Latin Square Design

1. A B C D D B C A C D A B D A B C Latin Square Design • If you can block on two (perpendicular) sources of variation (rows x columns) you can reduce experimental error when compared to the RBD • More restrictive than the RBD • The total number of plots is the square of the number of treatments • Each treatment appears once and only once in each row and column

2. Advantages and Disadvantages • Advantage: • Allows the experimenter to control two sources of variation • Disadvantages: • The experiment becomes very large if the number of treatments is large • The statistical analysis is complicated by missing plots and misassigned treatments • Error df is small if there are only a few treatments • This limitation can be overcome by repeating a small Latin Square and then combining the experiments - • a 3x3 Latin Square repeated 4 times • a 4x4 Latin Square repeated 2 times

3. Useful in Animal Nutrition Studies • Suppose you had four feeds you wanted to test on dairy cows. The feeds would be tested over time during the lactation period • This experiment would require 4 animals (think of these as the rows) • There would be 4 feeding periods at even intervals during the lactation period beginning early in lactation (these would be the columns) • The treatments would be the four feeds. Each animal receives each treatment one time only.

4. Mid Early Mid Late Early Late The “Latin Square” Cow A simple type of “crossover design” Are there any potential problems with this design?

5. ‘Row’ 1 2 3 4 1 2 3 4 A B C D B C D A C D A B D A B C ‘Column’ Uses in Field Experiments • When two sources of variation must be controlled • Slope and fertility • Furrow irrigation and shading • If you must plant your plots perpendicular to a linear gradient • Practically speaking, use only when you have more than four but fewer than ten treatments • a minimum of 12 df for error

6. Randomization First row in alphabetical order  A B C D E Subsequent rows - shift letters one position 4 3 5 1 2 A B C D E 2 C D E A B A B D C E B C D E A 4 A B C D E D E B A C C D E A B 1 D E A B C B C E D A D E A B C 3 B C D E A E A C B D E A B C D 5 E A B C D C D A E B Randomize the order of the rows: ie 2 4 1 3 5 Finally, randomize the order of the columns: ie 4 3 5 1 2

7. Linear Model • Linear Model: Yij =  + i + j +k + ij •  = mean effect • βi = ith block effect • j = jth column effect • k = kth treatment effect • ij = random error • Each treatment occurs once in each block and once in each column • r = c = t • N = t2

8. Analysis • Set up a two-way table and compute the row and column means and deviations • Compute a table of treatment means and deviations • Set up an ANOVA table divided into sources of variation • Rows • Columns • Treatments • Error • Significance tests • FT tests difference among treatment means • FR and FC test if row and column groupings are effective

9. Source df SS MS F Total t2-1 SSTot = Row t-1 SSR= SSR/(t-1) MSR/MSE Column t-1 SSC= SSC/(t-1) MSC/MSE Treatment t-1 SST = SST/(t-1) MST/MSE Error (t-1)(t-2) SSE = SSE/(t-1)(t-2) SSTot-SSR- SSC-SST The ANOVA

10. Means and Standard Errors Standard Error of a treatment mean Confidence interval estimate Standard Error of a difference Confidence interval estimate t to test difference between two means

11. Oh NO!!! not Missing Plots • If only one plot is missing, you can use the following formula: ^ Yij(k) = t(Ri + Cj + Tk)-2G [(t-1)(t-2)] • Where: • Ri = sum of remaining observations in the ith row • Cj = sum of remaining observations in the jth column • Tk = sum of remaining observations in the kth treatment • G = grand total of the available observations • t = number of treatments • Total and error df must be reduced by 1 • Alternatively – use software such as SAS Procedures GLM, MIXED, or GLIMMIX that adjust for missing values

12. Relative Efficiency • To compare with an RBD using columns as blocks RE = MSR + (t-1)MSE tMSE • To compare with an RBD using rows as blocks RE = MSC + (t-1)MSE tMSE • To compare with a CRD RE = MSR + MSC + (t-1)MSE (t+1)MSE

13. Numerical Examples To determine the effect of four different sources of seed inoculum, A, B, C, and D, and a control, E, on the dry matter yield of irrigated alfalfa. The plots were furrow irrigated and there was a line of trees that might form a shading gradient. A B D C E D E B A C C D A E B E A C B D B C E D A Are the blocks for shading represented by the row or columns? Is the gradient due to irrigation accounted for by rows or columns?

14. Data collection A B D C E 33.8 33.7 30.4 32.7 24.4 D E B A C 37.0 28.8 33.5 34.6 33.4 C D A E B 35.8 35.6 36.9 26.7 35.1 E A C B D 33.2 37.1 37.4 38.1 34.1 B C E D A 34.8 39.1 32.7 37.4 36.4

15. Source df SS MS F Total 24 296.66 Rows 4 87.40 21.85 7.13** Columns 4 16.56 4.14 1.35 Treatments 4 155.89 38.97 12.71** Error 12 36.80 3.07 ANOVA of Dry Matter Yield

16. Report of Statistical Analysis • Differences among treatment means were highly significant • No difference among inocula. However, inoculation, regardless of source produced more dry matter than did no inoculation • Blocking by irrigation effect was useful in reducing experimental error • Distance from shade did not appear to have a significant effect I source A B C D None SE Mean Yield 35.8 35.0 35.7 34.9 29.2 0.78 a a a a b LSD=2.4

17. Relative Efficiency • To compare with an RBD using columns as blocks RE = MSR + (t-1)MSE tMSE • To compare with an RBD using rows as blocks RE = MSC + (t-1)MSE tMSE • To compare with a CRD RE = MSR + MSC + (t-1)MSE (t+1)MSE (21.85+(5-1)3.07)/(5x3.07)=2.22 (2.22-1)*100 = 122% gain in efficiency by adding rows (4.14+(5-1)3.07)/(5x3.07)=1.07 (1.07-1)*100 = 7% gain in efficiency by adding columns (21.85+4.14+(5-1)3.07)/(6x3.07)=2.08 (2.08-1)*100 = 108% gain CRD would require 2.08*5 or 11 reps