CS代写 Mathematics and Statistics

Mathematics and Statistics –
Design and Analysis of Experiments
Week 4 – Randomized blocks and Latin Squares.

ANOVA → Regression
• Suppose we have the ANOVA model
• 𝒀 =𝝁+𝝉 +𝜺,𝑖=1,2,…,𝑘𝑎𝑛𝑑𝑗=1,2,…,𝑛 𝒊𝒋𝒊𝒋 𝑗
• Define 𝒙𝒊 = ቊ1, 𝑖𝑓 𝑡h𝑒 𝑜𝑏𝑗𝑒𝑟𝑣𝑎𝑡𝑖𝑜𝑛 𝑖𝑠 𝑓𝑜𝑟 𝑖 𝑙𝑒𝑣𝑒𝑙 0, 𝑜𝑡h𝑒𝑟𝑤𝑖𝑠𝑒
• Then𝒀𝒊𝒋 =𝜷𝟎 +𝜷𝟏𝒙𝟏 +𝜷𝟐𝒙𝟐 +⋯+𝜷𝒌𝒙𝒌 +𝜺𝒊𝒋
• It is easy to see that 𝝁 + 𝝉𝒊=𝜷𝟎 + 𝜷𝒊
• In matrix form the corresponding regression model is 𝐘=𝑿𝜷+𝜺

ANOVA → Regression Y  1 1 0 0  0
11 11 Y 11000 
 12       12 0 
1n    
   10100  21 
Y 10100 1 
Y=   X=  𝛽=2 =  
Y  10100  
2n22n2
    
Y  1 0 0 0  1 k  
 k1  1 0 0 0  1  
Y 10001 kn kn
• Thesolutionis:
Ifweset𝛽 =𝜇+ 𝛼 𝑡h𝑒𝑛 𝑖𝑖
we have that 𝐘 = 𝑿𝜷 + 𝜺, where
Observe that X’X is not invertible (sum of all columns = 1st column)
Add the assumption
n11 + n22 ++ nkk = 0 to have a solution to the system Ifweset𝛽 =𝜇+ 𝛼 𝑡h𝑒𝑛
we have that 𝐘 = 𝑿𝜷 + 𝜺, where X is as above without the first
column. So X will be invertible and
𝐘 = 𝑿𝜷 + 𝜺
X is as above without the last column. Model is:
𝒀𝒊𝒋 = 𝜷𝟎 + 𝜷𝟏𝒙𝟏 + 𝜷𝟐𝒙𝟐 + ⋯ + 𝜷𝒌−𝟏𝒙𝒌−𝟏 + 𝜺𝒊𝒋

• To measure the pressure 4 different metal bars can
hold the following data were collected
• 𝒀𝒊𝒋 =𝜷𝟎 +𝜷𝟏𝒙𝟏 +𝜷𝟐𝒙𝟐 +𝜷𝟑𝒙3 +𝜺𝒊𝒋

Analysis of variance
DF Regression 3 Residual 11
Sum of Squares .31887 .09836
Mean Square .10629 .00894
F = 11,88690
V ariable x3
-.33250 .06686 (Constant) 2.04500 .04728
Signif F = .0009 R Square = .76426 Variables in the Equation
B SE B -.20375 .06686
-.37667 .07222
Beta T -.54024 -3.047
-.90339 -5,215 -.88162 -4,973
Sig T .0111
.0004 43,252 .0000
and correspond to the standardized variables

Design of Engineering Experiments – The Blocking Principle
• Text Reference, Chapter 4
• Blocking and nuisance factors
• The randomized complete block design or the RCBD
• Extension of the ANOVA to the RCBD
• Other blocking scenarios: Latin square designs, Graeco – Latin squares, BIBs, …

The Blocking Principle
• Blocking is a technique for dealing with nuisance factors
• A nuisance factor is a factor that probably has some effect on the response, but it’s of no interest to the experimenter…however, the variability it transmits to the response needs to be minimized
• Typical nuisance factors include batches of raw material, operators, pieces of test equipment, time (shifts, days, etc.), different experimental units
• Many industrial experiments involve blocking (or should)
• Failure to block is a common flaw in designing an experiment (consequences?)

