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

$H_0: \mu_{experimental \; group} = \mu_{control \; group} \\ H_1: \mu_{experimental \; group} ≠ \mu_{control \; group}$

Experimental group:

$n_1 = 30 \\ \bar{x_1} = 344.5 \\ s_1 = 49.1$

Control group:

$n_2 = 30 \\ \bar{x_2} = 354.2 \\ s_2 = 50.9$

Test-statistic

$t = \frac{\bar{x_1} -\bar{x_2}}{s_p \sqrt{ \frac{1}{n_1} + \frac{1}{n_2} }} \\ s^2_p= \frac{(n_1-1)s^2_1 + (n_2-1)s^2_2}{n_1+n_2-2} \\ s^2_p = \frac{(30-1)49.1^2 + (30-1)50.9^2}{30+30-2} \\ s^2_p = 2500.8 \\ s_p = 50.00 \\ t = \frac{344.5 -354.2}{50.0 \sqrt{\frac{1}{30} + \frac{1}{30}}} \\ t = -0.75 \\ α = 0.05 \\ d.f. = n_1+n_2 -2 \\ = 30+30-2 \\ = 58$

Since, our test is two-tailed test, therefore we shall obtain two-tailed p-value for the test statistic.

The EXCEL formula to find the p-value for a two-tailed t-test and df=58 is

=tdist(0.75, 58, 2)

p-value = 0.456

p-value > α

We accept H₀ at 0.05 level of significance.

We can conclude that there was no difference in the test of two groups before the instruction began.