**13.12 Latin Squares (Optional)**

The randomized block design is very effective for reducing experimental error by removing one source of variation. Another esign that is particularly useful in controlling two sources of variation, while reducing tile required number of treatment combinations, is called the Latin square. Suppose that we are interested in the yields of 4 varieties of wheat using 4 ifferent fertilizers over a period of 4 years. The total number of treatment combinations for a completely randomized design would be 64. By selecting the same number of categories for all three criteria of classification, we may select a Latin square esign and perform the analysis of variance using the results of only 16 treatment combinations. A typical Latin square, selected at andom from all possible 4 x 4 squares, is the following:

Column Row 1 2 3 4 1 A B C D 2 D A B C 3 C D A B 4 B C D A

The four letters, , Ii, C, and D, represent the 4 varieties of wheat that are referred to as the treatments. The rows and columns, represented by the 4 fertilizers and the 4 years, respectively, are the two sources of variation that we wish to control. We now see that each treatment ccurs exactly once in each row and each column. With such a balanced arrangement the analysis of variance enables one to eparate the variation due to the different fertilizers and different years from the error sum of squares and thereby obtain a more ceurate test for differences in the yielding capabilities ~f the 4 varieties of wheat. When there • is interaction present between any f the sources of variation, the fvalues in the analysis of variance are no longer valid. In that case, the Latin square design would ( inappropriate.

**Generalization to the Latin Square **

We now generalize and consider an r x r Latin square where !lij” denotes n observation in the ith row and jth column corresponding to the kth letter .. Note that once iand i are specified for a particular atin square, we automatically know the letter given by k. For example, when i = 2 and i = 3 in the 4 x 4 Latin square above, we ave ::::2 B. Hence k is a function of i and j. If o, and {3j are the effects of the ith row aDd ;th column, Tk the effect of the kth treatment, . he grand mean, and £ijle the random error, then we can write Jlijle = P. + 0. + {3j + Tk + £ijle, where we impose the restrictions La. = LIJj = LTk =0 As before, the Yijk are assumed to be values of independent random variables having normal distributions with means J1.ijk = J1.+ Qi + {3j-+Tk and ommon variance q2. The hypothesis to be tested is as follows:

H o: Tl = T2 = Tr = 0,

HI: At least one of the Ti’S is not equal to ero. This test will be based on a comparison of independent estimates of q2 provided by splitting the total sum of squares of our ata into four components by means of the following identity. The reader is asked to provide the proof ill Exercise 13.37 on page 54.

Symbolically, we write the sum-of-squares identity as SST = SSR + sse + 8STr + SSE, where SSR and sse are called the row sum of quares and column sum of squares, respectively; SSTr is called the treatment sum of squares; and SSE is the error sum of squares. he degrees of freedom are partitioned according to the identity

r2 – 1 = (r – 1) + (r – 1) + (r – 1) +(r – l)(r – 2).

Dividing each of the um of squares on the right side of the sum-of-squares identity by their corresponding number of degrees of freedom, we obtain he four independent estimates .2 SSR51 = -r–l, 2 SSC 52 = r -1′ 2 SSTr 53 = r -1′ 2 SSE 5 = -:r-_—(=-I”‘r””:) (-r ~2)of q2. nterpreting the sums of squares as functions of independent random vari- . ables, it is not difficult to verify that

The analysis of variance (Table 13.11) indicates the appropriate F-test for rtemeets.