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l4 - Math 127B Notes for lecture 4 Mrinal Raghupathi...

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Math 127B, Notes for lecture 4 Mrinal Raghupathi Wednesday, January 26th, 2011 Administrivia The topics to be discussed today are the chance model for inference and chance errors in sampling. Most of the material is taken from chapter 20.3, 20.4 and 20.5. Reminders 1. Start reading chapter 21 for Monday. There is an online quiz posted. Details on how to access the quiz were sent by email. 2. Homework 2 is due Thursday, 1/27 in recitation. 3. There will be a quiz in recitation over chapter 19. 1 Using the normal curve In the box model the sum of draws follows the normal curve with an average given by the expected value and a standard deviation given by the standard error for the sum. In a similar way the percentage of draws from the box follows a normal curve with an average given by the expected value of the percentage and a standard deviation given by the standard error of the percentage. Example 1 . Let us return to our example about the number of seniors at Vanderbilt. Suppose that we draw a sample of 400 students from the student population of 7,000. Suppose that 25% of students are seniors. The box model for this example has been discussed before. We have 1,750 tickets labeled 1 and 5,250 labeled 0 . the average of this box is 0.25. The standard deviation is given by 0 . 25 × 0 . 75 0 . 43. If we make 400 draws from the box, then the expected value of the sum is 0 . 25 × 400 = 100 and the SE of the sum is 400 × 0 . 43 8 . 6. Now for the percentages we see that the EV of the percentage is 0 . 25 × 100% = 25%. The standard error for percentages is given by 8 . 6 / 400 × 100% = 2 . 15%. Thus, in the sample we should expect that the percentage of seniors is about 25% give or take 2%. o 1
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Example 2 . Let us now solve a related problem. In a sample of 400 students, what is the chance
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