vicki
02-28-2009, 12:43 AM
Suppose we have three different methods of teaching the same material. We wish to determine if one method is superior to the others. We can use SPSS to conduct an ANOVA test to answer that question.
First we sample 4 (in the interest of brevity) students from each of the different classes, test them, and get the following results:
Student Method Score
1 1 71
2 1 75
3 1 65
4 1 69
5 2 90
6 2 80
7 2 86
8 2 84
9 3 72
10 3 77
11 3 76
12 3 79
To conduct the analysis of variance test:
1. Build the above sheet in SPSS
2. Click ‘Analyze’
3. Click ‘Compare Means’
4. Click ‘One-Way ANOVA’
5. Since score depends on the method of instruction, Score is our dependent variable.
6. Click on Score and the upper arrow to move it to the ‘Dependent List’.
7. Click on Method and move it to the ‘Factor’ box.
8. Click on ‘Options’ and check the box next to ‘Discriptive’.
9. Click ‘Continue’.
10. Click ‘OK.
You should now have the output from the test. SPSS gives us the descriptive statistics for each of the methods as well as an analysis of variance summary.
Notice that our F statistic is 15.785.
Our null hypothesis in this test was that all the methods produce the same result (μ1 = μ2 = μ3).
To reject that hypothesis, there must be a significant difference in at least one of the means. To decide if there is we compare our F statistic to a critical F value we find in the F table in the book. To find that critical value we need to know the degrees of freedom from ‘between’ the groups and ‘within’ the groups.
From your output ANOVA table, note that the degrees of freedom (df) between the groups is 2, and within the groups is 9.
Looking at the tables in appendix E, Distribution of F, degrees of freedom ‘between’ are read across the top of table, and degrees of freedom ‘within’ are read down the left-hand side. Going across the top of the table to 2, then down that column to 9, we find the critical value of 4.26.
Since our F statistic of 15.785 is greater than the critical F of 4.26, we reject Ho and conclude that at least one of the methods produces results that are significantly different than the others.
First we sample 4 (in the interest of brevity) students from each of the different classes, test them, and get the following results:
Student Method Score
1 1 71
2 1 75
3 1 65
4 1 69
5 2 90
6 2 80
7 2 86
8 2 84
9 3 72
10 3 77
11 3 76
12 3 79
To conduct the analysis of variance test:
1. Build the above sheet in SPSS
2. Click ‘Analyze’
3. Click ‘Compare Means’
4. Click ‘One-Way ANOVA’
5. Since score depends on the method of instruction, Score is our dependent variable.
6. Click on Score and the upper arrow to move it to the ‘Dependent List’.
7. Click on Method and move it to the ‘Factor’ box.
8. Click on ‘Options’ and check the box next to ‘Discriptive’.
9. Click ‘Continue’.
10. Click ‘OK.
You should now have the output from the test. SPSS gives us the descriptive statistics for each of the methods as well as an analysis of variance summary.
Notice that our F statistic is 15.785.
Our null hypothesis in this test was that all the methods produce the same result (μ1 = μ2 = μ3).
To reject that hypothesis, there must be a significant difference in at least one of the means. To decide if there is we compare our F statistic to a critical F value we find in the F table in the book. To find that critical value we need to know the degrees of freedom from ‘between’ the groups and ‘within’ the groups.
From your output ANOVA table, note that the degrees of freedom (df) between the groups is 2, and within the groups is 9.
Looking at the tables in appendix E, Distribution of F, degrees of freedom ‘between’ are read across the top of table, and degrees of freedom ‘within’ are read down the left-hand side. Going across the top of the table to 2, then down that column to 9, we find the critical value of 4.26.
Since our F statistic of 15.785 is greater than the critical F of 4.26, we reject Ho and conclude that at least one of the methods produces results that are significantly different than the others.