= Solution
Minimize the <residual sum of squares> $\|Y-X\theta\|^2$. Differentiating gives the <normal equations> $X^TX\widehat\theta=X^TY$. <Full column rank> makes $X^TX$ positive definite and invertible, so the unique minimizer is
$$
\boxed{\widehat\theta=(X^TX)^{-1}X^TY.}
$$
Since $\mathbb EY=X\theta$, it is unbiased. The independent common-variance errors give $\operatorname{Cov}(Y)=\sigma^2I$, hence
$$
\boxed{\operatorname{Cov}(\widehat\theta)=\sigma^2(X^TX)^{-1}.}
$$
Independent <Gaussian errors> with common <variance> are a sufficient condition for this <estimator> also to be the maximum likelihood <estimator>: their log-likelihood differs from $-\|Y-X\theta\|^2/(2\sigma^2)$ by terms independent of $\theta$. Profiling an unknown common <variance> gives the same minimizer. Mean and <variance> assumptions alone do not specify a likelihood.
For the seven design points, $\bar x=0$ and $\overline{x^2}=4$. In increasing time order the <design matrix> is
$$
X=\begin{pmatrix}
1&-3&5\\1&-2&0\\1&-1&-3\\1&0&-4\\1&1&-3\\1&2&0\\1&3&5
\end{pmatrix}.
$$
Its columns are mutually <orthogonal>: the odd sums vanish and $\sum(x_i^2-4)=0$. Their squared lengths are $7,28,84$, respectively. Therefore
$$
X^TX=\operatorname{diag}(7,28,84),\qquad
\boxed{\operatorname{Var}(\widehat\alpha)=\frac{\sigma^2}7,\quad
\operatorname{Var}(\widehat\beta)=\frac{\sigma^2}{28},\quad
\operatorname{Var}(\widehat\gamma)=\frac{\sigma^2}{84}.}
$$
In particular $\widehat\gamma=\sum_i(x_i^2-4)Y_i/84$. Testing for a quadratic contribution means testing $H_0:\gamma=0$ against $\gamma\ne0$.
The cooling measurements are stated to have zero-mean errors and known common <variance>, but normality is not explicitly assumed for them. Under the independent-error <linear regression model> as stated, <Chebyshev's inequality> gives a valid conservative level-$\delta$ test: reject when
$$
\boxed{|\widehat\gamma|\ge\frac{\sigma}{\sqrt{84\delta}}.}
$$
Its null rejection <probability> is at most $\delta$ because the <estimator> has mean zero and the calculated <variance>. If <Gaussian errors> are additionally assumed, the sharper exact test uses $Z=\sqrt{84}\widehat\gamma/\sigma\sim N(0,1)$ under the null and rejects for $|Z|>z_{1-\delta/2}$. With only seven observations, exact normal calibration cannot be deduced from the printed moment assumptions alone.
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