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T Test for Paired Data When to use: in situations where the second of each set of scores is related to the first. The classic case is that of a study where there is no control group and we are having to gather pre-test and post-test scores from the same individuals. Example: A sample of 10 high school seniors has tried two methods of learning words in a second (used to be called foreign) language. In the first part of the study they studied in pairs, after which they were tested for the number of words known. In the second part of the study they tried it with a second method: alone. They were again tested for the number of words known. Did the different procedures result in different achievement? The null hypothesis says there is is no difference in the number of words known between the two conditions. Number of words learned __________________________ 2 Studied in pairs Studied alone 14 14 15 16 14 15 13 10 12 16 __ 9 10 9 8 14 7 8 10 13 9 __ 139=sum of pairs 97=sum of alones D 5 4 6 8 0 8 5 0 -1 7 __ 42 D 25 16 36 64 0 64 25 0 1 49 __ 280 The T test for paired (correlated, dependent) data uses the formula of _ D T= _ where D is the mean of the differences in scores between pairs of matched subjects S diff _ D is the sum of D divided by N, or S S = diff d _____ N 2 42/10 =4.2. S = d S 280 - (42) 10 ____________ 9 = 3.39 = diff = 3.39 3.16 280- 176.4 __________ 9 = = 103.6 _____ 9 = 11.51 = 3.39 1.07 10 T= 4.2 1.07 = 3.93 The T that is required for significance with N-1 or 9 pairs of data at the .05 level (two-tailed test) is 2.262. Our T of 3.93 is greater than the table value of T, so we reject the null hypothesis; studying in pairs is more effective in learning than studying alone. If we repeated the experiment 100 times, we could expect this trend in the data 95 times. Five times out of 100 studying alone would result in the retention of more words in a second language.