Download 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
S 
( yi  y ) estimates the variance 

n  1 i 1
2
7
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
8
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
9
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
10
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
11
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
14
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.
15
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
16
Checking Assumptions –
The Normal Probability Plot
17
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
18
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 )
19