Analyzing a Latin Square With One Missing Response

Why One Missing Response Changes the Analysis

A Latin square places each of $p$ treatments once in every row and once in every column. With a complete square, the row, column, and treatment effects are mutually orthogonal in the standard additive analysis. If one response is missing, however, the observed design is no longer balanced in the same way. Estimating the missing response can help describe the completed table, but it does not restore the exact orthogonality needed for the usual treatment F test.

The safe distinction is between estimating the missing cell and testing treatments. The least-squares estimate has a simple formula under the additive model. For an exact treatment test, fit the additive model to the observed responses and compare it with the model that includes rows and columns but omits treatment.

Model and Least-Squares Estimate

Assume a $p \times p$ Latin square with one missing response, where $p \geq 3$. Write the response in row $i$, column $j$, assigned treatment $k$, as $Y_{ijk}$. The additive model is

Y_{ijk}=\mu+\rho_i+\gamma_j+\tau_k+\varepsilon_{ijk}, \qquad \sum_{i=1}^p \rho_i= \sum_{j=1}^p \gamma_j= \sum_{k=1}^p \tau_k=0.

Here $\mu$ is the overall mean, $\rho_i$ is the effect of row $i$, $\gamma_j$ is the effect of column $j$, and $\tau_k$ is the effect of treatment $k$. For the usual exact F tests, assume the observed errors are independent and $N(0,\sigma^2)$, with a common variance, and that the additive model is appropriate. In particular, the model has no treatment-by-row or treatment-by-column interactions. The randomization and interpretation limits of that assumption matter: a missing-value calculation cannot correct a failure of additivity.

Definition: Let the missing response lie in row $i$, column $j$, and treatment $k$. Let $R_i$, $C_j$, and $T_k$ be the observed totals for that row, column, and treatment, and let $G$ be the grand total of all observed responses. Each of these totals excludes the missing response. The least-squares estimate of the missing response is $$ \widehat{Y}_{ijk} = \frac{p(R_i+C_j+T_k)-2G}{(p-1)(p-2)}. $$

The formula follows from the fitted value for a cell in a complete additive Latin square. If the missing entry is provisionally filled with $x$, the fitted value at its cell is the row mean plus the column mean plus the treatment mean minus twice the grand mean. Because the totals now include $x$, the fitted value is

\frac{R_i+x}{p}+\frac{C_j+x}{p}+\frac{T_k+x}{p} -2\frac{G+x}{p^2}.

At the least-squares estimate, $x$ equals its fitted value: including a value equal to its model prediction contributes zero residual, so it does not change the least-squares equations for the observed responses. Solving that equality gives the stated result:

x=\frac{R_i+C_j+T_k+3x}{p}-\frac{2(G+x)}{p^2} \quad\Longrightarrow\quad (p^2-3p+2)x=p(R_i+C_j+T_k)-2G,

and $p^2-3p+2=(p-1)(p-2)$. This derivation also shows why the totals in the formula must exclude the missing response.

Worked Example: Estimating the Missing Cell

A hypothetical $3 \times 3$ Latin square has the following treatment layout and observed responses. The missing cell is in row 1, column 1, and is assigned treatment A.

	Column 1	Column 2	Column 3
Row 1	A: missing	B: 9	C: 11
Row 2	B: 9	C: 11	A: 10
Row 3	C: 11	A: 8	B: 11

The relevant observed totals are $R_1=9+11=20$, $C_1=9+11=20$, and $T_A=10+8=18$. The grand total of the eight observed responses is $G=80$. Substituting $p=3$ gives

\widehat{Y}_{111} = \frac{3(20+20+18)-2(80)}{(3-1)(3-2)} = \frac{174-160}{2} =7.

Thus the additive-model least-squares estimate for the missing response is 7 response units. It is an estimate under the specified model, not an observed measurement. Inserting 7 gives a completed table whose fitted value at that cell is 7, but that completion alone does not make the ordinary treatment F test exact.

Degrees of Freedom After One Omission

There are $p^2-1$ observed responses. The additive model with rows, columns, and treatments has rank $3p-2$: one overall mean and $p-1$ independent effects for each of the three factors. Consequently, the residual degrees of freedom for the full model fitted to the observed data are

$$ df_E=(p^2-1)-(3p-2)=(p-1)(p-2)-1. $$

The treatment term still has $p-1$ degrees of freedom. The total degrees of freedom for the observed responses, after estimating their observed-data mean, are $p^2-2$. In a bookkeeping ANOVA table based on the completed square, the usual residual degrees of freedom $(p-1)(p-2)$ are reduced by one; the total degrees of freedom are also reduced by one. These adjustments describe the loss of one response, but they do not by themselves establish an exact treatment test.

Worked Example: Adjusting Degrees of Freedom

Suppose one response is missing from a hypothetical $5 \times 5$ Latin square. There are $25-1=24$ observed responses. The full additive model has rank $3(5)-2=13$, so its residual degrees of freedom are

$$ df_E=24-13=11. $$

Equivalently, the complete-square residual degrees of freedom would be $(5-1)(5-2)=12$, and one is lost, leaving 11. The treatment term has $5-1=4$ degrees of freedom. The observed-data total degrees of freedom are $24-1=23$. The model containing only rows and columns has rank $2(5)-1=9$, giving it $24-9=15$ residual degrees of freedom. Adding treatments therefore uses $15-11=4$ degrees of freedom, as it should.

