In: Statistics and Probability
A researcher has studied subjects’ ability to learn to translate words into Morse code. He has experimented with two treatment conditions: in one condition, the subjects are given massed practice; they spend 6 full hours on the task. In the other condition, subjects are given distributed practice; they also spend 6 hours, but their practice is spread over four days, practicing 2 hours at a time. After the practice, all subjects are given a test message to encode; the dependent variable is the number of errors made. The researcher has matched the subjects on intelligence. The results are in the following table. Decide which statistical test would be appropriate, carry out the test, and evaluate the outcome. Assume a significance level of .05and that the direction of the outcome has not been What are we dealing with here? T test…. what must you do with the null hypothesis guys (accept or reject)? Massed Practice Distributed Practice S1 6 S1 5 S2 4 S2 3 S3 3 S3 2 S4 5 S4 2
Since the researcher has matched the subjects on intelligence., both the groups are dependent, and we need to carry matched-pairs t-test.
Null hypothesis H0: Mean difference in errors between massed and distributed practice is 0. That is .
Alternative hypothesis H1: Mean difference in errors between massed and distributed practice is not 0. That is .
From the data, the differences in errors between two massed and distributed practice are,
6-5, 4-3, 3-2, 5-2
= 1, 1, 1, 3
Sample mean difference, = (1 + 1 + 1 + 3)/4 = 1.5
Standard deviation of difference , s = sqrt[((1 - 1.5)2 + (1 - 1.5)2 + (1 - 1.5)2 + (3 - 1.5)2 )/3] = 1
Standard error of mean difference , se = s / = 1 / = 0.5
Test statistic, t = ( - ) / se = (1.5 - 0) / 0.5 = 3
Degree of freedom = n-1 = 4-1 = 3
For two-tail test, p-value = p(t > 3) = 0.0577
Since, p-value is greater than 0.05 significance level, we fail to reject null hypothesis H0 and conclude that there is no strong evidence that the mean difference in errors between massed and distributed practice is not 0.