Question

A professor teaches an undergraduate course in statistics. He uses a lot of sports examples to explain key concepts. He is concerned that this may have biased his instruction to favor male students. To test this, he measures exam grades among women (n = 10) and men (n = 10). The mean score in the male group was 82 ± 4.0 (M ± SD); in the female group, it was 74 ± 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 .05 level of significance for a two-tailed test. Use denominator of 2.98.

Answer 1

Answer:

The null hypothesis is  rejected

Therefore there is sufficient evidence to conclude that the professors teaching method favored the male

Step-by-step explanation:

From the question we are told that

   The sample size for each population is  $n_1 = n_2 = n = 10$

   The  first sample  mean is  $\= x_1 = 82$

    The  second sample  mean is  $\= x_2 = 74$

    The  first standard deviation is $\sigma _1 = 4$  

    The second standard deviation is  $\sigma_2 = 8.0$

     The  level of significance is  $\alpha = 0.05$

The  null hypothesis is  $H_o : \mu_1 = \mu_2$

The  alternative  hypothesis is  $H_a : \mu_1 \ne \mu_2$

Generally the standard error is mathematically represented as

           $SE = \sqrt{ \frac{ \sigma_1^2}{n_1} + \frac{ \sigma_2^2}{n_2} }$

=>        $SE = \sqrt{ \frac{ 4^2}{10} + \frac{ 8^2}{10} }$

=>        $SE = 2.83$

Generally the test statistics is mathematically represented as

          $t = \frac{\= x_1 - \= x_2 }{SE}$

=>        $t = \frac{82 - 74}{2.83}$

=>        t =  2.83

Generally the p-value  mathematically represented as

        $p-value = 2 P(Z > 2.83)$

From the z table  

           $P(Z > 2.83) = 0.0023274$

So

    $p-value = 2 * 0.0023274$

   $p-value = 0.0047$

Since  

      $p-value < \alpha$

Hence the null hypothesis is  rejected

Therefore there is sufficient evidence to conclude that the professors teaching method favored the male