= Solution
The <random-slope linear mixed model> is
$$
Y_{ij}=\beta_0+\beta_1t_{ij}+b_jt_{ij}+\varepsilon_{ij},\qquad b_j\overset{\rm iid}{\sim}N(0,\tau^2),\quad \varepsilon_{ij}\overset{\rm iid}{\sim}N(0,\sigma^2),
$$
with <independence> between the <random effects> and errors. The estimates are
$$
\widehat\beta_0=24.32646,\quad\widehat\beta_1=0.70005,\quad\widehat\tau^2=0.00642,\quad\widehat\sigma^2=374.64823.
$$
Incubators are treated as exchangeable representatives of possible growth environments, so a <random slope> permits their growth rates to differ while estimating a population mean growth rate. The formula `0+hours` deliberately removes a <random intercept>: larvae are assigned only after hatching, so it is reasonable to assume no incubator-specific baseline at time zero. Random assignment supports this common-baseline assumption in <expectation>; it does not prove that observed initial sizes or other baseline differences are exactly identical.
For a new incubator, no observations are available to estimate its realized <random slope>, and its mean effect is zero. Under <squared-error loss> the best prediction integrates over that effect and the residual error:
$$
\boxed{\widehat Y(10,\text{new incubator})=24.32646+10(0.70005)=31.32696\ \mathrm{mm}.}
$$
This is the plug-in conditional <expectation> given the population parameters, rather than a prediction using a fitted effect from one of the four existing incubators. Ignoring parameter-estimation uncertainty, the prediction <variance> for a new individual is $10^2\widehat\tau^2+\widehat\sigma^2=375.29023$; it is not just the uncertainty in the population mean.
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