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Unformatted text preview: OneFactor BetweenSubjects Analysis of Subjects Variance Variance
Introduction to ANOVAs OneFactor BetweenSubjects OneFactor Analysis of Variance Analysis
New test statistic F = MSA MSerror R.A. Fisher developed test MS = Mean Square = Variance s2 = ∑ (XX)2 = SS N1 df Why F and not a ttest?
Used when IV (Factor) has more than two levels A1 A2 A3 or more! Why not multiple ttests? α A1 vs A2 .05 A1 vs A3 .05 A2 vs A3 .05 total α =.15 Use F to control total to .05 or .01 Mean Squares for F
F = MSA = MSbetween group MSerror = MSwithin group MSA derived from variation in sample means Sources of variance: 1. effect of IV 2. sampling error MSerror derived from variation in individual scores from the mean of their level of A. (individual differences) Statistical Hypothesis for F
H0: μA1= μA2 = μA3 H : not all μs are equal 1 If H is true 0 MSA ≈ MSerror F = MSA = 1.0 MSerror If H is true 1 MSA > MSerror: F>>1.0 Conceptual logic of ANOVA Analysis of Variance alysis Va
IV The type of drawings used to present 30 common objects DV The number of common objects verbally recalled (incomplete outline) A1 17 13 16 19 20 (outline) A2 (detailed) A3 14 13 16 12 10 13.0 2.24 15 10 11 12 12 12.0 1.87 nA= 5 N=15
XG=14.0 XA 17.0 2.74 sA Ftest requires
MSA = SSA dfA MS error = SSerror dferror or F = MSA = SSA MSerror dfA SSerror dferror Find the parts by partitioning (or analyzing) the total variance into two parts: MSA and MSerror General Notation
Xij i = subject reference 1, 2, … j = group reference A1, A2, … nA = # of subjects in each level of A N = Total # of subjects XG = Grand Mean of all scores A1 X11 X21 X31 X41 X51 XA1 A2 X12 X22 X32 X42 X52 XA2 A3 X13 X23 X33 X43 X53 XA3 Computing SS, SS, SS 1) 2) 3) 4) Steps Partition each score into 3 parts – all scores Compute numerical deviations (d) Square each deviation (d) Sum the squared deviations Sum the squared deviations
SSTotal = ∑ ∑ (Xij – XG)2
j=1 i=1 a na SSA = ∑ ∑ (XA XG)2
j=1 i=1 a na SSError = ∑ ∑ (Xij XA)2
j=1 i=1 a na Partitioning the Variance
Total variation = variation due + variation due in a score to treatment to error or IV (Xij – XG) = (XA – XG)+(Xij – XA)
Total Variance = MSA + MSError Example: X41 =19, XA1 =17, XG =14 (1914) = (1714) + (1917) 5 = 3 + 2 (Xij – XG) d d2 (XA – XG) d d2 (Xij – XA) d d2 X11 1714 3 9 1714 3 9 1717 0 0 X21 1314 1 1 1714 3 9 1317 4 16 X31 1614 2 4 1714 3 9 1617 1 1 X41 1914 5 25 1714 3 9 1917 2 4 X51 2014 6 36 1714 3 9 2017 3 9 X12 1414 0 0 1314 1 1 1413 1 1 X22 1214 1 1 1314 1 1 1313 0 0 X32 1614 2 4 1314 1 1 1613 3 9 X42 1214 2 4 1314 1 1 1213 1 1 X52 1414 0 0 1314 1 1 1013 3 9 X13 1514 1 1 1214 2 4 1512 3 9 X23 1014 4 16 1214 2 4 1012 2 4 X33 1114 3 9 1214 2 4 1112 1 1 X43 1214 2 4 1214 2 4 1212 0 0 X53 1214 2 4 1214 2 4 1212 0 0 Important General Points General Equation: ∑∑(Xij – XG)2 = ∑∑(XA – XG)2+ ∑∑(Xij – XA)2 SSTotal = SSA + SSError Data example 134 = 70+64 Determining Degrees of Freedom SSTotal = ∑ ∑ (Xij – XG)2 dfTotal= N1 df: the number of pieces of information that are free to vary in computing a statistic SSA = ∑ ∑ (XA XG)2 dfA =a1 SSError = ∑ ∑ (Xij XA)2 dfError= a(na1) = Na Degrees of Freedom computation dfTotal = N1 = 151 = 14 dfA = a1 = 31 = 2 dfError = Na = 153 = 12 Degrees of Freedom are additive dfTotal= dfA + dfError 14 = 2 + 12 Example: na=5, N=15, a=3 F for Example
Now know: SSTotal = 134 dfTotal = 14 SSA = 70 dfA = 2 SSError = 64 dfError = 12 F = MSA = SSA 70 MSError dfA 2 = SSError 64 dfError 12 F = 35.0 = 6.57 5.33 ANOVA Summary Tables
