1

Design of Engineering ExperimentsPart 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

2

Portland Cement Formulation (Table 2-1, pp. 24)

Observation (sample), j

Modified Mortar

(Formulation 1)

Unmodified Mortar (Formulation 2)

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

1 jy 2 jy

3

Graphical View of the DataDot Diagram, Fig. 2-1, pp. 24

4

Box Plots, Fig. 2-3, pp. 26

5

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 two-sample t-test

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The Hypothesis Testing Framework

• Sampling from a normal distribution• Statistical hypotheses:

0 1 2

1 1 2

:

:

H

H

7

Estimation of Parameters

1

2 2 2

1

1 estimates the population mean

1( ) estimates the variance

1

n

ii

n

ii

y yn

S y yn

8

Summary Statistics (pg. 36)

Formulation 1

“New recipe”

Formulation 2

“Original recipe”

10

248.0

061.0

04.17

2

2

22

2

n

S

S

y

10

316.0

100.0

76.16

2

1

21

1

n

S

S

y

9

How the Two-Sample t-Test Works:

28.004.1776.1621 yy

Use the sample means to draw inferences about the population means

Difference in sample means is

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:

2

22

1

21

21

nn

yyZo

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How the Two-Sample t-Test Works

2 2 2 21 2 1 2

1 2

2 21 2

1 2

2 2 21 2

2 22 1 1 2 2

1 2

Use and to estimate and

The previous ratio becomes

However, we have the case where

Pool the individual sample variances:

( 1) ( 1)

2p

S S

y y

S Sn n

n S n SS

n n

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How the Two-Sample t-Test Works:

• 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

1 20

1 2

The test statistic is

1 1

p

y yt

Sn n

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The Two-Sample (Pooled) t-Test

081.021010

)061.0(9)100.0(9

1

)1()1(

21

222

2112

nn

SnSnS p

20.2

101

101

284.0

04.1776.16

11

21

21

nnS

yyt

p

o

284.0pS

The two sample means are about 2 standard deviations apart. Is this a “large” difference?

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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

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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 ≥ 0.0411.

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Minitab Two-Sample t-Test ResultsTwo-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

• The 100(1- )% confidence interval on the difference in two means:

where ( ) 1 L U P L U

1 2

1 2

1 2 / 2, 2 1 2 1 2

1 2 / 2, 2 1 2

(1/ ) (1/ )

(1/ ) (1/ )

n n p

n n p

y y t S n n

y y t S n n