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Statistics 514: Design of Experiments
Topic 10
Topic Overview
This topic will cover
- Mixed Effects Designs
- Nested Factors
- Split-plot Designs
- Computing Standard Errors
- Repeated Measures Analysis
Factorial Experiments with Random Effects
- Much of the previous discussion has focused on fixed effects
- Always use MSE in denominator of F -test
- Use MSE in linear combinations and CI’s
- Not always true when random factors present
- May use interaction MS or combination of MS’s
- Will now use EMS as guide for tests
- Two models: Random model, Mixed model
Two-Factor Random Model
yi,j,k = μ + τi + βj + (τ β)i,j + i,j,k
i = 1, 2 ,... , a j = 1, 2 ,... , b k = 1, 2 ,... , n
τi ∼ N (0, σ^2 τ ) βj ∼ N (0, σ β^2 ) (τ β)i,j ∼ N (0, σ^2 τ β )
- Var(yi,j,k) = σ^2 + σ^2 τ + σ β^2 + σ^2 τ β
- Expected MS’s similar to one-factor random model
- E(MSA) = σ^2 + bnσ^2 τ + nσ^2 τ β
- E(MSB ) = σ^2 + anσ β^2 + nσ τ β^2
- E(MSAB ) = σ^2 + nσ^2 τ β
- EMS determine which MS to use in denominator
- H 0 : σ τ^2 = 0 → M SA/M SAB
- H 0 : σ β^2 = 0 → M SB /M SAB
- H 0 : σ τ β^2 = 0 → M SAB /M SE
- No hierarchical testing. Look at all tests.
Estimating Variance Components
- Using ANOVA method
- σˆ^2 = M SE
- σˆ^2 τ = (M SA − M SAB )/bn
- σˆ^2 β = (M SB − M SAB )/an
- σˆ^2 τ β = (M SAB − M SE )/n
- Sometimes results in negative estimates
- proc varcomp and proc mixed compute estimates
- Can use different estimation procedures
- ANOVA method – Method = type
- RMLE method – Method = reml(default)
- proc mixed
- Variance component estimates
- Hypothesis tests and confidence intervals
Gauge Capability Example in Text 13-
options nocenter ps=60 ls=80;
data randr; input part operator resp @@; cards; 1 1 21 1 1 20 1 2 20 1 2 20 1 3 19 1 3 21 2 1 24 2 1 23 2 2 24 2 2 24 2 3 23 2 3 24 3 1 20 3 1 21 3 2 19 3 2 21 3 3 20 3 3 22
Source Type III Expected Mean Square operator Var(Error) + 2 Var(operatorpart) + 40 Var(operator) part Var(Error) + 2 Var(operatorpart) + 6 Var(part) operatorpart Var(Error) + 2 Var(operatorpart)
Tests of Hypotheses Using the Type III MS for operator*part as an Error Term
Source DF Type III SS Mean Square F Value Pr > F operator 2 2.616667 1.308333 1.84 0. part 19 1185.425000 62.390789 87.65 <.
Tests of Hypotheses for Random Model Analysis of Variance
Dependent Variable: resp Source DF Type III SS Mean Square F Value Pr > F operator 2 2.616667 1.308333 1.84 0. part 19 1185.425000 62.390789 87.65 <. Error 38 27.050000 0. Error: MS(operator*part)
Source DF Type III SS Mean Square F Value Pr > F operator*part 38 27.050000 0.711842 0.72 0. Error: MS(Error) 60 59.500000 0.
Type 1 Analysis of Variance Sum of Source DF Squares Mean Square operator 2 2.616667 1. part 19 1185.425000 62. operator*part 38 27.050000 0. Residual 60 59.500000 0.
Type 1 Analysis of Variance Error Source Expected Mean Square Error Term DF operator Var(Residual) + 2 Var(operatorpart) MS(operatorpart) 38
- 40 Var(operator) part Var(Residual) + 2 Var(operatorpart) MS(operatorpart) 38
- 6 Var(part) operatorpart Var(Residual) + 2 Var(operatorpart) MS(Residual) 60 Residual Var(Residual)..
