F
(1–/2,n –1,n –1)
1
2
for H :  2 <  2
1
1
2
Reject H
0
s 2
2
if
2
s
1
> F
(,n –1,n –1)
2
1
s 2
1
if
s 2
2
> F
(,n –1,n –1)
1
2
2
2
for H1: 1 > 2
Reject H
C.I.
0
S12
12 S12
F1 / 2,n2 1,n1 1  2  2 F / 2,n2 1,n1 1 ,
S22
2 S 2
F1 / 2,,  1 / F / 2,,
The two d.f.
are
exchanged
48
Example:Two quality-control technicians measured the surface
finish of a metal part, obtaining the data shown below. Assume that
the measurements are normally distributed.
b. Construct a 95% confidence interval estimate of the ratio of the variances of
technician measurement error.
c. Construct a 95% confidence interval on the variance of measurement error for
Technician 2.
Technician 1
1.45
1.37
1.21
1.54
1.48
1.29
1.34
Technician 2
1.54
1.41
1.56
1.37
1.20
1.31
1.27
1.35
49
50
Text P700
51
Testing on Binomial Parameters
•
•
To test whether the parameter p of a binomial distribution equals a standard
value p0
The test is based on the normal approximation to the binomial distribution
H 0 : p  p0
H1 : p  p0
 ( x  0.5)  np0
 np (1  p )

0
0
Z0  
( x  0.5)  np0


 np0 (1  p0 )
if
if
x  np0
x  np0
Or using the central limit theorem
H 0 : p1  p2
H1 : p1  p2
Z0 
pˆ1  pˆ 2
;
1 1
pˆ (1  pˆ )(  )
n1 n2
pˆ 
H0 is rejected if | Z0 | Z / 2
Z0 
x  p0
p0 (1  p0 ) / n
n1 pˆ1  n2 pˆ 2
if
n1  n2
p1  p2
Example 4.5,4.6 on p122
52
Test on Poisson Distribution
•
A random sample of n observation is taken, say x1, x2, ..,xn. Each
{xi} is Poisson distributed with parameter . Then the sum x= x1+ x2
+...+xn is Poisson distributed with parameter n.
•
If n is large, x =x/n is approximately normal with mean  and
variance /n
Test hypothesis
x  0
H0:  =0
Z0 
0 / n
H1:  0
The null hypothesis would be rejected if |Z0|>Z/2.
•
•
53
Two Types of Hypothesis Test Errors
•
Type I error ( producer’s risk):
–  = P{type I error} = P{reject H0 |H0 is true}
=P{conclude bad | although actually good}
•
Type II error (consumer’s risk):
–  = P{type II error} = P{fail to reject H0 |H0 is false}
=P{conclude good | although actually bad}
•
Power of the test:
– Power = 1-  = P{reject H0 |H0 is false}
H0:  = 0
H1:  = 1 0
with known 2
f(x)
1
/2
/2
LCL
x
0
UCL

54
Summary of Type I and Type II Errors
You Conclude
:
Reality
"H is True"
0
"H is True"
1
"H0 is True"
"H1 is True"
Co nfidence
1–
Co nsumer
Erro r, 
P r o duc er
Erro r, 
P o w er
1–
55
Properties of Type I & Type II Errors


Both types of errors can be reduced by
increasing the sample size at the price of
increased inspection costs.
For a given sample size, one risk can only
be reduced at the expense of increasing
the other risk.
56
The Probability of Type II Error
— Detection of a mean shift with a known 
•
Type II error= =Pr{H0 |H1 |}=Pr{within the control limits| mean shift}
H0:  = 0
1  0  , if
0
H1:  = 1 0 with known 2
  Pr{H 0 | H1}
 Pr{ 0  Z / 2  / n  x   0  Z / 2  / n | H1}
 ( Z / 2 
 n
 n
)  ( Z / 2 
)


57
OC curve with =0.05
•
•
The larger the mean shift, the smaller the type II error
The larger the sample size, the smaller the type II error
d |  | / 
58
OC Curves
OC curve see Fig. 4.7 P126
• The larger the mean shift, the smaller the type II error
• The larger the sample size, the smaller the type II error
n 1
n2
n3
n4
=0.05
59
OC Curves
OC curve see Fig. 4.7 P126
• The larger the mean shift, the smaller the type II error
• The larger the sample size, the smaller the type II error
n 1
n2
n3
n4
=0.05
60
Example: Suppose we wish to test the hypotheses
H0: =15
H1: 15
where we know that 2=9.0. If the true mean is really 20,
what sample size must be used to ensure that the probability
of type II error is no greater than 0.10? Assume that =0.05.
61
Use OC Curve
n=4
62
Example 7-4: The mean contents of coffee cans filled on a particular production
line are being studied. Standards specify that the mean contents must be 16.0 oz,
and from past experience it is known that the standard deviation of the can
contents is 0.1 oz. The hypotheses are
H0: =16.0
H1: 16.0
A random sample of nine cans is to be used, and the type I error probability is
specified as =0.05. What is the type II error if the true mean contents are
1=16.1 oz?
63

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