Lecture 6.1: Quiz 3 prep • Use ANOVA Table and formulae to compute F from partial summaries. HW4, #G • Use algebra with: Y = b 0 + b 1 X 1 + b 2 X 2 + b 3 X 1 X 2 ; to discuss interaction with lin & non-lin terms; simple effects, etc. HW4, #A, B • Correlation, partial correlation, simple causal models, mediation, Sobel test in the linear, additive model. HW4, #C, E • Interpret & construct planned contrasts. HW4, #D, F 1
Q3 Topics; another version • Topics relevant to Quiz 3 & HW-4 • Partial correl, and its use in testing causal models with 3 variables. Drawing such causal models, e.g., X è Z è Y. • Sobel test of mediation • Design a contrast to test a specific effect; interpret a given contrast; orthogonal contrasts; t- or F- tests for a contrast • Linear, X.lin , & quadratic, X.quad , effects of a quantitative predictor, X ; interactions with a categorical predictor, G . • Compute missing entries in an ANOVA table. 2
HW-4 exercise • The stress level was measured for each of 24 research participants. Six participants were chosen from each of 4 groups: Group 1 was Low in Social Support (e.g., number of friends) and Low in Self-Concept ; Group 2 was High in Social Support and Low in Self-Concept ; Group 3 was Low in Social Support and High in Self- Concept ; and Group 4 was High in Social Support and High in Self-Concept . Shown below are edited sections from a Minitab analysis of the data. • Complete the ANOVA table ; interpret and test ( New question! ) a given contrast; calculate the s.e. of a sample mean; describe the effects of Social Support and Self-Concept on stress . 3
G. DV = stress level; N = 24; n = 6 in each of k = 4 groups. Group 1 was Low in Social Support and Low in Self- Concept ; Group 2 was (Hi, Lo); Group 3 was (Lo, Hi); Group 4 was (Hi, Hi). SelfConcept Group SocSupp SelfCon l � 1 � 2 All � Mean � m 1 Lo Lo 1 SS1 10.4 3.4 6.9 � m 2 Hi Lo -1 SS2 3.6 3.2 3.4 � m 3 Lo Hi -1 All 7.0 3.3 5.1 � m 4 Hi Hi 1 1. Complete the 1-way ANOVA table on next slide and describe the data (using plots and tables, not reg coeffs!). 2. Interpret the contrast, l = (m 1 - m 2 ) – (m 3 - m 4 ), in terms of SS & SC. Test if l is sig different from 0 ( new question !). 4
Analysis of Variance for Stress � Source DF SS MS F P � Group aa bb bb bb < bb � Error aa 49.80 bb � Total aa 272.60 � � Pooled StDev = 1.578 � � 1. DF: {aa, aa, aa} = { , , } � 2. SS(Group) = � 3. MS(Group) = � � ; MS(Error) = � 4. F = � 5. P via R or Tables � 5
Analysis of Variance for Stress � Source DF SS MS F P � Group aa bb bb bb < bb � Error aa 49.80 bb � Total aa 272.60 � � Pooled StDev = 1.578 � � 1. DF: {aa, aa, aa} = {3, 20, 23} � 2. SS(Group) = � 3. MS(Group) = � � ; MS(Error) = � 4. F = � 5. P via R or Tables � 6
Analysis of Variance for Stress � Source DF SS MS F P � Group aa bb bb bb < bb � Error aa 49.80 bb � Total aa 272.60 � � Pooled StDev = 1.578 � � 1. DF: {aa, aa, aa} = {3, 20, 23} � 2. SS(Group) = 272.6 – 49.8 = 222.8 � 3. MS(Group) = � � ; MS(Error) = � 4. F = � 5. P via R or Tables � 7
Analysis of Variance for Stress � Source DF SS MS F P � Group aa bb bb bb < bb � Error aa 49.80 bb � Total aa 272.60 � � Pooled StDev = 1.578 � � 1. DF: {aa, aa, aa} = {3, 20, 23} � 2. SS(Group) = 272.6 – 49.8 = 222.8 � 3. MS(Group) = 222.8/3 = 74.3 � 4. MS(Error) = 49.8/20 = 2.49; � 1. s = sqrt(MSE) = sqrt(2.49) = 1.578. � 5. F = MSB/MSE = 74.3/2.49 = 29.8, p < .00n � 6. P via R or Tables � 8
• Formal test of l = (m 1 - m 2 ) – (m 3 - m 4 ) : 2 2 = MS w a i t N − k = l ∑ ∑ l = 1, N − k = t N − k 2 a i x i , se l , or F . n i se l i l = (10.4 − 3.6) − (3.4 − 3.2 = 6.6, ⎛ ⎞ se 2 = (2.49)* 1 6 + 1 6 + 1 6 + 1 ⎟ = 1.66 ⎜ ⎝ ⎠ 6 t 20 = l se = 6.6 p = 5.2*10 − 5 . 1.29 = 5.12; The interaction between Soc Support and Self- Concept on the DV is sig; etc. (i.e., describe the interaction using a plot or table ). 9
Another HW-4 exercise • Analysis of SS: SS t = SS b + SS w , etc • Use ANOVA formulae to compute F from partial summaries (HW-4) 10
I . DV = satisfaction with dispute resolution 1. Complete the 1-way ANOVA. 2. Ans. What are the relevant quantities, and what formulae do we have for calculating them?
