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# 统计代写|多尺度模型代写Multilevel Models代考|UV9253 The adequacy of ordinary least squares estimates

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## 统计代写|多尺度模型代写Multilevel Models代考|The adequacy of ordinary least squares estimates

When a variance partition coefficient is small, we can expect reasonably good agreement between the multilevel estimates and the simpler OLS ones. While it is difficult to give general guidelines about when OLS is an adequate alternative, we can readily derive an explicit formula for the balanced 2-level variance component model using a simple regression equation with an intercept and a single explanatory variable
$$y_{i j}=\beta_0+\beta_1 x_{i j}+u_j+e_{i j}$$
Write $\rho_y \rho_x$ for the intra-unit correlations for $Y, X$ respectively and $n$ for the number of level 1 units in each level 2 unit. To obtain an estimate of the correct standard error for the estimate of $\beta_1$, we multiply the usual OLS estimate of the standard error by the quantity
$$\left{1+\rho_y \rho_x(n-1)\right}^{1 / 2}$$

## 统计代写|多尺度模型代写Multilevel Models代考|A 2-level example using longitudinal educational achievement data

We shall fit the simple 2-level variance components model (2.7) to the JSP data with the maths score at age 11 as the response and a single explanatory variable, the maths score at age 8 , in addition to the constant term, equal to 1 and defining the intercept. The parameter values are displayed in Table $2.1$ with the ordinary least squares estimates given for comparison.

Comparing the OLS with the multilevel estimates, we see that the fixed coefficients are similar, but that there is a variance partition coefficient value of $0.14$. The estimate of the standard error of the between school variance is less than a third of the variance estimate, suggesting a value highly significantly different from zero ${ }^2$. This comparison, however, should be treated cautiously, since the variance estimate does not have a normal distribution and the standard error is only estimated, although the size of the sample here will make the latter caveat less important. It is generally preferable to carry out a likelihood ratio test by estimating the ‘deviance’ for the current model and the model omitting the level 2 variance (see McCullagh and Nelder, 1989) and the next section will deal more generally with inference procedures. The difference between the deviances is $63.1$. This value would normally be referred to tables of the chi-squared distribution with one degree of freedom, and is highly significant. In the present case, however, the null hypothesis of a zero variance is on the ‘boundary’ of the feasible parameter space; we do not envisage a negative variance. In this case, the P-value to be used is half the one obtained from the tables of the chi-squared distribution (Shapiro, 1985). Note that if we use the standard error estimate given in Table $2.1$ to judge significance we obtain the corresponding value of $(3.19 / 1.0)^2=10.2$ which, in this case, is very much smaller than the likelihood ratio test statistic. Note also that if we use this test we would again use half the nominal P-value since only positive departures are possible.

A similar issue arises when simultaneously testing several parameters where one or more is constrained in this way, such as a variance plus covariances. In this case, the appropriate test statistic is a weighted mixture of chi-squared distributions and details are given by Shapiro (1985). In the common case where we are testing a set of covariances and a single associated variance, this reduces to the following.

Suppose that we have $r$ covariance parameters and adopt a $5 \%$ significance level. Then the critical value, $c$, for judging significance is given by the following formula for a mixture of two chi-squared distributions
$$0.5 \times\left[\operatorname{pr}\left(\chi_r^2 \geq c\right)+\operatorname{pr}\left(\chi_{r+1}^2 \geq c\right)\right]=0.05$$

## 统计代写|多尺度模型代写Multilevel Models代考|The adequacy of ordinary least squares estimates

$$y_{i j}=\beta_0+\beta_1 x_{i j}+u_j+e_{i j}$$

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## 统计代写|多尺度模型代写Multilevel Models代考|A 2-level example using longitudinal educational achievement data

$$0.5 \times\left[\operatorname{pr}\left(\chi_r^2 \geq c\right)+\operatorname{pr}\left(\chi_{r+1}^2 \geq c\right)\right]=0.05$$

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