Step 1

The null and alternative hypotheses are’

\(\displaystyle{H}_{{0}}:{m}={3.1}\)

\(\displaystyle{H}_{{1}}:{m}\ne{3.1}\)

The sample mean is 6.4 and the population mean is 3.1.

Here, \(SS=162\) and \(n=9\). The estimate of the standard deviation is

\(\displaystyle{s}=\sigma-\cap=\sqrt{{\frac{{{S}{S}}}{{{n}-{1}}}}}\).

Therefore, the \(\displaystyle{s}=\sqrt{{\frac{{162}}{{8}}}}={4.5}\).

Test statistic:

The test statistic for hypothesis test for mean is given by:

\(\displaystyle{t}=\frac{{{x}-\overline{-}\mu}}{{\frac{{s}}{\sqrt{{{n}}}}}}\)

\(\displaystyle=\frac{{{6.4}-{3.1}}}{{\frac{{4.5}}{\sqrt{{{9}}}}}}\)

\(= 2.2\)

The test statistic is 2.2.

Step 2

Critical value:

Here, sample size is 9.

Thus, the degrees of freedom for the test is \(8(=9-1)\).

Therefore, the df is 15.

The critical value is \(\displaystyle\pm{2.306}\) using the excel formula “=(T.INV.2T(0.05,8))”

Decision rule:

Denote t as test statistic value and \(\displaystyle{t}_{{\frac{\alpha}{{2}}}}\) as the critical value.

Decision rule based on critical approach:

If \(\displaystyle{t}\le–{t}_{{\frac{\alpha}{{2}}}}{\left({\quad\text{or}\quad}\right)}{t}\ge{t}_{{\frac{\alpha}{{2}}}}\), then reject the null hypothesis \(\displaystyle{H}_{{0}}\).

If \(\displaystyle-{t}_{{\frac{\alpha}{{2}}}}{<}{t}{<}{t}_{{\frac{\alpha}{{2}}}}\), then fail to reject the null hypothesis \(\displaystyle{H}_{{0}}\).

Conclusion:

The test statistic value is given as 2.2 and critical value is \(\displaystyle\pm{2.306}\).

Here, \(-2.306<2.2<2.306\).

By the rejection rule fail to reject \(\displaystyle{H}_{{0}}\).

The test statistic value is 2.2.

The critical/cut off values is \(\displaystyle\pm{2.131}\)