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AP StatisticsFebruary 28, 2012Narration starts on slide 41AP Statistics B warm-upTuesday, February 28, 2012Assume that after you draw from a deck of cards, you replace the card and reshuffle. What is the probability of your drawing: 3 aces?3 hearts2 hearts and 2 spades?Now assume that you do NOT replace the cards. Calculate the probabilities using the 3 distributions above. When you are finished, the answers are on the next slide.2Answers to warmupsTuesday, February 28, 2012Drawing WITH replacement3 aces?= (4/52) (4/52) (4/52) = (1/13) (1/13) (1/13)=not very much3 hearts =(13/52)(13/52)(13/52)= (1/4)(1/4)(1/4) = 1/642 hearts and 1 spades = same as 1.2 (why?)Drawing WITHOUT replacement3 aces?= (4/52) (3/51) (2/50) = hey, YOU guys are the ones with the calculators! Why am **I** doing the calculations????3 hearts =(13/52)(12/51)(11/50)=some small number2 hearts and 1 spades = (13/52)(12/51)(13/52)

3No narration today(original slide, now with narration)Sorry for that. The microphone has an echo on it that I havent been able to resolve. Ill be on the phone to tech support today to get it resolved.This is an opportunity, however, for you to work together as a class. Share in the reading of the slides. Make sure you understand the material before moving on.Let me know how this went today. I prefer to narrate, if only to clarify, but maybe you prefer silence, which is OK, too!4

Topics for today(Tuesday, February 28, 2012)Finish Chapter 15 (reversing the conditioning)Introduction to Chapter 16 on Random variables and the concept of expected valueI have an exercise on expected value that well do to introduce the conceptFeedback? Questions from last nights homework you want me to review with you tomorrow?Globe at Night tomorrow for 3rd period, so bring dataNo homework tonight, but turn in the homework from yesterday.5

5Reversing the ConditioningGo to tree diagrams on pp. 355-357, and final tree diagram on binge drinking on p. 357Probabilities here are probabilities of having an accident, given that you were a certain type of drinker, e.g.P(accident|binge)P(accident|moderate)P(accident|abstain)But what if we want to know P(binge|accident) instead? (i.e., the probability of being a binge drinker, give that youve had an accident?)This is reversing the conditioning6

Using the General Multiplication RuleOne way is to use and revise the general multiplication rule:

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But we were looking for P(A|B), not P(B|A)True, but its just a matter of substituting P(A|B) for P(B|A) in the equation:

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Doing the calculationWe already have P(accident), which is we calculate by adding up all the probabilities of having an accident:P(accident|binge)= 0.075P(accident|moderate)=0.033P(accident|abstain)=0P(accident), then, equals 0.075+0.033+0=0.108. This is the DENOMINATOR of the equation

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Doing the calculation (continued)How do we find ? We already have it as the probability of Binge and Accident, which is 0.075.So

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Application: false positives in tuberculosis (TB) testingNew vocabulary:False positiveFalse negativeExamples:In TSA-land when youre flying: False positive=your mother being erroneously identified as a terrorist; False negative=Osama bin Laden not being identified as a terroristExaminations: False positive: you pass when you should have failed; false negative: you fail when you should have passed

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Important point!Determining whether something is a false negative or false positive depends on how you word the questionFor TB, does the test correctly identify persons suffering from tuberculosis?For TSA, does the system accurately identify terrorists?For examinations, does the test correctly test the students mastery of the material?If you word the question differently, your false positives and false negatives will be different!12

TB example (p. 359)Demographically, only 5 out of 10,000 people have TB. Therefore the P(TB)= 5 out of 10,000, or 0.0005. The probability of NOT having Tb is the complement of having TB, i.e., P(no TB)=1-P(TB) = 1.0000-0.0005 = 0.9995Note: you can also calculate the non-TB rate directly from the statistics. If only 5 out of 10,000 people have TB, then 9,995 out of 10,000 do not, and 9,99510,000 = 0.999513

Calculating the false positivesFalse positive rate is 1% (1 out of 100 people who do not have TB will be incorrectly diagnosed as having it).The corresponds to a 99% accuracy rate.Read the last paragraph on p. 358 carefully, and then move to the next slide.14

Extra care for false negativesThis sounds like bait and switch, since theyre giving you TWO different success rates (1% for false positives, and 0.01% for false negatives)However, false negatives are MUCH more serious than false positives (though maybe not if you dont graduate from college as the result of a false positive), so tolerances are much narrower for false negativesFor us, this means that we have TWO sets of data to work with! 99%/1% for false positives, and 99.9%/0.1% for false negatives! Welcome to the real world, my children..15

Tree of outcomes on p.359You calculate a tree of probabilities just like we did last weekSummary:TB and tested positive: 0.0004995TB but tested negative: 0.0000005TB-free and tested negative: 0.989505TB-free but tested positive: 0.009995Heres were looking for POSITIVE outcomesWork out what the probabilities would be for positive outcomes, then move to the next slide16

