Answer: Using a two-sample pooled-variance \(t\)-test, \(t=2.62\) with \(12\) degrees of freedom. Since \(2.62\gt2.179\), reject \(H_0\). There is sufficient evidence at the \(5\%\) significance level to suggest that the mean tensile strengths of rods from the two machines are different.

Let \(\mu_A\) and \(\mu_B\) be the population mean tensile strengths for machines \(A\) and \(B\).

The hypotheses are

\(H_0:\mu_A=\mu_B\),

\(H_1:\mu_A\ne\mu_B\).

This is a two-tailed two-sample \(t\)-test. Since the population variances are assumed equal, use a pooled estimate of the common variance.

For machine \(A\):

\(n_A=8,\quad \sum x_A=3264,\quad \sum x_A^2=1331848\).

So

\(\bar{x}_A=\frac{3264}{8}=408\),

and the corrected sum of squares is

\(S_{xx,A}=1331848-\frac{3264^2}{8}=136\).

For machine \(B\):

\(n_B=6,\quad \sum x_B=2406,\quad \sum x_B^2=964964\).

So

\(\bar{x}_B=\frac{2406}{6}=401\),

and the corrected sum of squares is

\(S_{xx,B}=964964-\frac{2406^2}{6}=158\).

The pooled variance is therefore

\(s_p^2=\frac{S_{xx,A}+S_{xx,B}}{n_A+n_B-2}=\frac{136+158}{8+6-2}=\frac{294}{12}=24.5\).

Thus \(s_p=\sqrt{24.5}\). The test statistic is

\(t=\frac{\bar{x}_A-\bar{x}_B}{s_p\sqrt{\frac{1}{n_A}+\frac{1}{n_B}}}\).

Substituting the values gives

\(t=\frac{408-401}{\sqrt{24.5}\sqrt{\frac{1}{8}+\frac{1}{6}}}=2.62\) to 3 significant figures.

The degrees of freedom are

\(8+6-2=12\).

For a two-tailed test at the \(5\%\) significance level with \(12\) degrees of freedom, the critical value is \(2.179\).

Since \(2.62\gt2.179\), the test statistic lies in the critical region. Therefore reject \(H_0\).

There is sufficient evidence at the \(5\%\) significance level to suggest that the mean tensile strengths of steel rods produced by the two machines are different.