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Randomized Block Design
In chapter on paired t-tests, we learned to “match” subjects on variables that: influence performance but are not of interest. Matching gives a more sensitive test of H0 because it removes sources of variance that inflate 2. Lecture 17
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Randomized Block Design
In the analysis of variance, the matched subjects design is called the Randomized Block Design. subjects are first put into blocks a block is a group matched on some variable subjects in a block are then randomly assigned to treatments for p treatments, you need p subjects per block Lecture 17
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Question: does SSB come from SSE or from SST?
Sums of squares In the RBD, we compute SST as before. Compute SSB (SS for Blocks) analogously: Compute deviations of block means from grand mean. Square deviations, then add them up. Question: does SSB come from SSE or from SST? Lecture 17
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Where does SSB come from?
SST SSB SSTotal SSE Residual SSE Lecture 17
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Conceptual Formulas SST = Σb(XTi – XG)2 p-1 SSB = Σp(XBi – XG)2 b-1
SSTotal = Σ(Xi – XG)2 n-1 SSE = SSTotal – SST – SSB (b-1)(p-1) = n-b-p+1 MST = SST/(p-1) MSB = SSB/(b-1) MSE = SSE/(b-1)(p-1) = SSE/(n-b-p+1) Lecture 17
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Summary table Source df SS MS F Treat p-1 SST SST/(p-1) MST/MSE
Blocks b-1 SSB SSB/(b-1) MSB/MSE Error n-p-b+1 SSE SSE/(n-b-p+1) Total n-1 SSTotal Lecture 17
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Computational Formulas
CM = (ΣX)2/n SSTotal = ΣX2 – CM SST = ΣTi2/b – CM SSB = ΣBi2/p – CM SSE = SSTotal – SST – SSB p = # of samples b = # of blocks Ti = Total for ith treatment Bi = Total for ith block Lecture 17
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Randomized Block Design – Example 1
H0: 1 = 2 = 3 HA: At least two differ significantly Statistical test: F = MST/MSE Rej. region: Fobt > F(2, 8, .05) = 4.46 Lecture 17
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Randomized Block Design – Example 1
CM = SSTotal = ΣX2 – CM = … – = – = 363.6 Lecture 17
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Randomized Block Design – Example 1
SST = Σ(Ti2)/b – CM = 4012/ / /5 – = – = 98.8 SSB = 2442/3 + … /3 – = – = Note how much SSB is removing from SSE here. Lecture 17
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Randomized Block Design – Example 1
SSE = SSTotal – SST – SSB = – 98.8 – = 23.87 Lecture 17
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Randomized Block Design – Example 1
Source df SS MS F Treat Blocks Error Total Lecture 17
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Randomized Block Design – Example 1
H0: SN = GO HA: SN ≠ GO (Note: this is a post-hoc test. We’ll do N-K.) Statistical test: Q = Xi – Xj √MSE/n Lecture 17
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Randomized Block Design – Example 1
Rank order sample means: Sleepy Sneezy Grumpy Dopey Goofy Qcrit = Q(4, 8, .05) = 4.53 Lecture 17
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Randomized Block Design – Example 1
Qobt: 86 – = = 6.35 √(2.984)/ Reject H0. Goofy & Sneezy differ significantly. Lecture 17
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