Answer: There is not enough evidence, at the \(10\%\) significance level, to say that the mean mass differs from \(94\) kg.

We use a two-tailed \(t\)-test because the population variance is not known and the sample size is small.

The hypotheses are

\[H_0:\mu=94,\qquad H_1:\mu\ne94.\]

For the sample,

\[\sum x=681.6,\qquad n=7.\]

So the sample mean is

\[\bar x=\frac{681.6}{7}=97.371\ldots.\]

Also

\[\sum x^2=66536.1.\]

The unbiased estimate of the population variance is

\[s^2=\frac{66536.1-681.6^2/7}{6}=27.956\ldots.\]

So

\[s=5.287\ldots.\]

The test statistic is

\[t=\frac{\bar x-94}{s/\sqrt7}.\]

Substitute the values:

\[t=\frac{97.371\ldots-94}{5.287\ldots/\sqrt7}=1.69\quad\text{approximately}.\]

For a two-tailed test at the \(10\%\) level with \(6\) degrees of freedom, the critical value is about

\[1.94.\]

Since

\[1.69<1.94,\]

we do not reject \(H_0\). Therefore there is no significant evidence that the mean mass is different from \(94\) kg.