Transcript Design of Engineering Experiments Part 2

Design of Engineering Experiments
Part 2 – Basic Statistical Concepts
• Simple comparative experiments
– The hypothesis testing framework
– The two-sample t-test
– Checking assumptions, validity
• Comparing more than two factor levels…the
analysis of variance
–
–
–
–
ANOVA decomposition of total variability
Statistical testing & analysis
Checking assumptions, model validity
Post-ANOVA testing of means
• Sample size determination
1
Portland Cement Formulation (Table 2-1, pp. 24)
Observation
(sample), j
Modified Mortar
(Formulation 1)
1
16.85
16.62
2
16.40
16.75
3
17.21
17.37
4
16.35
17.12
5
16.52
16.98
6
17.04
16.87
7
16.96
17.34
8
17.15
17.02
9
16.59
17.08
10
16.57
17.27
y1 j
Unmodified Mortar
(Formulation 2) y2 j
2
Graphical View of the Data
Dot Diagram, Fig. 2-1, pp. 24
3
Box Plots, Fig. 2-3, pp. 26
4
The Hypothesis Testing Framework
• Statistical hypothesis testing is a useful
framework for many experimental
situations
• Origins of the methodology date from the
early 1900s
• We will use a procedure known as the twosample t-test
5
The Hypothesis Testing Framework
• Sampling from a normal distribution
• Statistical hypotheses: H :   
0
1
2
H1 : 1   2
6
Estimation of Parameters
1 n
y   yi estimates the population mean 
n i 1
n
1
2
2
2
S 
( yi  y ) estimates the variance 

n  1 i 1
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Summary Statistics (pg. 36)
Formulation 1
Formulation 2
“New recipe”
“Original recipe”
y1  16.76
y2  17.04
S12  0.100
S 22  0.061
S1  0.316
S 2  0.248
n2  10
n2  10
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How the Two-Sample t-Test Works:
Use the sample means to draw inferences about the population means
Difference in sample means is
y1  y2  16.76 17.04  0.28
The inference is drawn by comparing the difference with standard deviation of
the difference in sample means:
Difference in sample means
────────────────────────────────
Standard deviation of the difference in sample means
This suggests a statistic:
Zo 
y1  y2
 12
n1

 22
n2
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How the Two-Sample t-Test Works
Use S and S to estimate  and 
2
1
2
2
2
1
The previous ratio becomes
2
2
y1  y2
2
1
2
2
S
S

n1 n2
However, we have the case where     
2
1
2
2
2
Pool the individual sample variances:
(n1  1) S  (n2  1) S
S 
n1  n2  2
2
p
2
1
2
2
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How the Two-Sample t-Test Works:
The test statistic is
y1  y2
t0 
1 1
Sp

n1 n2
•
t0 is a “distance” measure-how far apart the averages are
expressed in standard deviation units
• Values of t0 that are near zero are consistent with the null
hypothesis
• Values of t0 that are very different from zero are consistent
with the alternative hypothesis
• Notice the interpretation of t0 as a signal-to-noise ratio
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The Two-Sample (Pooled) t-Test
(n1  1) S12  (n2  1) S 22 9(0.100)  9(0.061)
S 

 0.081
n1  n2  1
10  10  2
2
p
S p  0.284
y1  y2
16.76  17.04
to 

 2.20
1 1
1 1
Sp

0.284

n1 n2
10 10
The two sample means are about 2 standard deviations apart. Is
this a “large” difference?
12
The Two-Sample (Pooled) t-Test
• So far, we haven’t really
done any “statistics”
• We need an objective
basis for deciding how
large the test statistic t0
really is
• In 1908, W. S. Gosset
derived the reference
distribution for t0 …
called the t distribution
• Tables of the t
distribution - text, page
606
13
The Two-Sample (Pooled) t-Test
• A value of t0 between
–2.101 and 2.101 is
consistent with
equality of means
• It is possible for the
means to be equal and
t0 to exceed either
2.101 or –2.101, but it
would be a “rare
event” … leads to the
conclusion that the
means are different
• Could also use the
P-value approach
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The Two-Sample (Pooled) t-Test
• The P-value is the risk of wrongly rejecting the null hypothesis of equal
means (it measures rareness of the event)
• The P-value in our problem is P = 0.0411
• The null hypothesis Ho: 1 = 2 would be rejected at any level of
significance a ≥ 0.0411.
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Minitab Two-Sample t-Test Results
Two-Sample T-Test and CI: Form 1, Form 2
Two-sample T for Form 1 vs Form 2
N
Mean
StDev
SE Mean
Form 1
10
16.764
0.316
0.10
Form 2
10
17.042
0.248
0.078
Difference = mu Form 1 - mu Form 2
Estimate for difference:
-0.278000
95% CI for difference: (-0.545073, -0.010927)
T-Test of difference = 0 (vs not =): T-Value = -2.19
P-Value = 0.042 DF = 18
Both use Pooled StDev = 0.2843
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Checking Assumptions –
The Normal Probability Plot
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Importance of the t-Test
• Provides an objective framework for simple
comparative experiments
• Could be used to test all relevant hypotheses
in a two-level factorial design, because all
of these hypotheses involve the mean
response at one “side” of the cube versus
the mean response at the opposite “side” of
the cube
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Confidence Intervals (See pg. 43)
• Hypothesis testing gives an objective statement
concerning the difference in means, but it doesn’t
specify “how different” they are
• General form of a confidence interval
L    U where P( L    U )  1  a
a
• The 100(1- )% confidence interval on the
difference in two means:
y1  y2  ta / 2,n1  n2 2 S p (1/ n1 )  (1/ n2 )  1  2 
y1  y2  ta / 2,n1  n2 2 S p (1/ n1 )  (1/ n2 )
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