Source Factor A Error Total df SS MS SSA dfA SSError dfError F MSA MSError a1 ∑∑(XA–XG)2 Na ∑∑(Xij–XA)2 N1 ∑∑(Xij–XG)2 Example Values:
Source Memory Cond Error Total df 2 12 14 SS 70.0 64.0 134.0 MS 35.0 5.33 F 6.57 Sampling Distribution of F
F(2,120 =3.88 α = .05
0 1 2 3 3.88 Skewed: minimum = 0, max = ∞ Mode is 1.0: Expected value if H0 is true Critical region always in the positive tail: MSA >MSError Finding Critical Value for F Table A3a: α = .05 P. 492 Table A3b: α = .01 P. 493 F = MSA = dfMSa = dfnumerator MSError dfMSerror dfdenominator Table A3a
1 1
Degrees of Freedom for Denomina tor F(2,12) F(N,D)
2 199.5 19.00 9.55 3.98 3.88 Degrees of Freedom for Numerator 3 214.7 19.16 9.28 3.59 3.49 4… 224.6 19.25 9.12 3.36 3.26 161.4 18.51 10.13 4.84 4.75 2 3 ... 11 12 5Step Procedure H1: not all μs are equal 1. H0: μA1= μA2 = μA3 2. α = .05 3. Set Decision Stage a. use Ftest: 3 groups b. Sampling Distribution c. critical value of F(2,12)=3.88, α=.05 d. reject H0 if Fobs ≥3.88 4. Compute F (ANOVA)
Source Memory Cond Error Total df 2 12 14 SS 70.0 64.0 134.0 MS 35.0 5.33 F 6.57 5. Decision: Reject H0, accept H1 Fobs(2,12) = 6.57 ≥ Fcrit(2,12) = 3.88 Assumptions of FTest. 1) Each subject randomly and independently drawn from the population 2) Scores are distributed normally in the population 3) The variances of scores in the population are equal When violations of assumption When causes problems causes 1) unequal #s of subjects in conditions 2) shape of distribution for each sample are different and they are peaked or flat 3) α = .01 What we know when we reject H0? We know not all μs are equal… …However, we don’t know which means are different μA1≠ μA2 ? μA1≠ μA3 ? μA2 ≠ μA3 ? μA1≠ μA2 ≠ μA3 Following up the Overall FTest Omnibus FTest
Post hoc comparisons: Tukey HSD test Honestly Significant Differences Computes a CD Critical Difference: The minimum numerical difference between two treatment means that is statistically significant CD = q MSError nA Where q = an entry from the table of the Studentized Range Statistic Tukey HSD test for memory Tukey experiment experiment
CD = q MSError MSError = 5.33 nA nA = 5 CD = 3.77 5.33 q=3.77 for: 5 α = .05 CD = 3.77 1.067 a=3 CD = 3.9 dfError=12 Two means are significantly different from one another if their absolute difference is 3.9 units or more Comparison XA1 vs. XA2 (17.013.0) XA1 vs. XA3 (17.012.0) XA2 vs. XA3 (13.012.0) Absolute value of comparison 4.0 5.0 1.0 Statistical hypotheses H0: μA1= μA2 H1: μA1≠ μA2 H0: μA1= μA3 H1: μA1≠ μA3 H0: μA2= μA3 H1: μA2≠ μA3 Decisions Reject H0 Accept H1 Reject H0 Accept H1 Do not Reject H0 Do not Accept H1 CD = 3.9 Matrix Presentation of Tukey HSD
Memory Cond
A1 (17.0) A2 (13.0) A3 (12.0) Memory Cond A1 (17.0) A2 (13.0) A3 (12.0) __ 4.0* __ 5.0* 1.0 __ * p<.05 CD = 3.9 Situations in which violations of Situations assumptions probably don’t affect α (too much) (too
1. 2. 3. Equal numbers of subjects in all conditions Shape of the distributions from which scores were sampled are similar, and not very peaked or flat Use α = .05, rather than α =.01 Measuring Strength of Effect One Factor ANOVA
η2 =SSA OR η2 = (dfA)(Fobs) SSTotal (dfA)(Fobs)+dfError η2 = 70 OR η2 = (2)(6.57) 134 (2)(6.57)+12 η2 = .52 OR η2 = 13.14 13.14+12 η2 =.52 ...
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This note was uploaded on 01/16/2011 for the course PSYC 274 at USC.
 '07
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