A sample of 16 from a population produced a mean of 85.4 and a standard deviation of 14.8. a sample of 18 from another population produced a mean of 74.9 and a standard deviation of 16.0. assume that the two populations are normally distributed and the standard deviations of the two populations are equal. the null hypothesis is that the two population means are equal, while the alternative hypothesis is that the mean of the first population is different than the mean of the second population. the significance level is 1%.

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.