Question

An adviser is testing out a new online learning module for a placement test. They wish to test the claim that on average the new online learning module increased placement scores at a significance level of α = 0.05. For the context of this problem, μD=μnew–μold where the first data set represents the new test scores and the second data set represents old test scores. Assume the population is normally distributed. H0: μD = 0 H1: μD < 0 You obtain the following paired sample of 19 students that took the placement test before and after the learning module:

An adviser is testing out a new online learning module for a placement test. They wish to test the claim that on average the new online learning module increased placement scores at a significance level of α = 0.05. For the context of this problem, μ_D=μ_new–μ_old where the first data set represents the new test scores and the second data set represents old test scores. Assume the population is normally distributed.

H₀: μ_D = 0

H₁: μ_D < 0

You obtain the following paired sample of 19 students that took the placement test before and after the learning module:

Find the p value 4 decimal places

New LM	Old LM
57.1	55.8
58.3	51.7
83.6	76.6
50.5	47.5
51.5	48.6
20.6	15.5
35.2	29.9
46.7	54
23.5	21
48.8	58.5
53.1	42.6
76.6	61.2
29.6	26.3
14.5	11.4
43.7	56.3
57	46.1
66.1	72.9
38.1	43.2
44.4	51.1

Answer 1

New	old	d= new-old	(d-dbar)^2
57.1 58.3 83.6 50.5 51.5 20.6 35.2 46.7 23.5 48.8 53.1 76.6 29.6 14.5 43.7 57 66.1 38.1 44.4	55.8 51.7 76.6 47.5 48.6 15.5 29.9 54 21 58.5 42.6 61.2 26.3 11.4 56.3 46.1 72.9 43.2 51.1	1.3 6.6 7 3 2.9 5.1 5.3 -7.3 2.5 -9.7 10.5 15.4 3.3 3.1 -12.6 10.9 -6.8 -5.1 -6.7 M: 1.51	0.04 25.9 30.13 2.22 1.93 12.88 14.36 77.63 0.98 125.68 80.81 192.92 3.2 2.53 199.11 88.16 69.06 43.7 67.41 S: 1038.66

1) null and alternate hypothesis

H0:d=0

H1:d<0

2) level of significance = 0.05

3 ) test statistic

Mean: 1.51
μ = 0
S² = SS⁄df = 1038.66/(19-1) = 57.7
S²_M = S²/N = 57.7/19 = 3.04
S_M = √S²_M = √3.04 = 1.74

T-value Calculation

t = (M - μ)/S_M = (1.51 - 0)/1.74 = 0.87

4) p value = 0.1987