Expected mean squares

In obtaining Monte Carlo data such as shown in Figs. 2, 3, 5, it is also necessary to understand the statistical errors that are present because the number of states M — Mq over which we average (Eq. (24)) is finite. If the averages m, E, i/ are calculated from a subset of n uncorrelated observations m(Xy), E(Xy), ilf Xy), Standard error analysis applies and yields estimates for the expected mean square deviations, for n- cx),... 

From the mean square (MS) values and the formulas for the expected mean square (see Table 3.30) one can calculate the variance components. For the example of Figure 3.5 this gives ... 

Furthermore, it can be shown (just as for the one-way model) that the expected mean squares are ... 

These observations may be summarized conveniently in an analysis-of-variance table-. Table 26-7 illustrates this type of table for the above case. The overall variance (total mean square) Sj(N — 1) contains contributions due to variances within as well as between classes. The variation between classes contains both variation within classes and a variation associated with the classes themselves and is given by the expected mean square aj + not. Whether not is significant can be determined by the F test. Under the null hypothesis, = 0. Whether the ratio... 

Precision components are estimated using the random effects of the model. There are a number of ways to estimate the precision components (or variance components). Variance components are estimates of variances (variance = standard deviation squared) and consequently are restricted to being nonnegative. One approach is to conduct an ANOVA and solve for the variance components using the theoretical composition of the expected mean squares (called method of moments estimates or ANOVA estimates). The ANOVA method is shown in... 

Method of moments estimates (also known as ANOVA estimates) can be calculated directly from the raw data as long as the design is balanced. The reader is referred to Searle et al. (11) for a thorough but rather technical presentation of variance components analysis. The equations that follow show the ANOVA estimates for the validation example. First, a two-factor with interaction ANOVA table is computed (Table 7). Then the observed mean squares are equated to the expected mean squares and solved for the variance components (Table 8 and the equations that follow). 

The expected mean squares are equated to the observed mean squares (those from the data) and solved for the variance components. Thus, =MSE is the variance estimate... 

Table 8 Expected Mean Squares for the Two-Factor with Interaction Analysis of Variance Table Shown in Table 7...

According to the statistical theory, the mean squares SSB/(k respective expected mean squares (EMS) as (Box et al. 1978) ... 

How did we obtain the expected mean squares shown in Tables V and VI These are a consequence of the assumptions made concerning the nature of our observations. What are these assumptions It is customary to assume that any observation will consist of certain parts which when added together produce the value recorded. In our example there are only two parts one part due to the treatment employed and a second part due to the experimental unit (and all extraneous sources of variation). We... 

This analysis of variance provides another example for testing a null hypothesis and also gives further insight into the concept of components of variance. The expected mean square for between leaves on the same plant contains only one component of variance, since the only factor which affects (or produces) this variation is leaves. The expected mean square for among plants in same group contains two components of variance, since this source of variation reflects the variation among the... 

Source of variation Degrees of freedom Sum of squares Mean square Expected mean square... 

On examining the expected mean squares in Table XVI, it is apparent that an exact test of the null hypothesis H n = T2 = 0 is impossible. This unfortunate circumstance exists because Ci 9 C2 and this arises from the unequal number of determinations per sample and the unequal number of samples per treatment. This result brings home with great force the need for equal numbers of observations in the various subclasses. No difficulty was encountered when dealing with the data on ascorbic acid in turnip greens this was because there were equal frequencies in the various subclasses. 

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