Covariance Parameter Estimates Standard Z Cov Parm Estimate Error Value Pr Z Alpha Lower Upper operator 0.01491 0.03296 0.45 0.6510 0.05 -0.04969 0. part 10.2798 3.3738 3.05 0.0023 0.05 3.6673 16. operator*part -0.1399 0.1219 -1.15 0.2511 0.05 -0.3789 0. Residual 0.9917 0.1811 5.48 <.0001 0.05 0.7143 1.
The Mixed Procedure
Iteration History Iteration Evaluations -2 Res Log Like Criterion 0 1 624. 1 3 409.39453674 0. 2 1 409.39128078 0. 3 1 409.39127700 0. Convergence criteria met.
Covariance Parameter Estimates Standard Z Cov Parm Estimate Error Value Pr Z Alpha Lower Upper operator 0.01063 0.03286 0.32 0.3732 0.05 0.001103 3.737E part 10.2513 3.3738 3.04 0.0012 0.05 5.8888 22. operator*part 0...... Residual 0.8832 0.1262 7.00 <.0001 0.05 0.6800 1.
Confidence Intervals for Variance Components
- Can use asymptotic variance estimates to form CI
- Known as Wald’s approximate CI
- mixed: option cl = wald or method = type
Use standard normal → 95% CI uses 1. σˆ^2 β ± 1 .96(0.0330) = (− 0. 05 , 0 .08) σˆ^2 τ ± 1 .96(3.3738) = (3. 67 , 16 .89)
- In general proc mixed uses Satterthwaite CI
Default method – REML Versions < 6.12 computed Wald CI Current uses Satterthwaite’s Approximation Will discuss this CI construction later
Rules for Expected Mean Squares (13-5)
- In models so far, EMS fairly straightforward
- Could show EMS using brute force expectation method
- For mixed models, good to have formal procedure
- Montgomery describes procedure for restricted model
- Write the error term in the model as (i,j,..)m, where m represents the replication subscript.
- Write each variable term in the model as a row heading in a two-way table
F R R
a b n Expected Factor i j k Mean Square τi 0 b n σ^2 + nσ τ β^2 + bn^
∑ (^) τ 2 i a− 1 βj a 1 n σ^2 + anσ^2 β (τ β)i,j 0 1 n σ^2 + nσ τ β^2 (i,j),k 1 1 1 σ^2
3-Factor Mixed Model (A Fixed)
yi,j,k = μ + τi + βj + δk + (τ β)i,j + (τ δ)i,k + (βδ)j,k + i,j,k,`
F R R R a b c n Factor i j k Expected Mean Squares τi 0 b c n σ^2 + cnσ^2 τ β + bnσ^2 τ γ + nσ τ βγ^2 + bcn ∑^ τ (^) i^2 a− 1 βj a 1 c n σ^2 + anσ^2 βγ + acnσ β^2 γk a b 1 n σ^2 + anσ^2 βγ + abnσ^2 γ (τ β)i,j 0 1 c n σ^2 + nσ τ βγ^2 + cnσ τ β^2 (τ γ)i,k 0 b 1 n σ^2 + nσ τ βγ^2 + bnσ τ γ^2 (βγ)j,k a 1 1 n σ^2 + anσ βγ^2 (τ βγ)i,j,k 0 1 1 n σ^2 + nσ^2 τ βγ i,j,k, 1 1 1 1 σ^2
Construction of Hasse Diagram
- Described in Oehlert.
- Used for both restricted and unrestricted models
- Provide graphical view of design
- Shows nested/crossed and random/fixed structure
- Every term in model is a node
- Terms/nodes placed in layered structure
Term U is above term V if all terms in U are in V.
