Lecture 3. Experiments with a Single Factor: ANOVA Montgomery 3.1 through 3.3
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1 Lecture 3. Experiments with a Single Factor: ANOVA Montgomery 3.1 through 3.3 Fall, 2013 Page 1
2 Tensile Strength Experiment Investigate the tensile strength of a new synthetic fiber. The factor is the weight percent of cotton used in the blend of the materials for the fiber and it has five levels. percent tensile strength of cotton total average Fall, 2013 Page 2
3 Data Layout for Single-Factor Experiments treatment observations totals averages 1 y 11 y 12 y 1n y 1. ȳ 1. 2 y 21 y 22 y 2n y 2. ȳ a y a1 y a2 y an y a. ȳ a. Fall, 2013 Page 3
4 Analysis of Variance Interested in comparing several treatments. Could do numerous two-sample t-tests but this approach does not test equality of all treatments simultaneously. ANOVA provides a method of joint inference. Statistical Model: y ij = µ + τ i + ϵ ij i = 1, 2... a j = 1, 2,... n i µ - grand mean; τ i - ith treatment effect; ϵ ij N(0, σ 2 ) - error Constraint: a i=1 τ i = 0. Basic Hypotheses: H 0 : τ 1 = τ 2 =... = τ a = 0 vs H 1 : τ i 0 for at least one i Fall, 2013 Page 4
5 Derive Estimates: Partitioning y ij Notation y i. = n i j=1 y ij y i. = y i. /n i (treatment sample mean, or row mean) y.. = y ij y.. = y.. /N (grand sample mean) Decomposition of y ij : y ij = y.. + (y i. y.. ) + (y ij y i. ) Estimates for parameters: So y ij = ˆµ + ˆτ i + ˆϵ ij. ˆµ = y.. ˆτ i = (y i. y.. ) ˆϵ ij = y ij y i. ( residual ) It can be verified that a i=1 n iˆτ i = 0; ni j=1 ˆϵ ij = 0 for all i. Fall, 2013 Page 5
6 Test Basic Hypotheses: Partitioning the Sum of Squares Recall y ij y.. = ˆτ i + ˆϵ ij = (y i. y.. ) + (y ij y i. ) Can show i j (y ij y.. ) 2 = i n i(y i. y.. ) 2 + i j (y ij y i. ) 2 = i n iˆτ i 2 + i j ˆϵ2 ij. Total SS = Treatment SS + Error SS SS T = SS Treatments + SS E Derive test statistics for testing H 0 : τ 1 = = τ a = 0: look at SS Treatments = n iˆτ i 2 = n i (y i. y.. ) 2 Small if ˆτ i s are small If large then reject H 0, but how large is large? Standardize to account for inherent variability Fall, 2013 Page 6
7 Test Statistic F 0 = SS Treatments/(a 1) SS E /(N a) = MS Treatments MS E Mean squares: MS Treatments = a i=1 n iˆτ i /(a 1) MS E = i Can show that the expected values are E(MS E )=σ 2 j ˆϵ2 ij /(N a) E(MS Treatment ) = σ 2 + n i τ 2 i /(a 1) Hence, MS E is always a good (unbiased) estimator for σ 2, that is, ˆσ 2 =MS E. Under H 0, MS Treatment is also a good (unbiased) estimator for σ 2. Under H 0, F 0 is expected close to 1. Large F 0 implies deviation from H 0. What is the sampling distribution of F 0 under H 0? Fall, 2013 Page 7
8 Sampling Distribution of F 0 under H 0 Based on model: y ij N(µ + τ i, σ 2 ) implies y i. N(µ + τ i, σ 2 /n i ) implies j (y ij y i. ) 2 /σ 2 χ 2 n i 1 Therefore: SS E / σ 2 = i j (y ij y i. ) 2 /σ 2 χ 2 N a Under H 0 : SS Treatment /σ 2 = i n i(y i. y.. ) 2 /σ 2 χ 2 a 1 SS E and SS Treatment are independent. So F 0 = SS Treatment/σ 2 (a 1) SS E /σ 2 (N a) = χ2 a 1/(a 1) χ 2 N a /(N a) F a 1,N a F -distribution with numerator df a 1 and denominator df N a. Fall, 2013 Page 8
9 Analysis of Variance (ANOVA) Table Source of Sum of Degrees of Mean F 0 Variation Squares Freedom Square Between SS Treatment a 1 MS Treatment F 0 Within SS E N a MS E Total SS T N 1 If balanced: SS T = y 2 ij y2../n ; SS Treatment = 1 n y 2 i. y 2../N SS E =SS T - SS Treatment If unbalanced: SS T = y 2 ij y2../n ; SS Treatment = y 2 i. n i y 2../N SS E =SS T - SS Treatment Decision Rule: If F 0 > F α,a 1,N a then reject H 0 Fall, 2013 Page 9
10 Connection between F-Test and Two-Sample t-test a=2 Consider the square of the t-test statistic t 2 0 = (ȳ 1. ȳ 2. ) 2 (s pool 1/n1 + 1/n 2 ) 2 = n 1 n 2 n 1 +n 2 (ȳ 1. ȳ 2. ) 2 s 2 pool = n 1 n 2 n 1 +n 2 [(ȳ 1. ȳ.. ) (ȳ 2. ȳ.. )] 2 s 2 pool = n 1 n 2 n 1 +n 2 (ȳ 1. ȳ.. ) 2 + n 1n 2 n 1 +n 2 (ȳ 2. ȳ.. ) 2 2n 1n 2 n 1 +n 2 (ȳ 1. ȳ.. )(ȳ 2. ȳ.. ) s 2 pool Consider ȳ.. = 1 n 1 + n 2 ( n 1 j=1 y 1j + n 2 j=1 y 2j ) = n 1 n 1 + n 2 ȳ 1. + n 2 n 1 + n 2 ȳ 2. Fall, 2013 Page 10
