Country A has an average farm size of 191 acres, while country B has an average farm size of 199 acres. Assume the data were attained from two samples with standard deviations of 38 and 12 acres and sample sizes of 8 and 10, respectively. Is it possible to infer that the average size of the farms in the two countries is different at $α = 0.05$? Assume that the populations are normally distributed.

No, because the test statistic does not exceed the critical value.

The correct answer is Option (2) → No, because the test statistic does not exceed the critical value.

Given $\bar{x_1} = 191, \bar{x_2}= 199, S_1 = 38, S_2 = 12, n_1 = 8, n_2 = 10$ and $α = 0.05$

Let the hypothesis be given as

Null hypothesis $H_0: μ_1 =μ_2$

Alternative hypothesis $H_a: μ_1≠ μ_2$.

So, the test statistic $t=\frac{(\bar{x}_1 - \bar{x}_2)}{\sqrt{\frac{S_1^2}{n_1} + \frac{S_2^2}{n_2}}} = \frac{191-199}{\sqrt{\frac{(38)^2}{8} + \frac{(12)^2}{10}}}$

$⇒t=-0.57$

$df=\frac{\left( \frac{S_1^2}{n_1} + \frac{S_2^2}{n_2} \right)^2}{\frac{1}{n_1 - 1}\left( \frac{S_1^2}{n_1} \right)^2 + \frac{1}{n_2 - 1}\left( \frac{S_2^2}{n_2} \right)^2}=\frac{\left(\frac{(38)^2}{8} + \frac{(12)^2}{10}\right)^2}{\frac{1}{7}\left(\frac{(38)^2}{8}\right)^2+\frac{1}{9}\left(\frac{(12)^2}{10}\right)^2}$

$=\frac{(194.9)^2}{\frac{1}{7}(180.5)^2+\frac{1}{9}(14.4)^2}=\frac{37986.01}{4654.32+23.04}$

$=\frac{37986.01}{4677.36}=8.12$

$∴df=8$

Now, $t_{α/2}=t_{0.025} = 2.306$

$t>-t_{α/2}$, also, $t<t_{α/2}$

So, do not reject $H_0$.

There is not enough evidence to support the claim that the average size of the farms is different.