Answer:
[tex]t _{critical} = 1.760[/tex]
t = 2.2450
d. 0.264
Step-by-step explanation:
The null hypothesis is:
[tex]H_o: \mu_1 - \mu_2 = 0[/tex]
Alternative hypothesis;
[tex]H_a : \mu_1 - \mu_2 > 0\\[/tex]
The pooled variance t-Test would have been determined if the population variance are the same.
[tex]S_p^2 = \dfrac{(n_1-1)S_1^2+(n_2-1)S^2_2}{(n_1-1)+(n_2-1)}[/tex]
[tex]S_p^2 = \dfrac{(8-1)2.507^2+(8-1)2.8282^2}{(8-1)+(8-1)}[/tex]
[tex]S_p^2 = 7.14[/tex]
The t-test statistics can be computed as:
[tex]t= \dfrac{(x_1-x_2)-(\mu_1 - \mu_2)}{\sqrt{Sp^2 ( \dfrac{1}{n} +\dfrac{1}{n_2})}}[/tex]
[tex]t= \dfrac{(9-6)-0}{\sqrt{7.14 ( \dfrac{1}{8} +\dfrac{1}{8})}}[/tex]
[tex]t= \dfrac{3}{1.336}[/tex]
t = 2.2450
Degree of freedom [tex]df = (n_1 -1) + ( n_2 +1 )[/tex]
df = (8-1)+(8-1)
df = 7 + 7
df = 14
At df = 14 and ∝ = 0.05;
[tex]t _{critical} = 1.760[/tex]
Decision Rule: To reject the null hypothesis if the t-test is greater than the critical value.
Conclusion: We reject [tex]H_o[/tex] and there is sufficient evidence to conclude that the test scores for contact address s less than Noncontact athletes.
To calculate r²
The percentage of the variance is;
[tex]r^2 = \dfrac{t^2}{t^2 + df}[/tex]
[tex]r^2 = \dfrac{2.2450^2}{2.2450^2 + 14}[/tex]
[tex]r^2 = \dfrac{5.040025}{5.040025+ 14}[/tex]
[tex]r^2 = 0.2647[/tex]