Question-64

T-test

A psychologist wants to determine if there is a significant difference in the average test scores of two different teaching methods. The scores from Method $A$ ($n = 15$, mean $= 82$, standard deviation $= 8$) and Method $B$ ($n = 15$, mean $= 78$, standard deviation $= 6$) are compared. Given that the scores are normally distributed and the population variances are unknown, which of the following statements are true$?$ (Select all that apply)

The null hypothesis will be there is a significant difference between the means.
The null hypothesis will be there is no difference in means.
A t-test can be applied.
The degrees of freedom for the $t-$ test is 28.

Hint

Answer

The null hypothesis will be there is a significant difference between the means.
The null hypothesis will be there is no difference in means.
A t-test can be applied.
The degrees of freedom for the $t-$ test is 28.

Solution

For the given situation, null and alternative hypothesis will be

$H_0:$ There is no significant difference in the average test scores of the two teaching methods, i.e., $\mu_A = \mu_B$

$H_1:$ There is a significant difference in the average test scores of the two teaching methods, i.e., $\mu_A \neq \mu_B$.

Hence, option (B) is correct.

$t-$ test can be used when we have to test for means and the population variance is unknown.

Hence, option (C) is correct.

When conducting a t-test for two independent samples, the degrees of freedom can be calculated using the formula:

$n_A + n_B - 2 = 15 + 15 - 2 = 28$,

where $n_A $ is the sample size for method $A$ and $n_B$ is the sample size for method $B$.

Hence, option (D) is correct.