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27-08-2012                                                                                                                ...
27-08-2012                                            Practice Problem                                                    ...
27-08-2012                                                         Post Hoc Analysis  Assumptions in one-way ANOVA        ...
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Session 17

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Session 17

  1. 1. 27-08-2012 Shudhho Chinta: Part 3 Analysis Of VAriance Are the employees in the five different ANOVA offices equally bad? Session XVII Practice problem The Govt a/c office is interested in seeing whether similar-sized offices spend similar amounts on 1 way (and 2 way?) ANOVA personnel and equipment. Monthly expenses of 3 Comparing Means of multiple populations offices have been examined: 1 each in agriculture dept, state dept. and Interior dept. The data (expense in lakh Rs )from some past months are given below: Agriculture: 10 8 11 9 12 State: 15 9 8 10 13 13 Interior: 8 16 12 At 1% significance level, are there differences in expenses for the different offices? Splitting the SUM OF SQUARES(SS) Algebra in Splitting the SUM OF SQUARES: Practice Problem 6 mean s.d nAgriculture 10 8 11 9 12 10.00 1.58 5State 15 9 8 10 13 13 ( X ij − X ) = ( X ij − X i ) + ( X i − X ) 11.33 2.73 6Interior 8 16 12 12.00 4.00 3 Grand 11 2.602 14 ∑ ∑ ( X ij − X ) 2 = ∑ ∑ ( X ij − X i ) 2 + ∑ ∑ ( X i − X )2 2 4×1.582 + 5 ×2.732 + 2×42 = ∑ ( ni − 1) S i + ∑ ni ( X i − X ) 2 Total SS = within group SS + Between group SS Total SS χ 2 (if H0) χ 13×2.6022 5×(10-11)2 + 6 ×(11.33-11)2 +3×(12-11)2 n −1 within group SS χ 2 n− k Between group SS 2 k −1 2(after dividing by σ ) independent χ 2 n −1 χ 2 n−k χ 2 k −1After dividing by d.f. MSE MS due to ‘group’ 1
  2. 2. 27-08-2012 Practice Problem ANOVA TABLE Between group SS k -1 Test statistic is = Within group SS = Fk −1,n −k n -k ANOVA P-value VAR00001 At α=0.01, the C.R. is TS > 7.21 Sum of Squares df Mean Square F Sig. Between Groups 8.667 2 4.333 .601 .565Observed MSE = (4×1.582 + 5 ×2.732 + 2×42 )/11=7.2121 Within Groups 79.333 11 7.212MS due to group = [5×(10-11)2 + 6 ×(11.33-11)2 +3×(12-11)2 ]/2=4.33 Total 88.000 13So observed value of TS = 0.60So do not reject H0 at α =0.01 Observed value of TS ANOVA TABLE Three unbiased estimates of σ2Source ofvariation Sum of squares (SS) Degrees of freedom Mean SS F Statistic (under H0) (df)Between group k-1 ∑ n j ( x j − x ) 2 = SSb ˆ2 σb = SS b σ b2 ˆ = Fk −1,n− k • Usual S2 based on all n observation TSS/(n-1) k −1 ˆ2 σw • Pooled estimateWithin group n-k ∑ (n j − 1) S 2 = SS w j ˆ2 σw = SS w – SSE/(n-k) 2 n−k • Indirect estimate from σX – SS due to treatment/(k-1)TOTAL n-1 ∑∑ ( xij − x ) 2 = SST S2 = SST n −1 The last estimate is likely to be large compared to the second if the null hypothesis is not true This column does not add up! E (Mean square error/within group)=σ 2 ; k Usually not included in the ANOVA table 2 ∑ n (µ i =1 i i − µ )2 E (Mean square between group)=σ + . k −1 ANOVA: The Model in ANOVA and A comparison of three estimates of σ2 estimating the parameters µ3 µ2 µ1 in the model Sampling distribution of X X ij ֏ N ( µ + α i , σ 2 ) µ1=µ2=µ3 Assume temporarily that n1=n2=n3. µ = X .. ˆ( X A − X )2 + ( X S − X )2 + ( X I − X ) 2 σ2 α i = X i. − X .. ˆ is an estimate for σ2 = X 2 ni σ 2 = MSE ˆ 2 2 2n A ( X A − X ) + nS ( X S − X ) + nI ( X I − X ) is an estimate for σ2 2 2
  3. 3. 27-08-2012 Post Hoc Analysis Assumptions in one-way ANOVA Multiple pair-wise comparison Controversial• Each population is normal• Each population has equal variance • Post-Anova analysis: if one rejects H0, then• Samples are drawn independently from the what ? different population – Fisher’s LSD: Check if 1 1 | X i − X j | > tν × MSE × + ni n j Exercise• Complete the work on Shudhho Chinta: Pt2,3 3

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