The Blocking Principle
• If the nuisance variable is known and controllable, we use blocking
• If the nuisance factor is known and uncontrollable, sometimes we can use the analysis of covariance (see Chapter 15) to remove the effect of the nuisance factor from the analysis
• If the nuisance factor is unknown and uncontrollable (a “lurking” variable), we hope that randomization balances out its impact across the experiment
• Sometimes several sources of variability are combined in a block, so the block becomes an aggregate variable

The Hardness Testing Example
• Text reference, pg 139, 140, Also, see section 2.5.1 on page 53.
• We wish to determine whether 4 different tips produce different
(mean) hardness reading on a Rockwell hardness tester
• Gauge & measurement systems capability studies are frequent areas for applying DOX
• Assignment of the tips to an experimental unit; that is, a test coupon
• Structure of a completely randomized experiment
• The test coupons are a source of nuisance variability
• Alternatively, the experimenter may want to test the tips across coupons of various hardness levels
• The need for blocking

The Hardness Testing Example
• To conduct this experiment as a RCBD, assign all 4 tips to each coupon
• Each coupon is called a “block”; that is, it’s a more homogenous experimental unit on which to test the tips
• Variability between blocks can be large, variability within a block should be relatively small
• In general, a block is a specific level of the nuisance factor
• A complete replicate of the basic experiment is conducted
in each block
• A block represents a restriction on randomization
• All runs within a block are randomized

The Hardness Testing Example
• Suppose that we use b = 4 blocks:
• Notice the two-way structure of the experiment
• Once again, we are interested in testing the equality of treatment means, but now we have to remove the variability associated with the nuisance factor (the blocks)

Extension of the ANOVA to the RCBD
• Supposethatthereareatreatments(factor levels) and b blocks
• A statistical model (effects model) for the RCBD is y =+ + + i=1,2,…,a
ij i j ij j=1,2,…,b 
• The relevant (fixed effects) hypotheses are
H0 :1 =2 = =a wherei =(1/b)b (+i +j)=+i j=1

Extension of the ANOVA to the RCBD
ANOVA partitioning of total variability:
[(y −y)+(y −y) i. .. .j ..
+(y −y −y +y)]2 ij i. .j ..
(y −y)2 = ij ..
 i=1 j=1 i=1 j=1
ab =b(y−y)2+a (y−y)2
+(y −y −y +y)2
ij i. .j .. i=1 j=1
SST = SSTreatments + SSBlocks + SSE

Extension of the ANOVA to the RCBD
The degrees of freedom for the sums of squares in
SST = SSTreatments + SSBlocks + SSE are as follows:
ab −1 = a −1+ b −1+ (a −1)(b −1)
Therefore, ratios of sums of squares to their degrees of freedom result in mean squares and the ratio of the mean square for treatments to the error mean square is an F statistic that can be used to test the hypothesis of equal treatment means

ANOVA Display for the RCBD
Manual computing (ugh!)…see Equations (4-9) – (4-12), page 144 Use software to analyze the RCBD

Manual computing:

Estimation of the effects and multiple comparisons (Tukey)
ˆi =Yi−Y..,i=1,2,..,a,
where Yi is the average of the i th treatment, Y .. the average of all observations and
ˆi−ˆj =Yi−Y..−(Yj−Y..)=Yi−Yj γιαi,j=1,2,..,a,ij
We have that
E(Yi −Yj)=E(bYir −bYjr)=b(E(Yir)−E(Yjr))=
r=1 r=1 r=1 r=1
=b( (+i +r)− (+j +r))=b(bi −bj)=i −j
V(Yi −Yj)=V(Yi)+V(Yj)=12 +12 =22 bbb
1So,  − (ˆ −ˆ  w 2MSE/b) with w = q / i j i j T T a,(b−1)(a−1)
100(1-α)% confidence intervals (Tukey)
2 (obtained from Tables)

A Chemical engineer believes that the reaction time of a chemical process is affected by the type of the catalyst used. There are four different catalyst that can be used in the chemical reaction and the experiment is as follows: Four trained users input the material and the catalyst of the process in a reaction tube and they measure the reaction time. The engineer wish to eliminate the user effect and for this reason he uses the following RCBD.

