Answer to Question #333961 in Statistics and Probability for asiteru

Question

A course in Physics was taught to 10 students using the traditional method Another group of 11 students went through the same course using another method. At the end of the semester, the same test was administered to each group. The 10 students under method A got an average of 82 with a standard deviation of 5, while the 11 students

under method B got an average of 78 with a standard deviation of 6. Test the null hypothesis of no significant difference in the performance of the two groups of students at 5% level of significance

Answer 1

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

$$F = \frac{s_1^2}{s_2^2} = \frac{ 5^2}{ 6^2} = 0.694$$

The critical values are $F_L = 0.2523$ and $F_U = 3.779,$ and since $F = 0.694,$ then the null hypothesis of equal variances is not rejected.

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 $\alpha = 0.05,$ and the degrees of freedom are $df_{total} = 10-1+11-1=19.$

Hence, it is found that the critical value for this two-tailed test is $t_c =2.093024,$ for $\alpha = 0.05$ and $df = 19.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.093024\}.$

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

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

$$=\dfrac{82-78}{\sqrt{\dfrac{(10-1)5^2+(11-1)6^2}{10+11-2}(\dfrac{1}{10}+\dfrac{1}{11})}}$$

$$\approx1.649854$$

Since it is observed that $|t| = 1.649854 \le 2.093024=t_c,$ it is then concluded that the null hypothesis is not rejected.

Using the P-value approach:

The p-value for two-tailed, $df=19$ degrees of freedom, $t=1.649854$ is $p = 0.115407,$ and since $p = 0.115407\ge 0.05=\alpha,$ it is concluded that the null hypothes is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu_1$

is different than $\mu_2,$ at the $\alpha = 0.05$ significance level.