11 One has t 2 0 = [n 1 (ȳ 1. ȳ.. ) 2 + n 2 (ȳ 2. ȳ.. ) 2 ]/(2 1) [ n 1 j=1 (y 1j ȳ 1. ) 2 + n 2 j=1 (y 2j ȳ 2. ) 2 ]/(n 1 + n 2 2) = MS Treatment MS E When a = 2, t 2 0 = F 0 When a = 2, F -test and two-sample two-sided test are equivalent. Fall, 2013 Page 11
12 Example Twelve lambs are randomly assigned to three different diets. The weight gain (in two weeks) is recorded. Is there a difference between the diets? Diet Diet Diet N = 12, yij = 156 and ȳ.. = 156/12 = 13. n 1 = 3, y 1. = 33, ȳ 1. = 11; n 2 = 5, y 2. = 75, ȳ 2. = 15; n 3 = 4, y 3. = 48 and ȳ 3. = 12. ˆτ 1 = ȳ 1. ȳ.. = = 2; Similarly, ˆτ 2 = = 2 and ˆτ 3 = = 1. SS T = i j (y ij ȳ.. ) 2 = 246. SS Treatment = 3 ( 2) (2) ( 1) 2 = 36. SS E = = 210 Fall, 2013 Page 12
13 MS E = ˆσ 2 = 210/(12 3) = F 0 = (36/2)/(210/9) = 0.77; P-value > 0.25 Fail to reject H 0 : τ 1 = τ 2 = τ 3 = 0. Fall, 2013 Page 13
14 Using SAS (lambs.sas) option nocenter ps=65 ls=80; data lambs; input diet wtgain; datalines; ; Fall, 2013 Page 14
15 symbol1 bwidth=5 i=box; axis1 offset=(5); proc gplot; plot wtgain*diet / frame haxis=axis1; proc glm; class diet; model wtgain=diet; output out=diag r=res p=pred; proc gplot; plot res*diet /frame haxis=axis1; proc sort; by pred; symbol1 v=circle i=sm50; proc gplot; plot res*pred / haxis=axis1; run; Fall, 2013 Page 15
16 SAS Output The GLM Procedure Dependent Variable: wtgain Sum of Source DF Squares Mean Square F Value Pr > F Model Error Corrected Total R-Square Coeff Var Root MSE wtgain Mean Source DF Type I SS Mean Square F Value Pr > F diet Source DF Type III SS Mean Square F Value Pr > F diet Fall, 2013 Page 16
17 Side-by-Side Boxplot Fall, 2013 Page 17
18 Residual Plot Fall, 2013 Page 18
19 Tensile Strength Experiment options ls=80 ps=60 nocenter; goptions device=win target=winprtm rotate=landscape ftext=swiss hsize=8.0in vsize=6.0in htext=1.5 htitle=1.5 hpos=60 vpos=60 horigin=0.5in vorigin=0.5in; data one; infile c:\saswork\data\tensile.dat ; input percent strength time; title1 Tensile Strength Example ; proc print data=one; run; symbol1 v=circle i=none; title1 Plot of Strength vs Percent Blend ; proc gplot data=one; plot strength*percent/frame; run; proc boxplot; Fall, 2013 Page 19
20 plot strength*percent/boxstyle=skeletal pctldef=4; proc glm; class percent; model strength=percent; output out=oneres p=pred r=res; run; proc sort; by pred; symbol1 v=circle i=sm50; title1 Residual Plot ; proc gplot; plot res*pred/frame; run; proc univariate data=oneres normal; var res; qqplot res / normal (L=1 mu=est sigma=est); histogram res / normal; run; symbol1 v=circle i=none; title1 Plot of residuals vs time ; proc gplot; plot res*time / vref=0 vaxis=-6 to 6 by 1; run; Fall, 2013 Page 20
21 Tensile Strength Example Obs percent strength time : : : : Fall, 2013 Page 21
22 Scatter Plot Fall, 2013 Page 22
23 Side-by-Side Plot Fall, 2013 Page 23
24 The GLM Procedure Dependent Variable: strength Sum of Source DF Squares Mean Square F Value Pr > F Model <.0001 Error Corrected Total R-Square Coeff Var Root MSE strength Mean Source DF Type I SS Mean Square F Value Pr > F percent <.0001 Source DF Type III SS Mean Square F Value Pr > F percent <.0001 Fall, 2013 Page 24
25 Residual Plot Fall, 2013 Page 25
26 The UNIVARIATE Procedure Variable: res Moments N 25 Sum Weights 25 Mean 0 Sum Observations 0 Std Deviation Variance Skewness Kurtosis Uncorrected SS Corrected SS Coeff Variation. Std Error Mean Tests for Normality Test --Statistic p Value Shapiro-Wilk W Pr < W Kolmogorov-Smirnov D Pr > D Cramer-von Mises W-Sq Pr > W-Sq Anderson-Darling A-Sq Pr > A-Sq Fall, 2013 Page 26
27 Histogram of Residuals Fall, 2013 Page 27
28 QQ Plot Fall, 2013 Page 28
29 Time Serial Plot Fall, 2013 Page 29
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