Question

A professor teaches a certain section of material using a lot of examples with sports and cars to illustrate. He is concerned that this may have biased his instruction to favor male students. To test this, he measures exam grades from this section of material among women (n = 10) and men (n = 10). The mean score in the male group was 84 ± 4.0 (M ± SD); in the female group, it was 78 ± 8.0 (M ± SD) points. If the null hypothesis is that there is no difference in exam scores, then test the null hypothesis at a 0.05 level of significance.
A.Exam scores were significantly higher in the male group, t(18) = 2.83, p < .05B.Exam scores were the same between groups, t(18) = 2.12, p > .05.C.Exam scores were significantly higher in the male group, t(18) = 2.12, p < .05.D.Exam scores were the same between groups, t(18) = 1.00, p > .05.

Answer 1

Answer:

Step-by-step explanation:

Let X represent for males and Y for females

Given that

Group   Group One     Group Two  

Mean 84.00          78.00

SD        4.00                    8.00

SEM 1.26             2.53

N            10              10  

SEM is calculated as $\frac{s}{\sqrt{n} }$ for each group

$H_0: \bar x = \bar y\\H_a: \bar x \neq \bar y$

(Two tailed test at 5% significance level)

$t = 2.1213\\ df = 18\\ standard error of difference = 2.828\\p=0.0480$

Since p <0.05 we reject H0 and there is significant difference between the groups.

A.Exam scores were significantly higher in the male group, t(18) = 2.83, p < .05B