# 1 Randomized Block Design In chapter on paired t-tests, we learned to “match” subjects on variables that: – influence performance – but are not of interest.

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1 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 H 0 because it removes sources of variance that inflate  2. Lecture 17

2 Randomized Block Design (RBD) 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

3 Sums of squares in the RBD We compute SS Treat as before. Compute SS B (SS for Blocks) analogously: – Compute deviations of block means from grand mean. – Square deviations, then add them up. SS Total is composed of SS Treat + SS E. Does SS B come from SS Treat or from SS E ? Lecture 17

4 Where does SS B come from? Lecture 17 SSTotal Residual SS E SS B SS E SS T

5 Conceptual Formulas SS T = Σb(X Ti – X G ) 2 p-1 SS B = Σp(X Bi – X G ) 2 b-1 SS Total = Σ(X i – X G ) 2 n-1 SS E = SS Total – SS T – SS B (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

6 Summary table Sourcedf SSMS F Treatp-1 SS Treat SS T /(p-1) MS T /MS E Blocksb-1 SS B SS B /(b-1) MS B /MS E Error n-p-b+1 SS E SS E /(n-b-p+1) Totaln-1 SS Total Lecture 17

7 Computational Formulas CM = (ΣX) 2 SS Total = ΣX 2 – CM n SS Treat = ΣT 2 i – CMSS B = ΣB 2 i – CM b p SS E = SS Total – SS T – SS B p=# of samplesT i =Total for i th treatment b=# of blocksB i =Total for i th block Lecture 17

8 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: StudentFirst examMidtermFinal Grumpy788482 Sneezy818691 Dopey798083 Goofy778082 Sleepy869194 Lecture 17

9 Randomized Block Design – Example 1a 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). Lecture 17

10 Randomized Block Design – Example 1a H 0 :  1 =  2 =  3 H A : At least two differ significantly Statistical test:F = MST MSE Rej. region:F obt > F (2, 8,.05) = 4.46 Lecture 17

11 Randomized Block Design – Example 1a CM = 104834.4 SS Total = ΣX 2 – CM = 78 2 + 81 2 + … + 94 2 – 104834.4 = 105198 – 104834.4 = 363.6 Lecture 17

12 Randomized Block Design – Example 1a SS Treat = Σ(T i 2 ) – CM b = 401 2 + 421 2 + 432 2 – 104834.4 5 5 5 = 104933.2 – 104834.4 = 98.8 Lecture 17

13 Randomized Block Design – Example 1a SS B = ΣB 2 i – CM p SS B = 244 2 + … + 271 2 – 104834.4 3 3 = 105075.33 – 104834.4 = 240.93 Lecture 17

14 Randomized Block Design – Example 1a SS E = SS Total – SS Treat – SS B = 363.6 – 98.8 – 240.93 = 23.87 Lecture 17

15 Randomized Block Design – Example 1a SourcedfSSMSF Treat2 98.849.416.55 Blocks4240.9360.2320.18 Error8 23.87 2.98 Total14363.6 Decision: Reject H O – average scores do differ across exams. Lecture 17

16 Randomized Block Design – Example 1b H 0 :  w =  A H A :  W ≠  A (Note: this is a post-hoc test. We’ll do N-K.) Statistical test:Q = X i – X j √MSE/n Lecture 17

17 Randomized Block Design – Example 1b Rank order sample means: SleepySneezyGrumpyDopeyGoofy 90.3 86 81.3 80.679.67 Q crit = Q (4, 8,.05) = 4.53 Lecture 17 r = 4

18 Randomized Block Design – Example 1b Q obt : 86 – 79.67=6.33= 6.35 √(2.984)/30.997 Reject H O. Sneezy & Goofy differ significantly on their overall average on the 3 exams. Lecture 17

19 RBD 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. PersonStart3 Wks3 Months Mickey210201193 Minnie245240242 Hewey236228200 Dewey197190167 Louie340328290 Lecture 17

20 RBD 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. Lecture 17

21 Randomized Block Design – Example 2a H 0 :  1 =  2 =  3 H A : At least two differ significantly Statistical test:F = MS T MS E Rej. region:F obt > F (2, 8,.05) = 4.46 Lecture 17

22 Randomized Block Design – Example 2a CM = 3507 2 = 819936.6 15 SS Total = ΣX 2 – CM = 210 2 + 245 2 + … + 290 2 – 819936.6 = 855701 – 819936.6 = 35764.4 Lecture 17

23 Randomized Block Design – Example 2a SS Treat = Σ(T i 2 ) – CM b = 1228 2 + 1187 2 + 1092 2 – 819936.6 5 5 5 = 821883.4 – 819936.6 = 1946.8 Lecture 17

24 Randomized Block Design – Example 2a SS B = ΣB 2 i – CM p SS B = 604 2 + 727 2 … + 958 2 – 819936.6 3 3 3 = 852973.67 – 819936.6 = 33037.07 Lecture 17

25 Randomized Block Design – Example 2a SS E = SS Total – SS Treat – SS B = 35764.4 – 1946.8 – 33037.07 = 780.5 Lecture 17

26 Randomized Block Design – Example 2a SourcedfSSMSF Treat21946.8973.49.977 Blocks433037.078259.2784.656 Error8 780.597.563 Total1435764.4 Decision: Reject H O – weights do differ across the 3 time periods. Lecture 17

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