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.
"H_0: \\mu_{experimental \\; group} = \\mu_{control \\; group} \\\\\n\nH_1: \\mu_{experimental \\; group} \u2260 \\mu_{control \\; group}"
Experimental group:
"n_1 = 30 \\\\\n\n\\bar{x_1} = 344.5 \\\\\n\ns_1 = 49.1"
Control group:
"n_2 = 30 \\\\\n\n\\bar{x_2} = 354.2 \\\\\n\ns_2 = 50.9"
Test-statistic
"t = \\frac{\\bar{x_1} -\\bar{x_2}}{s_p \\sqrt{ \\frac{1}{n_1} + \\frac{1}{n_2} }} \\\\\n\ns^2_p= \\frac{(n_1-1)s^2_1 + (n_2-1)s^2_2}{n_1+n_2-2} \\\\\n\ns^2_p = \\frac{(30-1)49.1^2 + (30-1)50.9^2}{30+30-2} \\\\\n\ns^2_p = 2500.8 \\\\\n\ns_p = 50.00 \\\\\n\nt = \\frac{344.5 -354.2}{50.0 \\sqrt{\\frac{1}{30} + \\frac{1}{30}}} \\\\\n\nt = -0.75 \\\\\n\n\u03b1 = 0.05 \\\\\n\nd.f. = n_1+n_2 -2 \\\\\n\n= 30+30-2 \\\\\n\n= 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.
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