- Join nodes based on nested/cross structure
- Brackets placed around random terms
3-Factor Mixed Model
- Denominator for U is leading eligible random term(s)
- Leading: Closest connected random term below U
- Eligible:
- Unrestricted: Any random term possible
- Restricted: Any without fixed factor not in U M A (^) (B) (C)
(AB) (AC)^ (BC) (ABC)
(E)
PPP
QQ
aa (^) aa!Q!!!
aa a !! !
Restricted Model: A: Leading random terms are AB and AC → approximate test B: Leading random term is BC because AB has fixed factor A BC: Leading term is E because ABC has fixed factor A
Unrestricted Model: A: Leading random terms are AB and AC → approximate test B: Leading random terms is AB and BC → approximate BC: Leading term is ABC
Two-Factor Mixed Effects Model
- Same model but different parameter restrictions
- Assume A fixed and B random
τi = 0 and β ∼ N (0, σ^2 β ) usual assumptions 2 (τ β)i,j ∼ N (0, (a − 1)σ τ β^2 /a) (a − 1)/a simplifies EMS 3
j (τ β)i,j^ = 0 for^ β^ level^ j^ added restriction
- Due to added restriction
- Not all (τ β)i,j independent, Cov((τ β)i,j , (τ β)i′,j ) = − (^1) a σ^2 τ β
- Known as restricted mixed effects model
- This model coincides with EMS algorithm
Sample Size Calculations
Use Charts V and VI
Random Effects Model Factor λ dfnum dfden A
1 + bnσ
(^2) τ σ^2 +nσ^2 τ β^ a^ −^1 (a^ −^ 1)(b^ −^ 1)
B
anσ^2 β σ^2 +nσ nτ β^2 b^ −^1 (a^ −^ 1)(b^ −^ 1)
AB
nσ^2 τ β σ^2 (a^ −^ 1)(b^ −^ 1)^ ab(n^ −^ 1) Mixed Effects Model Factor λ or Φ dfnum dfden
A
bn ∑^ τ (^) i^2 a(σ^2 +nσ^2 τ β ) a^ −^1 (a^ −^ 1)(b^ −^ 1)
B
anσ^2 β σ^2 b^ −^1 ab(n^ −^ 1) AB
nσ^2 τ β σ^2 (a^ −^ 1)(b^ −^ 1)^ bn(n^ −^ 1) /* Gauge Capability Example in Text 12-3 */
options nocenter ps=40 ls=75;
data randr; input part operator resp @@; cards; 1 1 21 1 1 20 1 2 20 1 2 20 1 3 19 1 3 21 2 1 24 2 1 23 2 2 24 2 2 24 2 3 23 2 3 24 3 1 20 3 1 21 3 2 19 3 2 21 3 3 20 3 3 22 4 1 27 4 1 27 4 2 28 4 2 26 4 3 27 4 3 28 5 1 19 5 1 18 5 2 19 5 2 18 5 3 18 5 3 21 6 1 23 6 1 21 6 2 24 6 2 21 6 3 23 6 3 22 7 1 22 7 1 21 7 2 22 7 2 24 7 3 22 7 3 20 8 1 19 8 1 17 8 2 18 8 2 20 8 3 19 8 3 18 9 1 24 9 1 23 9 2 25 9 2 23 9 3 24 9 3 24 10 1 25 10 1 23 10 2 26 10 2 25 10 3 24 10 3 25 11 1 21 11 1 20 11 2 20 11 2 20 11 3 21 11 3 20 12 1 18 12 1 19 12 2 17 12 2 19 12 3 18 12 3 19 13 1 23 13 1 25 13 2 25 13 2 25 13 3 25 13 3 25 14 1 24 14 1 24 14 2 23 14 2 25 14 3 24 14 3 25 15 1 29 15 1 30 15 2 30 15 2 28 15 3 31 15 3 30 16 1 26 16 1 26 16 2 25 16 2 26 16 3 25 16 3 27 17 1 20 17 1 20 17 2 19 17 2 20 17 3 20 17 3 20 18 1 19 18 1 21 18 2 19 18 2 19 18 3 21 18 3 23 19 1 25 19 1 26 19 2 25 19 2 24 19 3 25 19 3 25 20 1 19 20 1 19 20 2 18 20 2 17 20 3 19 20 3 17;
proc glm; class operator part; model resp=operator|part; random part operatorpart / test; means operator / tukey lines E=operatorpart; lsmeans operator / adjust=tukey E=operator*part tdiff stderr;
proc mixed alpha=.05 cl covtest; class operator part; model resp=operator / ddfm=kr; random part operator*part; lsmeans operator / alpha=.05 cl diff adjust=tukey; run; quit;
Dependent Variable: resp Sum of Source DF Squares Mean Square F Value Pr > F Model 59 1215.091667 20.594774 20.77 <. Error 60 59.500000 0. Corrected Total 119 1274.