General Linear Model: time versus User; Catalyst
Factor Type Levels Values User fixed 4 1; 2; 3; 4 Catalyst fixed 4 1; 2; 3; 4
Analysis of Variance for time, using Adjusted SS for Tests
Source DF Seq SS Adj SS Adj MS FP User 3 202.250 202.250 67.417 6.84 0.011 Catalyst 3 140.750 140.750 46.917 4.76 0.030 Error 9 88.750 88.750 9.861
Total 15 431.750
S = 3.14024 R-Sq = 79.44% R-Sq(adj) = 65.74%

Descriptive Statistics: Time
Variable Catalyst
Time 1
ˆ −ˆ =−1.88+1.38=−0.5 2 1
ˆ −ˆ =−1.88+1.38=−0.5 31
ˆ −ˆ =5.12+1.38=6.5 4 1
ˆ3 −ˆ2 =−1.88+1.88=0 ˆ4 −ˆ2 =5.12+1.88=7 ˆ4 −ˆ3 =5.12+1.88=7
−(ˆ−ˆw 2MSE/b) 2121T
Descriptive Statistics: Time
Variable Mean
Time 72.88
q4,9,0.05 =4.41
wT =q4,9,0.05 / 2=3.118 b=4
MSE=9.861 s.e(ˆ −ˆ)= 2MSE/b=2.22
ˆ1 =71.5−72.88=−1.38ˆ2 =71−72.88=−1.88 Mean ˆ3 = 71 − 72.88 = −1.88 ˆ4 = 78 − 72.88 = 5.12
 (-0.5  3.118 (-7.42, 6.42)
2  9.861/4 )
i −j (ˆi −ˆj wT

Tukey 95.0% Simultaneous Confidence Intervals Response Variable time
All Pairwise Comparisons among Levels of Catalyst Catalyst = 1 subtracted from:
Catalyst Lower Center Upper –+———+———+———+—-
Catalyst =
-7.440 -0.5000 6.440 (———-*———–)
-7.440 -0.5000 6.440 (———-*———–)
-0.440 6.5000 13.440 (———–*———-)
–+———+———+———+—- -6.0 0.0 6.0 12.0
2 subtracted from:
Catalyst Lower Center Upper –+———+———+———+—- 3 -6.940 -0.00000 6.940 (———–*———–)
4 0.060 7.00000 13.940 (———–*———-)
–+———+———+———+—- -6.0 0.0 6.0 12.0
Catalyst Catalyst = 3 subtracted from:
Catalyst Lower Center Upper –+———+———+———+—- 4 0.06007 7.000 13.94 (———–*———-)
–+———+———+———+—- -6.0 0.0 6.0 12.0

Tukey Simultaneous Tests
Response Variable time
All Pairwise Comparisons among Levels of Catalyst
Catalyst = 1 subtracted from:
Catalyst 2
Difference SE of Adjusted
of Means Difference T-Value P-Value
-0.5000 2.220 -0.2252 0.9957
-0.5000 2.220 -0.2252 0.9957
6.5000 2.220 2.9273 0.0670
Catalyst = 2 subtracted from:
Catalyst 3
Difference SE of Adjusted
of Means Difference T-Value P-Value
-0.00000 2.220 -0.00000 1.0000
7.00000 2.220 3.15246 0.0478
Catalyst = 3 subtracted from:
Difference SE of Adjusted Catalyst of Means Difference T-Value P-Value 2 4 7.000 2.220 3.152 0.0478

Vascular Graft Example (pg. 145)
• To conduct this experiment as a RCBD, assign all 4 pressures to each of the 6 batches of resin
• Each batch of resin is called a “block”; that is, it’s a more homogenous experimental unit on which to test the extrusion pressures

Residual Analysis for the Vascular Graft Example

Residual Analysis for the Vascular Graft Example
• Basic residual plots indicate that normality, constant variance assumptions are satisfied
• Noobviousproblemswithrandomization
• No patterns in the residuals vs. block
• Can also plot residuals versus the pressure (residuals by factor)
• These plots provide more information about the constant variance assumption, possible outliers

Other Aspects of the RCBD See Text, Section 4.1.3, pg. 132
• The RCBD utilizes an additive model – no interaction between treatments and blocks
• Treatments and/or blocks as random effects
• Missingvalues
• What are the consequences of not blocking if we should have?
• Sample sizing in the RCBD? The OC curve approach can be used to determine the number of blocks to run..see page 133

BIBD ◼Example
◼Verify that l=2
Block 1234
AAAB BBCC CDDD

Random Blocks and/or Treatments

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