Anova Lecture

Engineering Experimentation II

Lecture 7

ME 311, Mechanical Engineering  University University of Kentucky 

Summar 

Regression Model 

Linear model coefficients



Model evaluation



Exploit contour and surface plots



Error bars for 22 example



Single factor multiple level



of Lecture 6



Make sense of your data



Model Linearization

Curve fitting  

R2 definition



Single factor example

ME 311, Mechanical Engineering  University University of Kentucky 

Summar 

Regression Model 

Linear model coefficients



Model evaluation



Exploit contour and surface plots



Error bars for 22 example



Single factor multiple level



of Lecture 6



Make sense of your data



Model Linearization

Curve fitting  

R2 definition



Single factor example

ME 311, Mechanical Engineering  University University of Kentucky 

  



Comparing more than two factor’s levels…the levels…the ana nalysis lysis of variance    



The hypothesis testing framework The two-sample t -test -test ec ng assump ons, va y

 ANOVA decomposition of total variability Statistical testing & analysis Checking assumptions, model validity Post-ANOVA testing of means

Samp Sa mple le size si ze determination

ME 311, Mechanical Engineering  University University of Kentucky 

Portl or tla and Cement Formu or mulati lation on (page 23)

ME 311, Mechanical Engineering  University University of Kentucky 

Dot Diagram, Fig. 2-1, pp. 24

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

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The Hypothesis Testing Framework 

 

Statistical hypothesis testing is a useful framework for many experimental situations

r g ns o

e me o o ogy a e rom

e ear y

s

We will use a procedure known as the two-sample t  -

ME 311, Mechanical Engineering  University of Kentucky 

The Hypothesis Testing Framework



Sampling from a normal distribution



Statistical h

otheses:

 H 0 : μ1 = μ 2  H 1 : μ1 ≠ μ 2 ME 311, Mechanical Engineering  University of Kentucky 

Estimation of Parameters

=

1 n

S = 2

n i =1 n

∑ n −1

( yi − y ) 2 estimates the variance σ  2

i =1

ME 311, Mechanical Engineering  University of Kentucky 

Summar Statistics

. 36

Modified Mortar

Unmodified Mortar

“

“Ori inal reci e”

”

 y1 = 16.76

 y1 = 17.04

= . S 1 = 0.316 n1 = 10

= . S 1 = 0.248 n1 = 10

1

2 1

ME 311, Mechanical Engineering  University of Kentucky 

How the Two-Sample t  -Test Works: Use the sam le means to draw inferences about the o ulation means  y1 − y2 = 16.76 − 17.04 = −0.28

Standard deviation of the difference in sample means 2 σ y

=

2

n This su

ests a statistic: Z0 =

y1 − y2 2 1

n1

+

2 2

n2 ME 311, Mechanical Engineering  University of Kentucky 

How the Two-Sample t  -Test Works: se

1

an

2

o es ma e σ 1 an

e prev ous ra o ecomes

σ 2

 y − y S12 n1

+

S 22 n2

However we have the case where σ 2 = σ 2 = σ  2 Pool the individual sample variances: S  p = 2

(n1 − 1) S12 + ( n2 − 1) S22   n +n −2 ME 311, Mechanical Engineering  University of Kentucky 

How the Two-Sample t  -Test Works: The test statistic is  y1 − y2

t 0 = S  p  





n1

+

n2

Values of t 0 that are near zero are consistent with the null hypothesis Values of t 0 that are very different from zero are consistent with the alternative h othesis t 0 is a “distance” measure-how far apart the averages are expressed in standard deviation units Notice the inter retation of t  as a si nal-to-noise ratio

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The Two-Sample (Pooled) t  -Test

S  = 2  p

(n1 − 1) S12 + (n2 − 1) S 22   9(0.100) + 9(0.061)

=

n1 + n2 − 2

10 + 10 − 2

= 0.081

S  p = 0.284

 

 y1 − y2

t 0 =  p

1 n1

= 1 n2

16.76 − 17.04 .

1 10

1 10

= −2.20

The two sample means are a little over two standard deviations apart Is this a "large" difference? ME 311, Mechanical Engineering  University of Kentucky 

The Two-Sample (Pooled) t  -Test 







So far, we haven’t reall done any “statistics” We need an objective basis for deciding how large the test statistic 0 really is In 1908, W. S. Gosset derived the reference distribution 0… distribution Tables of the t distribution ,

t 0

= -2.20

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The Two-Sample (Pooled) t  -Test   

 A value of t 0 between –2.101 and 2.101 is consistent with equality of means t 0 is exceeding the range of 2.101 or –2.101, leads to significant means difference Could also use the P  -value approach

=- .

