Question

In a journal, an article reported that the results of a peer tutoring program to help special children with disabilities learn to read. In the experiment, the children were randomly divided in two groups: the experimental group received peer tutoring along with regular instruction; and the control group received regular instruction with no peer tutoring. There were 30 children in each group. A test was given to both groups before instruction began. For experimental group, the mean score on the test was �𝑥�1� = 344.5 with sample standard deviation s₁ = 49.1. For the control group, the mean score on the same test was �𝑥�2 = 354.2 with sample standard deviation s₂ = 50.9. Use a 5% level of significance to test the hypothesis that there was no difference in the test of two groups before the instruction began.

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.

Testing for Equality of Variances

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

$$F=\dfrac{s_1^2}{s_2^2}=\dfrac{49.1^2}{50.9^2}=0.9305$$

The critical values for $\alpha=0.05, df_1=30-1=29$ degrees of freedom, $df_2=30-1=29$ degrees of freedom are $F_L=0.476,F_U=2.101,$ then the null hypothesis of equal variances is not rejected.

Based on the information provided, the significance level is $\alpha = 0.05,$ and the degrees of freedom are $df=n_1-1+n_2-1=58.$

Hence, it is found that the critical value for this two-tailed test is for $\alpha = 0.05, df=58$ are of freedom is $t_c = 2.001717.$

The rejection region for this two-tailed test is $R = \{t: |t| > 2.001717\}.$

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

$$t=\dfrac{\bar{X_1}-\bar{X_2}}{\sqrt{\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\dfrac{1}{n_1}+\dfrac{1}{n_2})}}$$

$$=\dfrac{344.5-354.2}{\sqrt{\dfrac{(30-1)49.1^2+(30-1)50.9^2}{30+30-2}(\dfrac{1}{30}+\dfrac{1}{30})}}$$

$$\approx-0.751237$$

Since it is observed that $|t|=0.751237<2.001717,$  it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value for two-tailed $df=58$ degrees of freedom, $t=-0.751237$ is $p = 0.455547,$ and since $p = 0.455547$

$>0.05=\alpha,$ 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.