Question

McAllister et al. (2012) compared varsity football and hockey players with varsity athletes from noncontact sports to determine whether exposure to head impacts during one season have an effect on cognitive performance. In the study, tests of new learning performance were significantly poorer for the contact sport athletes compared to the noncontact sport athletes. The following table presents data similar to the results obtained in the study.

Noncontact Athletes:
10, 8, 7, 9, 13, 7, 6, 12
Contact Athletes:
7, 4, 9, 3, 7, 6, 10, 2
a. Are the test scores significantly lower for the contact sport athletes than for the noncontact athletes? Use a one-tailed test with \alpha=.05
t-critical=
t=
b. Compute the value of (percentage of variance accounted for) for these data.
r^2= a. 0.123, b. 0.239, c. 0.138, d. 0.264

Answer 1

`t _(critical) = 1.760`

t = 2.2450

d. 0.264

Explanation:

The null hypothesis is:

`H_o: \mu_1 - \mu_2 = 0`

Alternative hypothesis;

`H_a : \mu_1 - \mu_2 > 0\\`

The pooled variance t-Test would have been determined if the population variance are the same.

`S_p^2 = ((n_1-1)S_1^2+(n_2-1)S^2_2)/((n_1-1)+(n_2-1))`

`S_p^2 = ((8-1)2.507^2+(8-1)2.8282^2)/((8-1)+(8-1))`

`S_p^2 = 7.14`

The t-test statistics can be computed as:

`t= \frac{(x_1-x_2)-(\mu_1 - \mu_2)}{\sqrt{Sp^2 ( (1)/(n) +(1)/(n_2))}}`

`t= \frac{(9-6)-0}{\sqrt{7.14 ( (1)/(8) +(1)/(8))}}`

`t= (3)/(1.336)`

t = 2.2450

Degree of freedom
`df = (n_1 -1) + ( n_2 +1 )`

df = (8-1)+(8-1)

df = 7 + 7

df = 14

At df = 14 and ∝ = 0.05;

`t _(critical) = 1.760`

Decision Rule: To reject the null hypothesis if the t-test is greater than the critical value.

Conclusion: We reject
`H_o` 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;

`r^2 = (t^2)/(t^2 + df)`

`r^2 = (2.2450^2)/(2.2450^2 + 14)`

`r^2 = (5.040025)/(5.040025+ 14)`

`r^2 = 0.2647`