A linguistic approach to the problemSometimes worrying about the formula leads us astray. Lets walk through it using ordinary language.What results test positive? Two: those who have TB and correctly test positive, and those who DONT have TB but still test positive. Remember that this is our denominatorHere, thats 0.0004995 (have TB and test positive) plus 0.009995 (dont have TB but test positive), or 0.010495.Numerator is those who have TB and test positive Answer is just under 5%: 0.04717

Interpreting the resultThe book phrases the results as the chance of having TB even after you test positive is less than 5%. Do you agree with this conclusion?Its still correct, because of the much higher false-negative rate (99.9%) than the false-positive rate (99%)18

ExerciseDo problem 44 on p 366. It follows these examples exactly. Draw a tree. Someone needs to share their result and show what conclusion they reached. When you have done so, move on to the next slide for the answer.19

Answer to problem 440.81420

Here endeth Chapter 15Give any questions you have to Ms. Thien; she will pass them on to me and Ill answer them tomorrow.Any problems on the homework, pass them on as well.Other questions or comments? You know the drill!21

Introduction to Chapter 16Here we are dealing with expected value, which usually proves a difficult concept to understand.To help you, Ive devised a game of sorts, sort of a lottery. Rules are on the next slide.22

RulesMs. Thien has a bag full of money. Well, not real money, but 20 bills, distributed as follows:14 bills of $52 bills of $104 bills of $20.The lottery is a pick with replacement. I.e., a person pays money to reach into the bag and selects a bill at random, which he/she/it can keep. The bill is then replaced so that you maintain the distribution I gave you in #1 above.What is a reasonable price to charge someone to play this game? Reasonable means a price that accurately reflects the likely outcome. Answering this question is your assignment for the day. Move on to the next page for specific instructions.23

AssignmentWork for 5-10 minutes at your table and come up with a figure. After all tables have reached a consensus, pick someone to report out your findings. Somebody needs to keep track of the figures on the board.After you have estimated the expected value that you would get, lets put it to the test. You will need your calculators and a scorekeeper for this part of the exercise.Ms. Thien will pass a paper bag of money around. Each student should draw a bill at random, show it to the class, and put it back in the bag (sorry, I only made so many copies). As each student selects a bill, someone needs to be tallying it on the board. Do not go on to the next slide until you have completed this activity.24

Work!Now that everybody has drawn and replaced a bill, calculate the average pay-out that a randomly-chosen student got.Compare this average to the reported suggestions of the expected value. (It is possible that youll have wildly divergent results. The Law of Large Numbers means large numbers, typically more than exist in a single class, and your figures may not converge to what we will calculate as an expected value.) 25

New activityMs. Thien will now give you a different bag of money, distributed as follows:15 $5-bills2 $10-bills3 $20-billsRepeat what you just did: calculating the expected value, and having each person draw and calculate the average payout. Be sure that somebody writes down the conclusions for both activities and gives it to Ms. Thien to report to me.When youre done, move on to the next slide and Ill show you how to do the calculations more easily.26

Calculating expected valueLets consider an easier game.Suppose you had 2 $20-bills in the bag, and 18 blanks pieces of paper. What is the probability that youd draw a $20 bill?Well, clearly, there are 2 bills out of 20, so your odds of drawing it are 2 out of 20, or 1/10, aka 10%. 27

Calculating expected value (part 2)Whats the average value of each time you draw?Well, what are the possible outcomes?On AVERAGE, youll get 0 for 18 of your draws, and $20 for 2. Thats $40 out of 20 (18 draws you net nothing, and 2 you get $20 each time)Add it up, and you get $40/20 turns = $2/turn as your average. So you should be willing to $2 per draw. 28

Calculating expected value (part 3)But theres an easier way of calculating itWe already know your probability of getting $20 is 0.10, and the probability of getting nothing is 0.90.Multiply by the probabilities and add. What do you get?I hope what you got is $2/draw: ($200.10 )+($0 0.90). Check the math to make sure we agree.You can use this same procedure to calculate the expected value in our exercise.29

Application to our first problemOn the first problem, we had three values--$5, $10, and $20. What you do is simply replace the 0 value we just did with THEIR value and probability.Distribution: $5/$10/$20= 14/2/4Probabilities: 0.7 for $5, 0.1 for $10, and 0.2 for $20 Expected value = ($50.7)+($100.1)+($200.2)=$3.50+$1+$4 = $8.50. 30

Application to the second problemDifferent distribution of $5/$10/$20: 15/2/3, or in probabilities 15/20 for $5, 2/20 for $10, and 3/20 for $20, i.e. 0.75/0.10/0.15.Same thing as before: multiply probabilities by outcomes, and you should get$7.75, right?31

Last slideIf you have time, work through the example in the book on pp.368-69. Somebody should volunteer to lead the discussion.Dont forget to bring your Globe at Night data tomorrow for entry.32