= Solution
For independent observations from a <normal distribution> with unknown mean and positive variance, the <maximum likelihood estimation> of the mean gives
$$
\boxed{\widehat\mu=\bar X=74.56}.
$$
Indeed $\sum_i(X_i-\mu)^2=S_{XX}+n(\bar X-\mu)^2$, which is minimized at $\mu=\bar X$. The <likelihood> is
$$
L(\mu,\sigma^2)=(2\pi\sigma^2)^{-n/2}\exp\!\left[-\frac{S_{XX}+n(\bar X-\mu)^2}{2\sigma^2}\right].
$$
For fixed $\mu$, differentiation in $\sigma^2$ gives the maximizer $[S_{XX}+n(\bar X-\mu)^2]/n$. Thus the unrestricted fit has $\widehat\sigma^2=S_{XX}/n$, whereas under the <null hypothesis> $\mu=\mu_0$ the fitted variance is $[S_{XX}+n(\bar X-\mu_0)^2]/n$. Both maximize rather than minimize the likelihood: it tends to zero at either variance endpoint when the residual sum of squares is positive.
The restricted-to-unrestricted <likelihood ratio> is therefore
$$
\Lambda=\left[\frac{S_{XX}}{S_{XX}+n(\bar X-\mu_0)^2}\right]^{n/2}=\left(1+\frac{T^2}{n-1}\right)^{-n/2},\qquad T=\frac{\sqrt n(\bar X-\mu_0)}{s},\quad s^2=\frac{S_{XX}}{n-1}.
$$
The <generalized likelihood-ratio test> rejects for small $\Lambda$, equivalently large $|T|$. Under the null, $Z=\sqrt n(\bar X-\mu_0)/\sigma$ is standard normal, $U=S_{XX}/\sigma^2$ has a <chi-squared distribution> with $n-1$ degrees of freedom, and they are independent. To see the independence, apply an <orthogonal matrix> to the standardized normal sample, taking its first row to be $(1,\ldots,1)/\sqrt n$: the transformed coordinates remain independent standard normal variables, and $U$ is the sum of squares of the other $n-1$ coordinates. Hence $T=Z/\sqrt{U/(n-1)}$ has <Student's t-distribution> with $n-1$ degrees of freedom.
For this two-sided test at $5\%$, the cutoff is the $97.5\%$ percentile of $t_9$, namely $2.26$. The data give
$$
s^2=\frac{12.824}{9},\qquad T=\frac{\sqrt{10}(74.56-75)}{\sqrt{12.824/9}}\simeq-1.166.
$$
Since $|T|<2.26$, \b[do not reject the manufacturer's mean-hardness claim at the $5\%$ level]. This is a failure to reject, not proof that the mean equals $75$.
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