An article reported the results of a peer tutoring program to help mildly mental

Question

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. Suppose there were n1 = n2 = 29 children in each group. A 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 = 47.4. For the control group, the mean score on the same test was x2 = 355.3, with sample standard deviation s2 = 42.0. Note: If a two-sample t-test is appropriate, for degrees of freedom d.f. not in the Student's t table, use the closest d.f. that is smaller. In some situations, this choice of d.f. may increase the P-value by a small amount and therefore produce a slightly more "conservative" answer.

(a) Use a 5% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began.

(i) What is the level of significance?______________

2. What is the value of the sample test statistic?___________ (Test the difference 1 2. Round your answer to three decimal places.)

(iii) Find the P-value. ____________(Round your answer to four decimal places.)

(b) Find a 95% confidence interval for 1 2. (Round your answers to two decimal places.) lower limit ________upper limit____________

Explanation / Answer

The output of t-test for difference of means is

Intermediate Calculations

Numerator of Degrees of Freedom

19127.4623

Denominator of Degrees of Freedom

346.5104

Total Degrees of Freedom

55.2003

Degrees of Freedom

55

Separate Variance Denominator

11.7602

Difference in Sample Means

-10.8

Separate-Variance t Test Statistic

-0.9184

Upper-Tail Test

Upper Critical Value

1.6730

p-Value

0.8188

Do not reject the null hypothesis

(a) Use a 5% level of significance to test the hypothesis that there was no difference in the vocabulary scores of the two groups before the instruction began.

Here the test statistic =-0.9184 and the p-value for alternative ypothesis the mean difference of experimental group is more than the control group is 0.8188

We accept the null hypothesis and conclude that there was no difference in the vocabulary scores of the two groups before the instruction began.

(i) The level of significance?_______5% =0.05_______

2. The value of the sample test statistic= -0.918

(iii) The P-value. 0.8188   (Round your answer to four decimal places.)

(b) A 95% confidence interval for 1 2. lower limit -34.37 upper limit_12.77

Intermediate Calculations

Numerator of Degrees of Freedom

19127.4623

Denominator of Degrees of Freedom

346.5104

Total Degrees of Freedom

55.2003

Degrees of Freedom

55

Separate Variance Denominator

11.7602

Difference in Sample Means

-10.8

Separate-Variance t Test Statistic

-0.9184

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