In the paper "The Relation of Sex and Sense of Direction to Spatial Orientation in an Unfamiliar Environment" (Journal of Environmental Psychology, Vol. 20, pp. 17-28), J. Sholl et al. published the results of examining the sense of direction of 30 male and 30 female students. After being taken to an unfamiliar wooded park, the students were given some spatial orientation tests, including pointing to the south, which tested their absolute frame of reference. The students pointed by moving a pointer attached to a $360°$protractor. Following are the absolute pointing errors, in degrees, of the participants.

At the $1\%$ significance level, do the data provide sufficient evidence to conclude that, on average, males have a better sense of direction and, in particular, a better frame of reference than females? (Note: ${\overline{x}}_1=37.6,s_1=38.5,{\overline{x}}_2=55.8,\text{and}{s}_2=48.3$.)

Sample data is supplied from two male and female groups.

${\overline{x}}_1=37.6,s_1=38.5$

${\overline{x}}_2=55.8,s_2=48.3$

Significance level is$5\%$.

Population $1$: Male, ${\overline{x}}_1=37.6,s_1=38.5,\text{and}{n}_1=30$.

Population $2$: Female, ${\overline{x}}_2=55.8,s_2=48.3,\text{and}{n}_2=30$.

The major goal is to come to the conclusion that, on average, males have a superior sense of direction, or frame of reference. This suggests that the male population's mean absolute pointing error should be lower than the female population's mean absolute pointing error.

Define null and alternate hypotheses.

Null hypotheses:$H_0={\mu }_1\ge {\mu }_2$

Alternate hypotheses:$H_a={\mu }_1<{\mu }_2$

Hypotheses is left-tailed.

We determine the significance level is $1\%$.

Pooled standard deviation,$s_p=\sqrt{\frac{\left(n_1-1\right)s_1{}^2+\left(n_2-1\right)}{n_1+n_2-2}}$

$\Rightarrow s_p=\sqrt{\frac{(30-1)(38.5{)}^2+(30-1\left)\right(48.3{)}^2}{30+30-2}}$

$\Rightarrow s_p=\sqrt{\frac{29(1482.25)+29(2352.25)}{58}}$

$\Rightarrow s_p=43.786$

Test statistic,$t_0=\frac{{\overline{x}}_1-{\overline{x}}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

$\Rightarrow t_0=\frac{37.6-55.8}{43.786\sqrt{\frac{1}{30}+\frac{1}{30}}}$

$\Rightarrow t_0=-1.138$

We need to decide the critical values

Here,localid="1651300781129" role="math" $$\begin{gathered} df=n_1+n_2-2 \\ =30+30-2 \\ =58 \end{gathered}$$

$\Rightarrow \mathrm{df}=58$

Using table IV

localid="1651300790563" role="math" $$\begin{gathered} \text{Critical value,}{t}_{\alpha }=t_{0.01} \\ =-2.392 \end{gathered}$$

The test statistic does not fall in the rejection zone of the left-tailed hypotheses test because $t_0>t_{\alpha }$. As a result, null hypotheses are not ruled out.