An adviser is testing out a new online learning module for a placement test. The

Question

An adviser is testing out a new online learning module for a placement test. They wish to test the claim that the new online learning module increased placement scores at a significance level of -0.01. For the context of this problem, HBef or e where the first data set represents the after test scores and the second data set represents before test scores. Assume the population is normally distributed. You obtain the following paired sample of 19 students that took the placement test before and after the learning module: After Before 694 67.8 50.4 55.9 80.1 70.8 47.3 51.1 264 20.2 56.1 56.7 70.570.8 59.2 57.2 32.839.1 54.346.9 68.6 59.3 58.453.6 67.260.8 49.557.6 46.445.8 36.6 30.2 44.6 45.5 46.2 43.2 40.1 37.7

Explanation / Answer

Here the same students who took the module so the test is paired data t - test for two dependent samples where

n = 19

degree of freedom = 18

H0 : d = 0

Ha : d > 0

here the difference table where d = after - vefore

Average difference in score dbar = 1.7842

Staandard deviation in score sd = 5.1832

Standard error of difference sed= sd/ sqrt(n) = 5.1832/sqrt(19) = 1.189

Test statistic

t = dbar /sed = 1.7842/1.189 = 1.5005

Critical test statistic for dF = 18, alpha = 0.01

tcritical = 2.5524

here t < tcritical

that leads us to a decision that we faild to reject the null hypothesis and final conclusion that we reject the claim that the new online learning module increased placement scores.

After Before Difference 69.4 67.8 1.6 50.4 55.9 -5.5 80.1 70.8 9.3 47.3 51.1 -3.8 26.4 20.2 6.2 56.1 56.7 -0.6 70.5 70.8 -0.3 59.2 57.2 2 32.8 39.1 -6.3 54.3 46.9 7.4 68.6 59.3 9.3 58.4 53.6 4.8 67.2 60.8 6.4 49.5 57.6 -8.1 46.4 45.8 0.6 36.6 30.2 6.4 44.6 45.5 -0.9 46.2 43.2 3 40.1 37.7 2.4 Average 1.7842 Std. Dev. 5.1832

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