In the journal Mental Retardation , an article reported the results of a peer tu
ID: 3160557 • Letter: I
Question
In the journal Mental Retardation, an article reported the results of a peer tutoring program to help mildly mentally retarded children learn to read. In the experiment, the mildly retarded children were randomly divided into 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 n1 = n2 = 36 children in each group. The Gates-MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was x1 = 344.5, with sample standard deviation s1 = 46.7. For the control group, the mean score on the same test was x2 = 356.5, with sample standard deviation s2 = 49.9. Use a 1% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began. (a) What is the level of significance? (B) What is the value of the sample test statistic? (Test the difference 12. Round your answer to three decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
In the journal Mental Retardation, an article reported the results of a peer tutoring program to help mildly mentally retarded children learn to read. In the experiment, the mildly retarded children were randomly divided into 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 n1 = n2 = 36 children in each group. The Gates-MacGintie Reading Test was given to both groups before instruction began. For the experimental group, the mean score on the vocabulary portion of the test was x1 = 344.5, with sample standard deviation s1 = 46.7. For the control group, the mean score on the same test was x2 = 356.5, with sample standard deviation s2 = 49.9. Use a 1% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began. (a) What is the level of significance? (B) What is the value of the sample test statistic? (Test the difference 12. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(a) What is the level of significance? (B) What is the value of the sample test statistic? (Test the difference 12. Round your answer to three decimal places.)
(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Explanation / Answer
a)
As given,
alpha = 0.01 [ANSWER]
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b)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 0
Ha: u1 - u2 =/ 0
At level of significance = 0.01
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 344.5
X2 = 356.5
Calculating the standard deviations of each group,
s1 = 46.7
s2 = 49.9
Thus, the standard error of their difference is, by using sD = sqrt(s1^2/n1 + s2^2/n2):
n1 = sample size of group 1 = 36
n2 = sample size of group 2 = 36
Thus, df = n1 + n2 - 2 = 70
Also, sD = 11.39066382
Thus, the t statistic will be
t = [X1 - X2 - uD]/sD = -1.053494352 [ANSWER, TEST STATISTIC]
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c)
Also, using p values, as this is two tailed,
p = 0.295737392 [ANSWER]
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As P > 0.01, WE FAIL TO REJECT THE NULL HYPOTHESIS.
There is no significant difference in the vocabulary scores of the two groups before the instruction began. [CONCLUSION]
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Hi! If you use z distribution in your class for this sample size, the the P value (part c) is Pvalue = 0.292114481. [ALTERNATE ANSWER FOR C]
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