Solution: The test statistic under the null hypothesis is:
[tex]t=\frac{\bar{x_{1}}-\bar{x_{2}}}{s_{p}\sqrt{(\frac{1}{n_{1}})+(\frac{1}{n_{2}})}}[/tex]
Where:
[tex]s_{p} = \sqrt{\frac{(n_{1}-1)s^{2}_{1}+(n_{2}-1)s^{2}_{2}}{n_{1}+n_{2}-2} }[/tex]
[tex]s_{p} = \sqrt{\frac{(16-1)14.8^{2}+(18-1)16^{2}}{16+18-2} }[/tex]
[tex]=15.45[/tex]
[tex]\therefore t=\frac{85.4-74.9}{15.45\sqrt{(\frac{1}{16})+(\frac{1}{18})}}[/tex]
[tex]=\frac{10.5}{5.31}[/tex]
[tex]=1.98[/tex]
Now, to find the critical values, we need to use the t distribution table at 0.01 significance level for [tex]df=n_{1} + n_{2} -2 = 16+18-2=32[/tex] and is given below:
[tex]t_{critical}=-2.738,2.738[/tex]
Since the t statistic is less than the t critical value, we therefore fail to reject the null hypothesis and conclude that the two population means are equal.
