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. Suppose that in the experiment, a reading test was administered to both an experimental group and a control group after 6 weeks of instruction, score on the vocabulary portion of the test was x1 368.4, with group was x2·346 0 with sa ple standard deviation s2-52.4 Nede: tf a two-sample t-test is appropriate, for degrees of freedom df. not in the Student's r table, use the dasest d, that is s mair In some situations, this choice of dif, may ncrease the value by a small amount and therefore produce a slightly more-conservative" answer. group after 6 weeks of instruction, during which the experimental group received peer tutoring and the control group did not. For the experimental group 30 chldren, the mean test was x1 -368.4, with sample standard deviation s-38.6. The average score on the vocabulary portion of the test for the n2 30 subjects in the control (a) use a 1% level of spificance to test the claim that the experimental group performed better than the control group (0) What is the level of significance? 01 State the null and alternate hypotheses @ Ho: ja.-P2, Hl: 1" 2 )What sampling distribution will you use? What assumptions are you making? The Student's t. Both sample sizes are large with known standard deviations The Student's t. Both sample sizes are large with unknown standard deviations The standard normal. Both sample sizes are large with known standard deviations. The standard normal. Both sample sizes are large with unknown standard deviations what is the value of the sample test statistic, (Test the dmerencep.-P2. Round your answer to three deo napen) (e) Find the P value. (Round your answer to four decimal places) Sketch the sampling distribution and show the area corresponding to the P-value

Explanation / Answer

(a) Level of signficance = 0.01

H0 : 1 = 2

Ha : 1 >2

(b) Pooled Standard Deviation sp = sqrt [(n1 -1)s12 + [(n2 -1)s22)/ (n1 + n2 -2) ]

sp = sqrt [(38.62 + 52.42)/2  ] = 46.02

standard error = sp * sqrt [1/n1 + 1/n2 ] = 46.02 * sqrt [1/30 + 1/30] = 11.88

t = (1 - 2) / se0 = (368.4 -346.0)/ 11.88 = 1.886

t58,0.05 = 1.6716

P - value = 0.0321 is greater than 0.01 so it is not significant.

sampling distribution graph

(b) 98% confidence interval for 1 - 2 = (1 - 2) +- t58,0.01 * se0

=(368.4 - 346.0) + 2.392 * 11.88

= (-6.02, 50.82)

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