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

The null hypothesis is rejected

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

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 \\e \mu_2`

Generally the standard error is mathematically represented as

`SE = \sqrt{ ( \sigma_1^2)/(n_1) + ( \sigma_2^2)/(n_2) }`

=>
`SE = \sqrt{ ( 4^2)/(10) + ( 8^2)/(10) }`

=>
`SE = 2.83`

Generally the test statistics is mathematically represented as

`t = (\= x_1 - \= x_2 )/(SE)`

=>
`t = (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