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Unformatted text preview: + + Chapter 3 Section 3.1: The probability histogram Below is the probability histogram for the IS (p. 80 of text): 0.5 1.0 1.5 1.5755 0.7500 . 90 . 30 0.30 0.90 How is it determined? + 81 + + On a horizontal number line, list all possible values of x ; in this case: . 90 , . 70 , . 50 , . 30 , . 10 , . 10 , . 30 , . 50 , . 70 , . 90. Draw rectangles: Each x serves as the center of the base of its rectangle. The base of each rectangle equals (this means the rectangles touch, but do not overlap). The height of the rectangle centered at x is: P ( X = x ) / . Below is the verification of two heights given in our picture: P ( X = 0 . 10) / . 20 = 0 . 3151 / . 20 = 1 . 5755 P ( X = 0 . 30) / . 20 = 0 . 1500 / . 20 = 0 . 7500 + 82 + + Why this strange definition of height? B/c we want to consider areas. The area of the triangle centered at x is, of course, its base times its height: [ P ( X = x ) / ] = P ( X = x ). Here is the main thing to remember: In a probability histogram, the area of a rect angle equals the probability of its center value. Once we know this, we can see the symme try in the sampling distribution for the IS. We are now going to add another adjective (remember actual?). The sampling distribution of Chapter 2 will be called the exact sampling distribution and it yields the exact Pvalue. We say this b/c in Chapter 3 we will learn two aways to approximate a sampling distribution: computer simulation and fancy math. + 83 + + Section 3.2: Computer simulation In the text, I talk about a Colloquium Study (CQS). Its data are below: Treat. S F Total 1 7 7 14 2 1 13 14 Total 8 20 28 It can be shown that there are 40,116,600 possible assignments, and that 12,932,920 of these give x = 0 ( a = 4). Thus, P ( X = 0) =12,932,920 / 40,116,600= 0 . 3224. The idea of the computer simulation approxi mation is quite simple: Perhaps we can obtain a good approximation to any probability by looking at some , not all of the assignments. + 84 + + For example, I looked at 10,000 assignments for the CQS and discovered that 3267 of them gave a = 4 and x = 0. Thus, the relative frequency (RF) of occurrence of 0 is 0.3267, which is very close to its probability, 0.3224. But, I am ahead of myself. Once we decide to look at only some of the assignments, two questions arise. 1. How many should we look at? The answer is called the number of runs of the com puter simulation. As we shall see, 10,000 is a good choice for the number of runs. 2. Which ones should we look at? Well, to avoid bias, we select assignments at ran dom. If you want to see more details on this, read pages 84 and 85 in the text. But you dont need to understand this....
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This note was uploaded on 10/23/2009 for the course STAT STATS 301 taught by Professor Professorwardrop during the Fall '08 term at Wisconsin.
 Fall '08
 ProfessorWardrop
 Statistics, Probability

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