1 randomized block design 1.randomized block design 2.sums of squares in the rbd 3.where does ss b...

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Randomized Block Design

1. Randomized block design2. Sums of squares in the RBD3. Where does SSB come from?

4. Conceptual formulas5. Summary table6. Computational formulas7. Examples

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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.

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Randomized Block Design (RBD)

• In the analysis of variance, the matched subjects/within 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

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Sums of squares in the RBD

• We compute SSTreat as before. Compute SSB (SS for Blocks) analogously: • Compute deviations of block means from grand

mean.• Square deviations, then add them up.

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Sums of squares in the RBD

• SSTotal is composed of SSTreat + SSE

• SSB must come out of SSTotal

• Does SSB come from SSTreat or from SSE?

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Where does SSB come from?

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SSTotal

ResidualSSE

SSB

SSE

SST

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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+1MST = SST/(p-1)MSB = SSB/(b-1)MSE = SSE/(b-1)(p-1) = SSE/(n-b-p+1)

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Summary table

Source df SS MS F

Treat p-1 SSTreat 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

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Computational Formulas

CM = (ΣX)2 SSTotal = ΣX2 – CM

nSSE = SSTotal – SST – SSB

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Computational Formulas

SSTreat = ΣT2i – CM SSB = ΣB2

i – CM

b p

Ti=Total for ith treatment Bi=Total for ith block

b=# of blocks p=# of samples

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Randomized Block Design – Example 1a

1. Five Grade 10 high school students in an advanced math program are tested at the beginning of the term. Later in the term, they write a midterm and then a final exam. Each test/exam contains similar mathematics problems, and a comparison will be done to see whether significant differences exist between mean scores on the 3 exams. The students’ scores on the exams are: Student First exam Midterm Final Grumpy78 84 82Sneezy 81 86 91Dopey 79 80 83Goofy 77 80 82Sleepy 86 91 94

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a. Is there an overall significant difference between mean scores on the 3 exams ( = .05).

b. Although no specific predictions were made

beforehand, after inspecting the data it could be seen that Sneezy consistently obtained higher exam scores than Goofy. Regardless of the results of your analysis in part (a), perform a post-hoc test to determine whether Sneezy and Goofy differ significantly on their overall average on the 3 exams ( = .05).

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H0: 1 = 2 = 3

HA: At least two differ significantly

Statistical test: F = MSTMSE

Rej. region: Fobt > F(2, 8, .05) = 4.46

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CM = 104834.4

SSTotal = ΣX2 – CM

= 782 + 812 + … + 942 – 104834.4 = 105198 – 104834.4 = 363.6

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SSTreat = Σ(Ti2) – CM

b= 4012 + 4212 + 4322 – 104834.4 5 5 5= 104933.2 – 104834.4= 98.8

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SSB = ΣB2i – CM

p

SSB = 2442 + … + 2712 – 104834.4

3 3= 105075.33 – 104834.4= 240.93

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SSE = SSTotal – SSTreat – SSB

= 363.6 – 98.8 – 240.93

= 23.87

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Source df SS MS F

Treat 2 98.8 49.4 16.55Blocks 4 240.93 60.23 20.18Error 8 23.87 2.98Total 14 363.6

Decision: Reject HO – average scores do differ across exams.

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Randomized Block Design – Example 1b

H0: w = A

HA: W ≠ A

(Note: this is a post-hoc test. We’ll do N-K.)

Statistical test: Q = Xi – Xj

√MSE/n

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Rank order sample means:

Sleepy Sneezy Grumpy Dopey Goofy 90.3 86 81.3 80.6 79.67

Qcrit = Q(4, 8, .05) = 4.53

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Qobt:

86 – 79.67 = 6.33 = 6.35√(2.984)/3 0.997

Reject HO. Sneezy & Goofy differ significantly in their overall average on the 3 exams.

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Randomized Bloc Design – Example 2

2. People in a weight loss program are weighed at the beginning of the program, 3 weeks after starting, and 3 months after starting. The following are the weights (in pounds) of a random sample of 5 participants at each of these time periods.

PersonStart 3 Wks 3 Months

Mickey 210 201 193Minnie 245 240 242Hewey 236 228 200Dewey 197 190 167Louie 340 328 290

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Randomized Bloc Design – Example 2

a. Are there significant differences between weights across the 3 time periods? ( = .05)

b. Regardless of your answer in part a., perform the appropriate tests to determine at which time periods the participants’ mean weights differ significantly.

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H0: 1 = 2 = 3

HA: At least two differ significantly

Statistical test: F = MST

MSE

Rejection region: Fobt > F(2, 8, .05) = 4.46

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CM = 35072 = 819936.6 15

SSTotal = ΣX2 – CM

= 2102 + 2452 + … + 2902 – 819936.6 = 855701 – 819936.6 = 35764.4

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SSTreat = Σ(Ti2) – CM

b= 12282 + 11872 + 10922 – 819936.6 5 5 5= 821883.4 – 819936.6= 1946.8

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SSB = ΣB2i – CM

p

SSB = 6042 + 7272 … + 9582 – 819936.6

3 3 3= 852973.67 – 819936.6= 33037.07

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SSE = SSTotal – SSTreat – SSB

= 35764.4 – 1946.8 – 33037.07

= 780.5

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Source df SS MS F

Treat 2 1946.8 973.4 9.977Blocks 4 33037.07 8259.27

84.656Error 8 780.5 97.563Total 14 35764.4

Decision: Reject HO – weights do differ across the 3 time periods.

Lecture 17

1 randomized block design 1.randomized block design 2.sums of squares in the rbd 3.where does ss b...

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