In: Statistics and Probability
You are interested in enhancing learning. You decide to test the hypothesis that continual self-testing as part of the learning process can benefit the long-term retention of information. In this case, testing refers to learning the material to a 100% accurate level and then continuing to be tested on the material during later learning trials. You decide to compare participants in the self-test condition to participants who learned the material to the 100% level and continued to study the material but did not engage in ongoing self- testing. You test the long-term retention of the information for the two groups 1 week later. The data are shown on the right. Is there support for your hypothesis?
Use the four-step hypothesis testing procedure described in class, formatted as in the sample assignment. In addition, calculate 1) the effect size and 2) confidence interval associated with the data. Show your work.
Mean recall accuracy (%) | |
Self-test condition | No self-test condition |
78.6 | 56.3 |
79.5 | 53.3 |
65.8 | 55.5 |
59.8 | 51.1 |
73.6 | 59.4 |
75.7 | 63.9 |
62.8 | 54.6 |
61.8 | 66.3 |
77.1 | 48.7 |
75.2 | 61.3 |
79.6 | 52.0 |
68.7 | 48.1 |
65.2 | 53.3 |
70.0 | 60.2 |
81.9 | 50.1 |
58.5 | 61.4 |
65.0 | 50.5 |
75.7 | 62.8 |
73.6 | 50.2 |
77.2 | 70.5 |
64.5 | 68.6 |
52.7 | 56.4 |
74.5 | 61.2 |
69.5 | 64.6 |
67.4 | 65.1 |
68.9 | 61.2 |
75.6 | 61.6 |
67.4 | 48.8 |
52.1 | 72.0 |
73.1 | 52.7 |
66.3 | |
70.5 | |
66.8 |
EFFECT SIZE = (Xbar1 - Xbar2)/Sp
= (69.53 - 58.06)/7.1972
= 1.5937
N Mean StDev SE Mean
self-test 33 69.53 7.47 1.3
no self-test 30 58.06 6.89 1.3
Difference = μ (self-test) - μ (no self-test)
Estimate for difference: 11.48
DF = 61 , t = =T.INV.2T(0.05,61) = 1.999624
Both use Pooled StDev (sp) = 7.1972
95% confidence interval = (7.846, 15.107)