Why the Ordinary Completed-Table Treatment Test Is Not Exact

For a complete Latin square, the usual treatment sum of squares is orthogonal to the row and column effects. After a cell is omitted, treatment contrasts are generally no longer orthogonal to those effects. A treatment sum of squares computed from the completed table is therefore not generally the same as the reduction in residual sum of squares obtained by adding treatment to a model that already contains rows and columns.

This distinction matters for the reference distribution. Under independent normal errors with common variance, a partial F test comparing nested linear models has an exact F distribution. But those assumptions do not make a completed-table treatment mean square divided by the residual mean square an exact F statistic: the needed orthogonality and equal-variance numerator structure are not generally present. Completed-table component sums of squares can still be reported descriptively, or used in an explicitly approximate procedure, but they should not be presented as an exact treatment test.

Key result: For an exact treatment test, calculate the partial regression sum of squares $$ SS_{\text{treatment}\mid\text{rows, columns}} = SSE_R-SSE_F, $$ where $SSE_R$ is the residual sum of squares from the observed-data model with rows and columns, and $SSE_F$ is the residual sum of squares from the observed-data model with rows, columns, and treatments. The exact statistic is $$ F= \frac{(SSE_R-SSE_F)/(p-1)} {SSE_F/\{(p-1)(p-2)-1\}}, $$ with numerator degrees of freedom $p-1$ and denominator degrees of freedom $(p-1)(p-2)-1$, provided the stated normal linear-model assumptions hold.

Under $H_0$, all treatment effects are equal (equivalently, all $\tau_k=0$ under the sum-to-zero constraint). The reduced and full models are nested: the reduced model contains rows and columns, and the full model adds treatments. The upper-tail p-value is the probability, assuming $H_0$ and the model assumptions, of obtaining an F statistic at least as large as the observed one. It is not the probability that $H_0$ is true.

Worked Example: An Exact Treatment Test on Observed Responses

Use the hypothetical $3 \times 3$ square shown earlier, with the missing response in row 1, column 1. The eight observed responses, in the order (row 1 column 2, row 1 column 3, row 2 column 1, row 2 column 2, row 2 column 3, row 3 column 1, row 3 column 2, row 3 column 3), are

\mathbf y=(9,11,9,11,10,11,8,11).

Fit the reduced model with row and column effects, then the full additive model with row, column, and treatment effects, using only these eight observations. For this particular data vector, least-squares projection gives $SSE_R=22/3$ for the reduced model and $SSE_F=2/3$ for the full model. Thus the treatment partial sum of squares is $22/3-2/3=20/3$. The full-model rank is $3(3)-2=7$, so its residual degrees of freedom are $8-7=1$. The exact partial F statistic is

F= \frac{(22/3-2/3)/(3-1)}{(2/3)/1} =5.

The reference distribution is $F_{2,1}$. Its upper-tail probability at 5 is approximately 0.3015. At a conventional significance level such as $\alpha=0.05$, this result does not provide evidence against equal treatment means. The very small denominator degrees of freedom also make clear that this $3 \times 3$ example provides little precision for estimating residual variation.

The calculation uses residual sums of squares from two models fitted to the observed data, not a treatment component sum of squares from the completed table. The imputed value 7 is useful for describing the cell under the additive model, but it is not inserted as an additional observed response for this test.

A Practical Analysis Sequence

Keep the estimation step and the inferential step separate. A disciplined sequence prevents the imputed value from being treated as if it carried the same information as a measured response.

Specify the model.
State the additive Latin-square model, its identifying constraints, and the assumptions about independent, common-variance normal errors.

Estimate the missing response.
Use the row, column, treatment, and grand totals formed from observed responses only. Label the resulting value as an estimate.

Adjust the degrees of freedom.
For the full observed-data model, use $p-1$ treatment degrees of freedom and $(p-1)(p-2)-1$ residual degrees of freedom.

Test treatments with nested models.
Fit the rows-and-columns model and the rows-columns-treatments model to the observed responses, then calculate the partial F statistic from their residual sums of squares.

The method is useful because it preserves the intended comparison: treatment effects are assessed after accounting for row and column variation. The imputation formula is not a general remedy for multiple missing responses, interactions, or an incorrectly specified model. With one missing cell, it provides a model-based point estimate; the exact treatment test still comes from the observed-data model comparison.

Common Mistake

Wrong reasoning: “I estimated the missing response, filled the square, and the completed-table treatment sum of squares is orthogonal to the error, so the usual F test is exact.”
Correction: With the cell omitted, treatment contrasts need not remain orthogonal to row and column effects. The completed-table component calculation is not generally the partial treatment sum of squares. For an exact test under the normal linear-model assumptions, compare the rows-and-columns model with the full model using the observed responses and their model ranks.

Key takeaway: Estimate one missing Latin-square response by least squares under the additive model, but do not mistake that imputation for restored balance. Adjust the full-model residual degrees of freedom by one, and use the observed-data partial F test for exact treatment inference.

Check Your Understanding

Try these questions before reviewing the worked examples.

Which totals enter the least-squares formula for one missing response, and should each include the missing value?
For a $6 \times 6$ Latin square with one missing response, calculate the observed-data full-model residual degrees of freedom.
Why does normality and equal error variance alone not make the completed-table treatment F test exact?
Which two models are compared for an exact treatment test, and what is the numerator degrees of freedom?
Explain why the imputed response is not counted as an additional observed response in the exact model comparison.