Chapter12 Pooled Wald test

The significance level for the pooled OR is derived by using the pooled Wald test. The pooled Wald test is calculated as:

$$Wald_{Pooled} =\frac{-0.067}{0.046}=$$

This Wald pooled value follows a t-distribution with degrees of freedom (df) according to Formula 5.9.

$$df_{Old} = \frac{m-1}{lamda^2}$$

For this Formula we need information of lambda, which is calculated as:

$$lambda = \frac{V_B + \frac{V_B}{m}}{lamda^2}$$

Using the values of the regression coefficients and standard errors, estimated in each imputed dataset of (Figure 5.10) we can calculate the following values for the between imputation and the total variance.

$$V_B= \frac{(-0.090+0.067)^2 + (-0.061+0.067)^2 +(-0.051+0.067)^2}{2}=\frac{0.000821}{2}=0.0004105$$

To calculate the total variance also the within imputation variance is needed. The within imputation variance can be calculated using Formula 5.2:

$$V_W= \frac{0.039^2 + 0.040^2 + 0.039^2}{3}=0.001547333$$

The total variance becomes:

$$V_{Total} = 0.001547333+0.0004105+ \frac{0.0004105}{3}=0.002094666$$

Now we can calculate lambda using:

$$lambda = \frac{0.0004105 + \frac{0.0004105}{m}}{0.002094666}=0.2612986$$

The lambda value is not presented by SPSS, but only in R using mice. Now we know the value for lambda, we can calculate the degrees of freedom to derive the p-value:

$$df_{Old} = \frac{m-1}{lamda^2}=\frac{2}{0.2612986^2}=29.29246$$

Which results in a p-value of: 0.1502045.