Expected mean squares analysis

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),... 

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... 

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). 

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

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... 

This model has previously been shown (Hll, K12) to have a residual mean square comparing favorably with that expected from pure error, as discussed in Section IV. It is to be noted that we have been led logically from one model to another within the small class of models for which n — 3 by the above analysis. For these data, adsorbed methane is not required however, for data with higher methane concentrations, the adsorbed-methane term may be needed. 

K, the static disorder is certainly maintained. The results are presented as plots of formula in Fig. 7. The deviations from linearity of the plots is small enough to support such method of analysis. The slopes of the curves give the 5a values tabulated in Table 4. It follows that in the (1 x l)Co/Cu(lll) case the anisotropy of surface vibrations clearly appears in the measured values of 8a and 5aT There are two reasons for such anisotropy the first is a surface effect due to the reduced coordination in the perpendicular direction. cF is a mean-square relative displacement projected along the direction of the bond Enhanced perpendicular vibrational amplitude causes enhanced mean-square relative displacement along the S—B direction. The second effect is due to the chemical difference of the substrate (Fig. 8). S—B bonds are Co—Cu bonds and the bulk Co mean-square relative displacement, cr (Co), is smaller than the bulk value for Cu, aJ(Cu). Thus for individual cobalt-copper bonds, the following ordering is expected ... 

The Debye Waller analysis of the S—B bonds gives A05 g2(HO) = 2.9 x 10 A. This value is lower than the pure Co value (3.6 x 10 A ). Due to the low density of the (110) face, one mi t have expected a large mean-square relative displacement. The measured small value reveals a stiffening of the force constant of the Co—Cu bond. This is consistent with the large eontraction of the Co—Cu interlayer distance (sell % see above). The stiffening in strongly relaxed surfaces has been observed before and overcompensate the effect of the reduced surface coordination in the perpendicular direction. Reversed surfaee anisotropy of the mean square relative atomic displacements has also been found on an other low-density surfaee C2 x 2 Cl/Cu(l 10) i.e. one half density of Cl vs. Cu(l 10) in plane density where the Cl atoms moves with amplitudes parallel to the surface eomparable with those of the Cu subtrate, but with a much reduced amplitude in the perpendicular direction... 

If we desire to study the effects of two independent variables (factors) on one dependent factor, we will have to use a two-way analysis of variance. For this case the columns represent various values or levels of one independent factor and the rows represent levels or values of the other independent factor. Each entry in the matrix of data points then represents one of the possible combinations of the two independent factors and how it affects the dependent factor. Here, we will consider the case of only one observation per data point. We now have two hypotheses to test. First, we wish to determine whether variation in the column variable affects the column means. Secondly, we want to know whether variation in the row variable has an effect on the row means. To test the first hypothesis, we calculate a between columns sum of squares and to test the second hypothesis, we calculate a between rows sum of squares. The between-rows mean square is an estimate of the population variance, providing that the row means are equal. If they are not equal, then the expected value of the between-rows mean square is higher than the population variance. Therefore, if we compare the between-rows mean square with another unbiased estimate of the population variance, we can construct an F test to determine whether the row variable has an effect. Definitional and calculational formulas for these quantities are given in Table 1.19. 

If we suppose that two arrays of time delays Mi and M2 correspond to the normal distribution law and they are not homogeneous then the attack is carried out using intermediate computers [12], Therefore, it is necessary to fulfill the appropriate analysis in order to determine the attack source status. For this purpose it is necessary to calculate mathematical expectations xv for an array Mi(f) and yv for an array M2(0, dispersions Dvx, Dvy and mean-square errors crvx and [Pg.197]

Within each of the assays A, B, C, and D, least squares linear regression of observed mass will be regressed on expected mass. The linear regression statistics of intercept, slope, correlation coefficient (r), coefficient of determination (r ), sum of squares error, and root mean square error will be reported. Lack-of-fit analysis will be performed and reported. For each assay, scatter plots of the data and the least squares regression line will be presented. 

When discussing the analysis of the fermentation data it was noted that no exact test of the null hypothesis H ti = rj = 0 was available owing to the unequal frequencies at the various stages of the subsampling. Such a difficulty may be circumvented when it occurs by using our estimates of the components of variance. By these means we shall synthesize one (or two) mean squares which will have the same expected valve if the null hypothesis is true. These synthesized mean squares will then be used to form a ratio which is approximately distributed as F. This method, therefore, provides only an approximate solution. If the expression... 

As an example, let s say that Chi-square analysis of your data gives a p-value of 0.17. This means that there is a 17% probability that the difference between the observed and the expected values is due to chance. It also means that there is an 83% (100% -17% = 83%) probability that the difference is not due to chance the difference is real. 

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