Use of formulae to compute F from partial summaries • Journal articles sometimes give the means and s.d.s of k groups without F and p -values. Without the raw data you cannot use software to check the statistics; but you can do it ‘ by hand ’ using the appropriate formulae. • Need to calculate: SS b and SS w ; the df of these 2 sums of squares; use the df to compute MS b and MS w ; compute the F ratio; and test its significance. 12
n j k k ∑ ∑ ∑ C = T 2 / N T j = ; T = ; N = x ij T j n j ; i = 1 j = 1 j = 1 ∑ ∑ SS tot = − C = SS b + SS w 2 x ij j i k k ∑ ∑ SS j = ( n j − 1) s j 2 , SS w = ; SS b = / n j ) − C 2 SS j ( T j j = 1 j = 1 MS w = SS w /( N − k ); MS b = SS b /( k − 1) F = MS b / MS w with ( k − 1, N − k ) df. 13
2 T j = n j x j SS j = ( n j − 1) s j 2 / n j 2 Group n j x j s j T j Manuf 12 25.2 3.6 302.4 39.6 7620.48 Markt 14 32.6 4.8 Resrch 11 28.1 3.6 ----------------------------------------------- .... Total 37 -- -- 1067.9 ... C = T 2 /N = 30821.9, SS w = 155, … , F = 8.53, p < .001. 14
• Exercise : The manufacturing group is the 'standard' group, and the other two groups are (i) possibly different from the standard, and (ii) possibly different from each other. Set up two orthogonal contrasts (involving the group means) to test these hunches. Carry out the appropriate tests. • Check that the contrasts l 1 = (2, -1, -1) and l 2 = (0, -1, 1) for (man, mar, res), respectively, test the 2 hunches, and are orthogonal. • For a formal test , we compute the values of l 1 and l 2 , their s.e.’s, and t or F . ( Assume the DV = ‘satisfaction’ ) 15
• Formal test for l 1 : 2 2 = MS w a i t N − k = l ∑ ∑ l = 1, N − k = t N − k 2 a i x i , se l , or F . n i se l i l 1 = 2*25.2 − 1*32.6 − 1*28.1 = − 10.3, ⎛ ⎞ 12 + ( − 1) 2 + ( − 1) 2 2 = (21.264)* 2 2 ⎟ = 10.54 se 1 ⎜ ⎝ ⎠ 14 11 = − 10.3 t 34 = l 1 3.247 = − 3.172; p = .0032. se 1 Satisfaction is sig lower in 'manuf' than the other groups. 16
• Formal test for l 2 : 2 2 = MS w a i t N − k = l ∑ ∑ l = 1, N − k = t N − k 2 a i x i , se l , or F . n i se l i l 2 = 0*25.2 − 1*32.6 + 1*28.1 = − 4.5, ⎛ ⎞ 2 = (21.264)* ( − 1) 2 + (1) 2 ⎟ = 3.452 se 2 ⎜ ⎝ ⎠ 14 11 = − 4.5 t 34 = l 2 1.858 = − 2.422; p = .021. se 2 Satisfaction is sig higher in 'marketing' than in 'research'. 17
HW-4: Interaction between categorical, train , and quantitative, difficulty • Performance (‘score’) on a task is expected to depend on the ‘difficulty’ ( D , quant: 1, … , 5) of the task, and the level of expertise of the participant (‘train’, T , = novice or expert). • Can examine linear and quadratic effects of D , and their interactions with T in a number of ways: lm(score ~ train * (difficulty + lm(score ~ train * (difficulty + I(difficulty^2)), I(difficulty^2)), or lm(score ~ train * (scale(difficulty) + lm(score ~ train * (scale(difficulty) + I(scale(difficulty)^2)), I(scale(difficulty)^2)), or lm(score ~ train * poly(difficulty, 2)). lm(score ~ train * poly(difficulty, 2)). 18
Describing interactions • Is there a significant interaction between train and d.lin, or between train and d.quad? If so, use the plot of means to describe it. (The regression coeffs are usually not reliable by themselves.) • For some cases, such an informal description is sufficient. At other times, we want to be more precise in our description of the interaction; e.g., does d.lin have a sig effect for novices, or does train have a sig effect when ‘difficulty’ = 3? For this purpose, we need to test formally the specific simple effects . 19
• When there is no interaction, main effects are easily interpreted. When there is an interaction, main effects can be misleading, and simple effects are more useful for description. • The preceding 3 models of lm() differ in how ‘difficulty’ or ‘train’ is coded; but they do not differ in substance . How do the models compare in their convenience for testing simple effects ? • The 1 st model uses D (> 0) and D 2 , which are correlated. This can lead to the problem of multicollinearity, namely, ‘large’ s.e’s for the regression coefficients. But sometimes D 2 is just the predictor we want! 20
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