Two factors, factorial experiment, both factors

Transcription

1 Extension to Factorial Treatment Structure Two factors, factorial experiment, both factors random (Section , pg. 490) i 1,2,..., a yijk μ+ τi + β j + ( τβ) ij + εijk j 1,2,..., b k 1,2,..., n V( τ ) σ, V( β ) σ, V[( τβ) ] σ, V( ε ) σ i τ j β ij τβ ijk V( y ijk ) στ + σβ + στβ + σ The model parameters are all normally distributed, independent random variables with mean 0 and variances given above Random effects model

2 Testing Hypotheses - Random Effects Model Once again, the standard ANOVA partition is appropriate Relevant hypotheses: More Statistics tutorial at H : σ 0 H : σ 0 H : σ τ 0 β 0 τβ H : σ > 0 H : σ > 0 H : σ > τ 1 β 1 τβ Form of the test statistics depend on the expected mean squares: A ( A) σ + στβ + στ 0 AB E n bn F ( ) σ + σ + σ EB n τβ an β F0 E 2 2 E ( AB ) σ + nσ τβ F0 2 E ( ) σ B AB AB E

3 Recall E for two-way way fixed effect

4 Estimating the Variance Components Two Factor Random model As before, use the ANOVA method; equate expected mean squares to their observed values: bn an n 2 ˆ σ 2 A AB ˆ σ τ 2 B AB ˆ σ β 2 AB E ˆ σ τβ Potential problems with these estimators E

5 Example 13-2 (pg. 492) A Measurement Systems Capability Study Gauge capability (or R&R) is of interest The gauge is used by an operator to measure a critical dimension i on a part This is a two-factor factorial (completely randomized) with both factors (operators, parts) random a random effects model

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7 #####obtain ANOVA table ###### aov1aov(y~part*operator) sum1summary(aov1)[[1]] sum1 Df Sum Sq Mean Sq F value Pr(>F) part <2e-16 *** operator part:operator Residuals Signif. codes: 0 '***' '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Only the F test for interaction is valid!!!!!

8 # The F test for Apart and Boperator ##### ###sum[1:2,3] are the values for A and B#### ###sum1[3,3] are the value for E##### FAB F.A.Bsum1[1:2,3]/sum1[3,3] 1[123]/ 1[33]##F-value for A and B respectively p.a.b1-pf(f.a.b,sum1[1:2,1],sum1[3,1]) ###p-value for A and B respectively > F.A.B [1] > p.a.b [1] No significant part-operator interaction Effect of parts is large Operators may have a small negligible effect

9 # The variance estimates sigma2.esum1[4,3] asum1[1,1]+1 bsum1[2,1]+1 nsum1[4,1]/a/b+1 #df for SSE is ab(n-1) sigma2.ab(sum1[3,3]-sigma2.e)/n 3]-sigma2 sigma2.a(sum1[1,3]-sum1[3,3])/b/n sigma2.b(sum1[2,3]-sum1[3,3])/a/n c(sigma2.a,sigma2.b,sigma2.ab,sigma2.e) bn an n 2 ˆ σ 2 A AB ˆ σ τ 2 B AB ˆ σ β 2 AB E ˆ σ τβ E > c(sigma2.a,sigma2.b,sigma2.ab,sigma2.e) 2A i 2B i 2AB i 2E) [1] Not reasonable. But unavoidable if use ANOVA method of estimation set to 0 (reasonable in this case) since interaction is not significant + Additional variability in the measurement system resulting from use of the instrument by the operator Variation observed when the same part is measured by the same operator

10 Reduced Model aov2aov(y~part+operator) p sum2summary(aov2)[[1]] sum2 Df Sum Sq Mean Sq F value Pr(>F) part <2e-16 *** operator Residuals The two F tests are valid!!!!!

11 # The variance estimates asum2[1,1]+1 1 bsum2[2,1]+1 n(sum2[3,1]+a+b-1)/a/b sigma2.esum2[3,3] sigma2.a.b(sum2[1:2,3]-sum2[3,3])/n/c(b,a) 2AB 2[123] 2[3 3])/ / (b ) c(sigma2.a.b,sigma2.e) [1] Additional variability in the measurement system resulting from use of the instrument by the operator Variation i observed when the same part is measured by the same operator