Source DF Type III SS Mean Square F Value Pr > F operator 2 2.616667 1.308333 1.32 0. part 19 1185.425000 62.390789 62.92 <. operator*part 38 27.050000 0.711842 0.72 0.
Source Type III Expected Mean Square operator Var(Error) + 2 Var(operatorpart) + Q(operator) part Var(Error) + 2 Var(operatorpart) + 6 Var(part) operatorpart Var(Error) + 2 Var(operatorpart)
Tests of Hypotheses for Mixed Model Analysis of Variance
Dependent Variable: resp Source DF Type III SS Mean Square F Value Pr > F operator 2 2.616667 1.308333 1.84 0. part 19 1185.425000 62.390789 87.65 <. Error 38 27.050000 0. Error: MS(operator*part)
Source DF Type III SS Mean Square F Value Pr > F operator*part 38 27.050000 0.711842 0.72 0. Error: MS(Error) 60 59.500000 0.
Alpha 0. Error Degrees of Freedom 38 Error Mean Square 0. Critical Value of Studentized Range 3. Minimum Significant Difference 0.
Means with the same letter are not significantly different.
Residual 1.
Fit Statistics -2 Res Log Likelihood 409. AIC (smaller is better) 413.
Type 3 Tests of Fixed Effects
Num Den Effect DF DF F Value Pr > F operator 2 98 1.48 0.2324 *** KR adjustment
Restricted model
Var(¯y 1 ..) = (σ^2 + nσ τ β^2 + nσ^2 β )/bn = (0.8832 + 2(10.2513))/ 40 Var(¯y 1 .. − y¯ 2 ..) = 2(σ^2 + nσ τ β^2 )/bn = 0. 8832 / 20
Least Squares Means Standard Effect operator Estimate Error DF t Value Pr > |t| Alpha operator 1 22.3000 0.7312 20.1 30.50 <.0001 0. operator 2 22.2750 0.7312 20.1 30.46 <.0001 0. operator 3 22.6000 0.7312 20.1 30.91 <.0001 0.
Least Squares Means Effect operator Lower Upper operator 1 20.7752 23. operator 2 20.7502 23. operator 3 21.0752 24.
Differences of Least Squares Means
Standard Effect operator _operator Estimate Error DF t Value Pr > |t| operator 1 2 0.02500 0.2101 98 0.12 0. operator 1 3 -0.3000 0.2101 98 -1.43 0. operator 2 3 -0.3250 0.2101 98 -1.55 0.
Differences of Least Squares Means Effect operator _operator Adjustment Adj P Alpha operator 1 2 Tukey-Kramer 0.9922 0. operator 1 3 Tukey-Kramer 0.3308 0. operator 2 3 Tukey-Kramer 0.2739 0.
Differences of Least Squares Means Adj Adj Effect operator _operator Lower Upper Lower Upper operator 1 2 -0.3920 0.4420 -0.4751 0. operator 1 3 -0.7170 0.1170 -0.8001 0. operator 2 3 -0.7420 0.09201 -0.8251 0.
