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# week13 - Lecture 32 Nancy Pfenning Stats 1000 Chapter 16...

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Unformatted text preview: Lecture 32 Nancy Pfenning Stats 1000 Chapter 16: Analysis of Variance Example Suppose your instructor administers 3 different forms of a final exam. When scores are posted, you see the observed mean scores for those 3 different forms—82, 66, and 60—are not the same. Is this due to chance variation, or do the 3 exams not share the same level of difficulty? Version 1: 65, 73, 78, 79, 86, 93, 100 Version 2: 39, 58, 63, 67, 69, 74, 92 Version 3: 39, 52, 62, 64, 66, 77 [Note: if there were only 2 different exams, a two-sample t test could be used.] It would be unrealistic to expect 3 identical sample mean scores, even if the exams were all equally difficult: in other words, there’s bound to be some variation among means in the 3 groups. Also, of course there will be variation of scores within each group. If the ratio of variation among groups to variation within groups is large enough, we will have evidence that the population mean scores for the 3 groups actually differ. Picture two possible configurations for the data, both of which represent 3 data sets with means 82, 66, and 60. Thus, variation among means (from the overall mean of 70) would be the same for both configurations. 1. There could be large within-group variation, such as we see in Exams 1a, 2a, and 3a, in which case the ratio among to within is small; in this case, the data could be coming from populations that share the same mean. 2. There could be small within-group variation, such as we see in Exams 1b, 2b, and 3b, in which case the ratio among to within is large; in this case, the population means probably differ. 141 Note that sample size, although not yet mentioned, plays a role, too. If there were 100 students in each group, we would set more store by the differences than if there were only 2 students in each group....
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week13 - Lecture 32 Nancy Pfenning Stats 1000 Chapter 16...

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