Answer to Question #256827 in Statistics and Probability for Arvie

Question

A course in Mathematics is taught to 12 students by a conventional class room procedure. A second group of 10 students was given the same course by means of programmed materials. At the end of the semester the same examination was given to each group. The 12 students meeting in the classroom made an average grade of 85 with a standard deviation of 4. While the 10 students using programmed materials made an average of 81 with standard deviation of 5. Test the hypothesis using 0.01 level of significance. (Assuming Normal Distribution)

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 $\alpha=0.01,$ $df=n_1-1+n_2-1=12-1+10-1=20$ degrees of freedom, and the critical value for a two-tailed test is $t_c=2.84534.$

The rejection region for this two-tailed test is $R=\{t:|t|>2.84534\}.$

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

$$t=\dfrac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$ $$=\dfrac{85-81}{\sqrt{\dfrac{(12-1)(4)^2+(10-1)(5)^2}{12+10-2}(\dfrac{1}{12}+\dfrac{1}{10})}}$$ $$\approx2.08633$$

Since it is observed that $t=2.08633<2.84534=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for two-tailed test $\alpha=0.01,$ $df=20$ degrees of freedom, $t=2.08633$ is $p=2P(T>2.08663)=0.049963,$ and since $p=0.049963>0.01=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$ is different than $\mu_2,$ at the $\alpha = 0.01$ significance level.