ME 311, Mechanical Engineering  University of Kentucky 

The Two-Sample (Pooled) t  -Test

t 0





= -2.20

The P- v alue is the risk of wron l re ectin the null h means (it measures rareness of the event) The P- value in our problem is P = 0.042

othesis of e ual

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The Normal Probabilit Plot

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Im ortance of the t  -Test 



Provides an objective framework for simple comparative experiments ou e use o es a re evan ypo eses n a wolevel factorial design, because all of these hypotheses “ ” versus the mean response at the opposite “side” of the cube

ME 311, Mechanical Engineering  University of Kentucky 

What If There Are More Than Two Factor Levels?  







The t -test does not directly apply There are lots of practical situations where there are either more than two levels of interest, or there are several factors of simultaneous interest The analysis of variance (ANOVA) is the appropriate analysis “engine” for these types of experiments – Chapter 3, textbook e was eve ope y s er n t e ear y initially applied to agricultural experiments

s, an

Used extensivel toda for industrial ex eriments

ME 311, Mechanical Engineering  University of Kentucky 

 An Exam le See 

 



. 60

 An engineer is interested in investigating the relationship . objective of an experiment like this is to model the relationship between etch rate and RF power, and to specify the power . The response variable is etch rate. She is interested in a particular gas (C2F6) and gap (0.80 cm), and wants to test four levels of RF power: 160W, 180W, 200W, and 220W. She decided to test five wafers at each level of RF power. The experimenter chooses 4 levels of RF power 160W, 180W, 200W, and 220W  –  order 

ME 311, Mechanical Engineering  University of Kentucky 

 An Example (See pg. 62)





Does changing the power change the mean etch rate? Is there an optimum level for ower?

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The Analysis of Variance (Sec. 3-2, pg. 63)







In general, there will be a levels of the factor, or a treatments, and n re licates of the ex eriment run in random order …a completely randomized design (CRD) N = an total runs … will be discussed later  Objective is to test hypotheses about the equality of the a ME 311, Mechanical Engineering  University of Kentucky 

The Analysis of Variance 

The name “analysis of variance” stems from a partitioning of are consistent with a model for the experiment



The basic single-factor ANOVA model is

= +τ + ε 

μ εij

⎧ i = 1,2,..., a ⎩ j = 1,2,..., n

= an overall mean, τ i = ith treatment effect, = exper men a error,

,σ 

ME 311, Mechanical Engineering  University of Kentucky 

Models for the Data There are several ways to write a model for the data:

 yij = μ + τ i + ε ij is called the effects model

= i,  y = μ + ε  is called the means model i

Regression models can also be employed 

ME 311, Mechanical Engineering  University of Kentucky 

 The Analysis of Variance 

Total variability is measured by the total sum of squares:



The basic ANOVA partitioning is:

SST =

a

n

∑∑

( yij − y.. ) 2

i =1 j =1 a

n

∑∑ ( y i =1 j =1

a

n

2 y y y y y − ) = [( − ) + ( − )] ∑∑ .. .. ij i. ij i. 2

i =1 j =1

=n

a

y . − y..

2

+

i =1

a

n

y − y.

2

i =1 j =1

SST = SSTreatments + SSE    ME 311, Mechanical Engineering  University of Kentucky 

The Analysis of Variance

SST = SSTreatments + SS E    





 A large value of SS Treatments reflects large differences in treatment means  A small value of SS Treatments likely indicates no differences in treatment means Formal statistical hypotheses are:

 H  :

=

=L =

a

 H 1 : At least one mean is different ME 311, Mechanical Engineering  University of Kentucky 

The Analysis of Variance 



While sums of squares cannot be directly compared to test the hypothesis of equal means, mean squares can be compared.  A mean square is a sum of squares divided by its degrees of freedom:

= reatments rror   an − 1 = a − 1 + a ( n − 1) ota

 MS Treatments = 



Treatments

a −1

, MS E  =

E 

a ( n − 1)

If the treatment means are equal, the treatment and error mean squares will be (theoretically) equal. rea men means er, e rea men mean square w e arger the error mean square.

