AP Statistics February 28, 2012

AP Statistics B warm-up Tuesday, February 28, Assume that after you draw from a deck of cards, you replace the card and reshuffle. What is the probability of your drawing: 1. 3 aces? 2.3 hearts 3.2 hearts and 2 spades? 2.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.

Answers to warmups Tuesday, February 28, Drawing WITH replacement 1.3 aces?= (4/52) (4/52) (4/52) = (1/13) (1/13) (1/13)=not very much 2.3 hearts =(13/52)(13/52)(13/52)= (1/4)(1/4)(1/4) = 1/ hearts and 1 spades = same as 1.2 (why?) 2.Drawing WITHOUT replacement 1.3 aces?= (4/52) (3/51) (2/50) = hey, YOU guys are the ones with the calculators! Why am **I** doing the calculations???? 2.3 hearts =(13/52)(12/51)(11/50)=some small number 3.2 hearts and 1 spades = (13/52)(12/51)(13/52)

No narration today Sorry for that. The microphone has an echo on it that I haven’t been able to resolve. I’ll 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!

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 value” I have an exercise on expected value that we’ll do to introduce the concept Feedback? Questions from last night’s homework you want me to review with you tomorrow? Globe at Night tomorrow for 3 rd period, so bring data No homework tonight, but turn in the homework from yesterday.

Reversing the Conditioning Go to tree diagrams on pp , and final tree diagram on binge drinking on p. 357 Probabilities 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 you’ve had an accident?) This is “reversing the conditioning”

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

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

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

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

Application: false positives in tuberculosis (TB) testing New vocabulary: – False positive – False negative Examples: – In TSA-land when you’re flying: False positive=your mother being erroneously identified as a terrorist; False negative=Osama bin Laden not being identified as a terrorist – Examinations: False positive: you pass when you should have failed; false negative: you fail when you should have passed

Important point! Determining whether something is a false negative or false positive depends on how you word the question – For 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 student’s mastery of the material? If you word the question differently, your false positives and false negatives will be different!

TB example (p. 359) Demographically, only 5 out of 10,000 people have TB. Therefore the P(TB)= 5 out of 10,000, or The probability of NOT having Tb is the complement of having TB, i.e., P(no TB)=1-P(TB) = = Note: 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 9,995÷10,000 =

Calculating the false positives False 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.

Extra care for false negatives This sounds like “bait and switch,” since they’re 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 don’t graduate from college as the result of a false positive), so tolerances are much narrower for false negatives For 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…..

Tree of outcomes on p.359 You calculate a tree of probabilities just like we did last week Summary: – TB and tested positive: – TB but tested negative: – TB-free and tested negative: – TB-free but tested positive: Here’s we’re looking for POSITIVE outcomes Work out what the probabilities would be for positive outcomes, then move to the next slide

A linguistic approach to the problem Sometimes worrying about the formula leads us astray. Let’s walk through it using ordinary language. What results test positive? Two: those who have TB and correctly test positive, and those who DON’T have TB but still test positive. Remember that this is our denominator Here, that’s (have TB and test positive) plus (don’t have TB but test positive), or Numerator is those who have TB and test positive Answer is just under 5%: 0.047

Interpreting the result The 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? It’s still correct, because of the much higher false-negative rate (99.9%) than the false- positive rate (99%)

Exercise Do 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.

Answer to problem

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

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

Rules 1.Ms. Thien has a bag full of money. Well, not real money, but 20 bills, distributed as follows: 1.14 bills of $5 2.2 bills of $ bills of $20. 2.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. 3.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.

Assignment 1.Work 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. 2.After you have estimated the “expected value” that you would get, let’s put it to the test. You will need your calculators and a scorekeeper for this part of the exercise. 3.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). 4.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.

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 you’ll 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.)

New activity Ms. Thien will now give you a different bag of money, distributed as follows: – 15 $5-bills – 2 $10-bills – 3 $20-bills Repeat 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 you’re done, move on to the next slide and I’ll show you how to do the calculations more easily.

Calculating expected value Let’s consider an easier game. Suppose you had 2 $20-bills in the bag, and 18 blanks pieces of paper. What is the probability that you’d 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%.

Calculating expected value (part 2) What’s the average value of each time you draw? Well, what are the possible outcomes? On AVERAGE, you’ll get 0 for 18 of your draws, and $20 for 2. That’s $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.

Calculating expected value (part 3) But there’s an easier way of calculating it We already know your probability of getting $20 is 0.10, and the probability of getting nothing is Multiply by the probabilities and add. What do you get? I hope what you got is $2/draw: ($20×0.10 )+($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.

Application to our first problem On 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= 14/2/4 Probabilities: 0.7 for $5, 0.1 for $10, and 0.2 for $20 Expected value = ($5×0.7)+($10×0.1)+($20×0.2)=$3.50+$1+$4 = $8.50.

Application to the second problem Different 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?

Last slide If you have time, work through the example in the book on pp Somebody should volunteer to lead the discussion. Don’t forget to bring your Globe at Night data tomorrow for entry.