Question

1.      A course in elementary statistics was taught using two different methods. One group of 12 students was taught using the conventional classroom procedure. A second group of 10 students was taught the same course but also used individualized programmed material at the end of the semester the same examination was given to each group. The first group had a mean grade of 87 with s=5. Based on the above data can we conclude that there is a significant difference between the two methods of instruction at 5% significance level. 

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu_1=\mu_2$

$H_1:\mu_1\not=\mu_2$

This corresponds to a two-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used.

Based on the information provided, the significance level is $\alpha=0.05,$ $df=n_1-1+n_2-1=12-1+10-1=20$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.086.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.086\}.$  

Since it is assumed that the population variances are equal, the t-statistic is computed as follows:

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{87-81}{\sqrt{\dfrac{(12-1)(5)^2+(10-1)(4)^2}{12+10-2}(\dfrac{1}{12}+\dfrac{1}{10})}}$$

$$\approx3.0615$$

Since it is observed that $t=3.0615>2.086=t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach: The p-value for two-tailed test $\alpha=0.05,$ $df=20$ degrees of freedom is $p=0.006159,$ and since $p=0.006159<0.05=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.05$ significance level.