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| English | Other | Distance (mm) |
| 2.3 | ||
| 0.5 | ||
| 0.7 | ||
| 1.7 | ||
| 2.2 | ||
| 1.8 |
Hirsch reported (or at least any reasonable reader would say that she reported) that there was no difference between the centroids of activation for the two languages in early bilingual subjects. Run a t test to test that hypothesis. Do you agree with Hirsch? Think about the nature of this test and whether you would expect to retain the null hypothesis, and why or why not.
It is important that you be able to solve for t both using SPSS and by hand. The following is the standard one-sample formula for t. Make the calculations using the calculator on your PC (or any other calculator) and draw the appropriate conclusion.
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The critical value of t for (6 1) df = 2.571
Warning: This question is not as simple as it seems. What would it take for the null hypothesis to be true?
A decent psychology journal would insist that Hirsch go beyond a simple t test here--especially if the difference had been significant. They would like to know what the confidence limit on this mean distance is. Compute the confidence limits and interpret them carefully.
2. The most interesting part of Hirschs study was the fact that she compared early and late bilingual speakers. The data below represent the distance between centroids of activation in Brocas area when people are thinking in two different languages.
| Early | Late |
| 2.3 | 7.9 |
| 0.5 | 4.5 |
| 0.7 | 9.0 |
| 1.7 | 7.5 |
| 2.2 | 4.7 |
| 1.8 | 7.0 |
| 11.2 |
First solve this problem by hand, using the calculator on the PC or one that you have. Here the formula is
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and the critical value on 11 df is approximately 2.201
Next solve the problem using SPSS and paste in the output table. Annotate that table by telling me what each of the entries means.
Again, we want to know something about the "true" mean distances (or the difference in distances). Again compute confidence limits and interpret them with respect to this study.
3. Everitt (1994) compared two different treatments for anorexia. One group represented a family therapy intervention, while the other was a control group. The dependent variable was the amount of weight gain over the course of the experiment. He had earlier shown that girls in the Family Therapy condition gained weight, but that could simply be because the girls were growing older. Comparison with a Control Group would help to clarify the results. The data shown below have been taken from Everitt and are the weight gains in the two treatment conditions.
Weight Gain as a Function of Treatment Condition
| Control Condition (N = 26) | Family Therapy Group (N = 17) |
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-0.5 -9.3
-5.4 12.3 -2.0 -10.2 -12.2 11.6 -7.1 6.2 -0.2 -9.2 8.3 3.3 11.3 0.0 -1.0 -10.6 -4.6 -6.7 2.8 0.3 1.8 3.7 15.9 -10.2 Mean1 = -0.45 s1 = 7.99 s12 = 63.82 |
11.4 11.0 5.5
9.4 13.6 -2.9 -0.1 7.4 21.5 -5.3 -3.8 13.4 13.1 9.0 3.9 5.7 10.7
Mean2= 7.26 s2 = 7.16 s12 = 51.23 |
Enter these data into an SPSS file. Then examine the data for the two groups separately, getting descriptive statistics and, perhaps, histograms or box-plots. Use SPSS to draw a set of conclusions based on a t test. SPSS will give you two different values for t. From your reading of the text you should be able to explain why this is so. Use a calculator to reproduce the values that SPSS produces.
Now draw a bar graph, using SPSS, to plot the means of the two groups. Include limits for the standard error and for confidence limits (both are used in the literature, and it is not clear when to prefer one over the other). I will demonstrate this in class.
I have just put a copy of the calculations for Question 2 on a web page. Click here.
Last revised: 10/17/01