 Solution

The comparison removes the random slope and tests $H_0:{\tau }_1^2=0$ against $H_1:{\tau }_1^2>0$, while retaining the same fixed intercept and slope. A variance cannot be negative, so zero is a boundary point of the parameter space. The regular Wilks theorem assumptions behind an ordinary ${\chi }_1^2$ likelihood-ratio test fail. Consequently the default chi-squared p-value from anova is not reliably calibrated for this variance-component likelihood-ratio test at a boundary. This is different from the fixed-slope test in part (c). The reference to strength_model1 means the earlier object strength_model; no separate model 1 object was defined.

A useful alternative is a parametric bootstrap. Fit the null random-intercept model, then simulate data on the actual predictor values and furnace grouping by generating null intercept effects and observation errors from its fitted Gaussian distribution. Refit null and alternative models to each simulated dataset using the same likelihood method and compute the same likelihood-ratio statistic. If $T_{\mathrm{obs}}$ is the observed statistic, estimate its upper-tail probability byThis incorporates the zero-variance boundary and the actual small group count. It is a fitted-null simulation approximation because nuisance parameters are estimated, not a universally exact test. Since both models have the same fixed-effect space, a similarly calibrated restricted-likelihood statistic is also possible. Under suitable asymptotics for a single identifiable added variance component the familiar limiting law is $\frac{1}{2}{\chi }_0^2+\frac{1}{2}{\chi }_1^2$, but using the bootstrap avoids relying on that approximation with only three groups. There is no supplied statistic or calibrated bootstrap result here from which to decide whether the slope variance is required.