Question

HW#9 - Chapter 18 (20pts for turned in on time) O. (12pts) To evaluate the effectiveness of a short MATH review Wor to take the remedial MATH course. A researcher selecte math test before the workshop. Another test with t workshop. And below are the test scores. Provide a statistical test to se effective or not, use 0.05 significance level. Before 65 72 54 After 73 70 67 80 Diff: 5 x = 10.4 ne5 ATH review workshop for students who need searcher selected a SRS of 5 students, and gave them a culty level was given after the a statistical test to see if the workshop is 63 76

Answer 1

Here we have paired data so paired t test will be used. Let d = before- after

If test is effective then true average difference should be less than zero.

Following table shows the calculations:

Before	After	d=test1-test2	(d-mean)^2
65	70	-5	29.16
72	67	5	237.16
54	80	-26	243.36
60	73	-13	6.76
63	76	-13	6.76
Total		-52	523.2

Sample size n = 5 Sample sum of difference: d = -52 Sample mean of differences: a-2a --52/5 = -10.4 Sample standard deviation of differences: E(d-d)? Savn-1 -=sqrt(523.2/(5-1)) =11.4368

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Hypotheses are: Ho: la =0 H:Hg <0 Level of significance: a =0.05 Test is one tailed (left tailed). Degree of freedom: df =n-1=4 Test Statistics: d d t (-10.4- = = (-10.4-0)/(11.4368/sqrt(5)) Sg/n = -2.033 Critical value: CV = -2.132 Rejection Region: Ift <-2.132, Reject HO Decision: Since test statistics does not lie in rejection region so we fail to reject the null hypothesis P-value: P-value=0.056 Decision: Since p-value is not less than level of significance so we fail to reject the null hypothesis Excel function for critical value: Excel function for p-value: ETINV(0.1,4) =TDIST(2.033,4,1)

Conclusion: There is no evidence to conclude that test is effective.