Other Mixed Models
- SAS uses different mixed model in analysis
- Reduce parameter restrictions ∑ τi = 0 and β ∼ N (0, σ β^2 ) (τ β)i,j ∼ N (0, σ τ β^2 )
- Known as unrestricted mixed model
- For two-factor model
E(M SE ) = σ^2 E(M SA) = σ^2 + bn
τ (^) i^2 /(a − 1) + nσ τ β^2 E(M SB ) = σ^2 + anσ β^2 + nσ^2 τ β E(M SAB ) = σ^2 + nσ^2 τ β
- random statement is SAS also gives these results
- Differences
- Test H 0 : σ^2 β = 0 using M SAB in denominator
- Often more conservative test, ˆσ^2 β = (M SB − M SAB )/an
To decide which is appropriate, suppose you ran experiment again and sampled (by chance) the same random effects levels. Should this mean you also have the same set of interaction effects?
Yes: Restricted No: Unrestricted
General Mixed Effects Model
Y = Xβ + Zδ +
β is a vector fixed-effect parameters δ is a vector of random-effect parameters is the error vector
- δ and assumed uncorrelated
- means 0
- covariance matrices G and R (allows correlation)
- Cov(Y ) = ZGZ′^ + R
run; quit;
Covariance Parameter Estimates Standard Z Cov Parm Estimate Error Value Pr Z Alpha Lower subject 14.2086 6.7767 2.10 0.0180 0.05 6. subject*lotion 0.2660 0.1579 1.68 0.0460 0.05 0. Residual 0.1320 0.04174 3.16 0.0008 0.05 0.
Type 3 Tests of Fixed Effects Num Den Effect DF DF F Value Pr > F lotion 1 9 6.76 0.
- Significant source of variation due to combination
- one lotion may not be best for all subjects
- Significant subject-to-subject variability
- Lotion 2 ‘‘on average’’ offers more protection
- Is difference practically significant?
Least Squares Means Standard Effect lotion Estimate Error DF t Value Pr > |t| Alpha lotion 1 7.8200 1.2058 9.21 6.49 0.0001 0. lotion 2 7.1500 1.2058 9.21 5.93 0.0002 0.
Least Squares Means Effect lotion Lower Upper lotion 1 5.1015 10. lotion 2 4.4315 9.
Differences of Least Squares Means Standard Effect lotion _lotion Estimate Error DF t Value Pr > |t| lotion 1 2 0.6700 0.2577 9 2.60 0.
For restricted model – Var(¯yi..) = (σ^2 + (a − 1)nσ^2 τ β /a + nσ β^2 )/bn = (0.1320 + 0.2660 + 2(14.2086))/20 = 1.44, which is slightly larger than the unrestricted model.
Approximate F -tests and Confidence Intervals
- For some models, no exact F -test exists
- Recall 3-Factor Mixed Model (A - fixed)
- No exact test for A based on EMS
Assume a = 3, b = 2, c = 3, n = 2 and following MS were obtained
Source DF MS EMS F p A 2 0.7866 φA + 6σ AB^2 + 4σ^2 AC + 2σ ABC^2 + σ^2?? B 1 0.0010 18 σ B^2 + 6σ BC^2 + σ^2 0.33 0. AB 2 0.0056 6 σ AB^2 + 2σ^2 ABC + σ^2 2.24 0. C 2 0.0560 12 σ C^2 + 6σ BC^2 + σ^2 18.87 0. AC 4 0.0107 4 σ AC^2 + 2σ^2 ABC + σ^2 4.28 0. BC 2 0.0030 6 σ BC^2 + σ^2 10.00 0. ABC 4 0.0025 2 σ ABC^2 + σ^2 8.33 0. Error 18 0.0003 σ^2
Could assume some variances negligible: not recommended without “conclusive” evidence
Examples
- If assume σ^2 ABC and σ^2 AB equals 0
Source DF MS EMS F p A 2 0.7866 φA + 2σ^2 ABC + σ^2 314.64 0. B 1 0.0010 18 σ B^2 + 6σ^2 BC + σ^2 0.33 0. C 2 0.0560 12 σ C^2 + 6σ^2 BC + σ^2 18.87 0. BC 2 0.0030 6 σ^2 BC + σ^2 1.2 0. ABC 4 0.0025 2 σ^2 ABC + σ^2 1.0 0. Error 24 0.0025 σ^2
Could test interactions and then possibly remove
AC and AB found insignificant. Test A over ABC → F = 314.64 and p < 0 .001.