an

ME 311, Mechanical Engineering  University of Kentucky 

 Analysis of Variance: Summarized

  

Computing…see text, pp 66-70 The reference distribution for F 0 is the F a -1, a (n- 1) distribution e nu ypo es s equa rea men means e ec

F0 > F α  a −1 a

n −1 ME 311, Mechanical Engineering  University of Kentucky 

 ANOVA Table: Example 3-1

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The Reference Distribution:

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 ANOVA calculations are usually done via computer 



Calculations can be done on Minitab, NCSS, Excel, Matlab, Scilab, …etc

ME 311, Mechanical Engineering  University of Kentucky 

Model Adequacy Checking in the ANOVA Text reference, Section 3-4, pg. 75 

Checking assumptions is important



Normalit



Constant variance



Inde endence



Have we fit the right model?



Later we will talk about what to do if some of these assumptions are violated

ME 311, Mechanical Engineering  University of Kentucky 

Model Adequacy Checking in the ANOVA

residuals (see text, Sec. 3-4, pg. 75)

eij = yij − yˆij ij 





−

i.

NCSS enerates the residuals Residual plots are very Normal probability plot of residuals ME 311, Mechanical Engineering  University of Kentucky 

Other Important Residual Plots

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Post-ANOVA Comparison of Means

 



 

means  Assume that residual analysis is satisfactory a ypo es s s re ec e , we on now w c spec c means are different Determining which specific means differ following an ANOVA is called the multiple comparisons problem There are lots of ways to do this…see text, Section 3-5, pg. 87 We will use airwise t -tests on means…sometimes called Fisher’s Least Significant Difference (or Fisher’s LSD) Method

ME 311, Mechanical Engineering  University of Kentucky 

Two-Factor Multi le levels Ex eriment

a  levels

of factor A; b  levels of factor B ; n  replicates

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Extension of the ANOVA to Factorials a

b

n

( yijk − y... ) = bn

i =1 j =1 k =1

a

( yi.. − y... ) + an

i =1 a

 

b

( y. j. − y... )

j =1 b

a

b

n

+ n∑∑ ( yij . − yi .. − y. j . + y... ) +∑∑∑ ( yijk − yij. )2 2

i =1 j =1

SST = SS A + SS B + SS AB + SS E 

i =1 j =1 k =1

 

 

df   breakdown: abn − 1 = a − 1 + b − 1 + a − 1 b − 1 + ab n − 1

ME 311, Mechanical Engineering  University of Kentucky 

 –

NCSS and Minitab will perform the computations

Text gives details of manual computing – see pp. 169 & 170 ME 311, Mechanical Engineering  University of Kentucky 

 An aly si s of Varian ce Table Source Sum of Term DF Squares  A: C2 2 900801.2 B: C3 2 420599.2  AB 4 809992.1 S 18 3162.667 Total Ad usted 26 Total 27 * Term significant at alpha = 0.05 Means and Effect s Section

Mean Square 450400.6 210299.6 202498 175.7037 2134555

F-Ratio 2563.41 1196.90 1152.50

Prob Level 0.000000* 0.000000* 0.000000*

Power   (Alpha=0.05) 1.000000 1.000000 1.000000

Standard  All  A: C2 1 2 B: C3 1 2 3  AB: C2,C3 1,1 1,2 1,3 , 2,2 2,3 3,1 3,2 ,

27

478.2592

9 9

468.7778 706.5555 .

4.418442 4.418442 .

-9.481482 228.2963 .

9 9 9

305.4445 595.7778 533.5555

4.418442 4.418442 4.418442

-172.8148 117.5185 55.2963

3 3 3

16.33333 796.6667 593.3333 . 708 873 361.3333 282.6667 .

7.652967 7.652967 7.652967 . 7.652967 7.652967 7.652967 7.652967 .

-279.6296 210.3704 69.25926 . -116.0741 111.1481 274.7037 -94.2963 .

3 3 3 3

478.2592

ME 311, Mechanical Engineering  University of Kentucky 

Factorials with More Than Two Factors treatment combinations are run in random order  

…

 ANOVA identity is also similar:

SST = SS A + SS B + L + SS AB + SS AC  + L

+ SS 

+ L + SS

L

+ SS  

Complete three-factor example in text, Example 5-5

ME 311, Mechanical Engineering  University of Kentucky