- Both Type I and Type II errors possible
- What level to test insignificance?
Pooling Mean Squares with Error
- Variation of previous approach
- Works well when df for error is small (< 6)
- Often test significance at α = 0. 25
- May pool together something that is different from zero
- Use high α to protect against that
- Interpolation needed Pr(F 2 , 4 > 57) = 0. 0011 Pr(F 2 , 5 > 57) = 0. 0004
p = 0.85(0.0011) + 0.15(0.0004) = 0. 001
- SAS can be used to compute p-values and quantile values for F and χ^2 values with non-integer degrees of freedom.
p-values: probf(x, df1, df2) and probchi(x, df) Quantiles: finv(p, df1, df2) and cinv(p, df)
data pvalue; p = 1-probf(57, 2.0, 4.15); f = finv(0.95, 2.0, 4.15); c1 = cinv(0.025, 18.57); c2 = cinv(0.975, 18.57);
proc print p f c1 c2;
Obs p f c1 c 1 .000959732 6.71564 8.61485 32.
Example
For the 3 factor model (avoiding subtraction),
M SA + M SABC M SAB + M SAC
p =
= 2. 01 q = 0.^0163 2
- 01072 /4+0. 00562 / 2 = 6.^00
This is again found significant.
Confidence Intervals
- Use Satterthwaite’s pseudo F -tests to create CI
- Recall dfE M SE/σ^2 ∼ χ^2 dfE M SE χ^2 α/ 2 ,dfE
≤ σ^2 ≤
dfE M SE χ^21 −α/ 2 ,dfE
- Use pseudo F -tests, ˆσ^2 = M S′^ − M S′′
- Both MS are independent and have similar χ^2 distribution
- Assume linear combination of χ^2 is χ^2 with df (M Sr +... + M Ss − M Su −... − M Sv)^2 M S r^2 /fr +... + M S s^2 /fs + M S u^2 /fu +... + M S^2 v /fv
- Use same CI formula as above
Random Effects Example 13-
Sum of Source DF Squares Mean Square F Value Pr > F Model 59 1215.091667 20.594774 20.77 <. Error 60 59.500000 0. Corrected Total 119 1274.
Source DF Type III SS Mean Square F Value Pr > F operator 2 2.616667 1.308333 1.32 0. part 19 1185.425000 62.390789 62.92 <. operator*part 38 27.050000 0.711842 0.72 0.
df =
(62. 39 − 0 .71)^2
CI: (18.57(10.28)/ 32. 28 , 18 .57(10.28)/ 8 .61) = (5. 91 , 22 .17)
σˆ^2 β = (1. 31 − 0 .71)/40 = 0. 015
df =
(1. 31 − 0 .71)^2
CI: (0.413(0.015)/ 3. 079 , 0 .413(0.015)/ 2. 29 × 10 −^8 ) = (0. 002 , 270781)
Example of Restricted vs Unrestricted Models
There are various ways to compact a gold filling to make it harder. Fillings need to be hard in order to wear well. There are three standard ways to do this:
- Condensing: The dentist uses a special hand tool (a condenser to pack the gold into the cavity).
- Hand-malleting: The dentist holds the condenser in place, and an assistant taps it with a small hammer.
- Mechanical malleting: Like Method 2, except the hammer is built into the condenser, and tapping is done by machine.
Five dentists are chosen from the UCLA School of Dentistry, and the factors are crossed: each dentist uses each of the three methods to pack gold into a small cavity drilled into a block of ivory. Hardness was measured by pushing a pyramid-shaped diamond into the filling and recording the size of the indentation. Each method is used twice, and order is assumed not to be a factor.
Method is fixed, while dentist is random.
