1. MM207 Statistics Welcome to the Unit 7 Seminar With Ms. Hannahs

2. Unit 7 Assignments and Final Exam Most (if not all) of your MSL StatCrunch Homework and Quiz work in Unit 7 is about looking at the data tables given and interpreting them. See Help on Contingency tables for computing probabilities. Expected value problems and Section 6.5 (and/or/conditional probability) problems will be computed with pencil and paper and not StatCrunch, best I can tell. Get started on the final as I will not take any late final exams (due date is Tuesday, August 7). By the end of Unit 7, you should be able to have problems 1-6 completed!! Show all work on computations and paste StatCrunch results when needed.

3. Statistical Significance A set of measurements or observations in a statistical study is said to be statistically significant if it is unlikely to have occurred by chance. Is the probability that the observed difference occurred by chance less than or equal to 0.05 (or 1 in 20)? If the answer is yes (the probability is less than or equal to 0.05), then we say that the difference is statistically significant at the 0.05 level. If the answer is no, the observed difference is reasonably likely to have occurred by chance, so we say that it is not statistically significant. The choice of 0.05 is somewhat arbitrary, but it’s a figure that statisticians frequently use. Nevertheless, other probabilities are often used, such as 0.1 or 0.01.

4. Example: An event is considered “significant” if its probability is less than or equal to 0.05. a.Yesb.No Is it significant to be dealt an ace when you are dealt one card from a standard 52-card deck? (There are four aces in the deck.)

5. Example: An event is considered “significant” if its probability is less than or equal to 0.05. a.Yesb.No Is it significant to be dealt an ace when you are dealt one card from a standard 52-card deck? (There are four aces in the deck.) Note: this is an example of theoretical probability

6. Three Approaches to Finding Probability A theoretical probability is based on assuming that all outcomes are equally likely. It is determined by dividing the number of ways an event can occur by the total number of possible outcomes. A relative frequency probability is based on observations or experiments. It is the relative frequency of the event of interest. A subjective probability is an estimate based on experience or intuition.

7. Theoretical Probability Experiment: Rolling a single die Sample Space: All possible outcomes from experiment S = {1, 2, 3, 4, 5, 6} Outcomes are the most basic possible results of observations or experiments Event: a collection of one or more outcomes (denoted by capital letter) Event A = {3} Event B = {even number} Probability = (number of favorable outcomes) / (total number of outcomes) P(A) = 1/6 P(B) = 3/6 = 1/2

8. Theoretical Probability – OR problems Section 6.5 Page 272 green box ALWAYS works. P(A or B) = P(A) + P(B) – P(A and B) Do not use Page 271 green box. Let’s look at examples and see this is better. Recall:S = {1, 2, 3, 4, 5, 6} Event A = {3} Event B = {even number} Event C = {4, 6} Find P(B or C) P(B or C) = P(B) + P(C) – P(B and C) = 3/6 + 2/6 – 2/6 = 3/6 = 1/2 Find P(A or B) P(A or B) = P(A) + P (B) – P(A and B) = 1/6 + 3/6 – 0/6= 4/6 =2/3

9. Counting Possible Outcomes Suppose process A has a possible outcomes and process B has b possible outcomes. Assuming the outcomes of the processes do not affect each other, the number of different outcomes for the two processes combined is: a * b This idea extends to any number of processes. If a third process C has c possible outcomes, the number of possible outcomes for the three processes combined is: a * b * c. Tree Diagrams are good for this. See page 241

10. Applying the Counting Rule How many outcomes are there if you roll a fair die and toss a fair coin? The first process, rolling a fair die, has six outcomes The second process, tossing a fair coin, has two outcomes (H, T). Therefore, there are 6 × 2 = 12 outcomes for the two processes together (1H, 1T, 2H, 2T,..., 6H, 6T). 1 2 3 4 5 6 H T H T H T H T H T H T Sample Space is 1H 1T 2H 2T 3H 3T 4H 4T 5H 5T 6H 6T Note: Probability of getting a 1 and then an H is 1/12

11. And Probability Problems And Probability for Independent Events The key to And Problems is interpreting when there is “with replacement” (meaning independent) as well as when there is “no replacement” (meaning dependent) Two events are independent if the outcome of one event does not affect the probability of the other event. Consider two independent events A and B with probabilities P(A) and P(B). The probability that A and B occur together is P(A and B) = P(A) * P(B)

12. Example “And” - Independent P(A and B) = P(A) * P(B) Setup: Two cards are to be selected with replacement from a deck of cards. Example 1: Find the probability that two red cards will be selected. P(Red and Red) = P(Red) * P(Red) = (26/52) * (26/52) = 1/4 Example 2: Find the probability that a red card and then a black card will be selected. P(Red and Black = P(Red) * P(Black) = (26/52) * (26/52) = 1/4

13. And Probability for Dependent Events Two events are dependent if the outcome of one event affects the probability of the other event. The probability that dependent events A and B occur together is P(A and B) = P(A) * P(B given A) where P(B given A) means the probability of event B given the occurrence of event A. This is often written as P(A) * P(B|A)  fancy notation

14. Example “And” - Dependent P(A and B) = P(A) * P(B given A) = P(A) * P(B|A)  fancy notation Setup: Two cards are to be selected without replacement from a deck of cards. Example 1: Find the probability that two red cards will be selected. P(Red and Red) = P(Red) * P(Red) = (26/52) * (25/51) = 25/102. Example 2: Find the probability that a red card and then a black card will be selected. P(Red and Black) = P(Red) * P(Black) = (26/52) * (26/51) = 26/102

15. Relative Frequency (Empirical) Probability

16. Probability of an Event Not Occurring If the probability of an event A is P(A), then the probability that event A does not occur is P(not A). Because the event must either occur or not occur, we can write: P(A) + P(not A) = 1 or P(not A) = 1 – P(A) The event not A is called the complement of the event A; the “not” is often designated by a bar, so Ā means not A. Example: A quarterback completes 67% of his passes, what is the probability that he will NOT complete his next pass? Answer: P(not A) = 1 – P(A) = 1 -.67 =.33. There is a 33% chance he will not complete the next pass.

17. Probability Distributions A probability distribution represents the probabilities of all possible events. Do the following to make a display of a probability distribution: 1.List all possible outcomes. Use a table or figure if it is helpful. 2.Identify outcomes that represent the same event. Find the probability of each event. 3.Make a table in which one column lists each event and another column lists each probability. The sum of all the probabilities must be 1. Here is the probability distribution for our Sample Space from earlier 1H1/12 1T1/12 2H1/12 2T1/12 3H1/12 3T1/12 4H1/12 4T1/12 5H1/12 5T1/12 6H1/12 6T1/12 Total = 12/12 = 1 Note: this is a uniform distribution

18. Slide 6.2- 18 Copyright © 2009 Pearson Education, Inc. If we toss three coins, we have a total of 2 × 2 × 2 = 8 possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT, as shown in Figure 6.4 b. Figure 6.4 b Tree diagram showing the outcomes of tossing three coins. Creating a Probability Distribution - Another example

19. Creating a Probability Distribution What is the probability distribution for the number of heads that occurs when three coins are tossed simultaneously? The number of different outcomes when three coins are tossed is 2 × 2 × 2 = 8. Question. What is the probability of getting exactly 2 heads? Answer 3/8. Question. What is the probability of getting 2 or more heads? Answer 4/8 = 1/2.

20. 3 Coin Probability Distribution

21. Law of Large Numbers The law of large numbers (or law of averages) applies to a process for which the probability of an event A is P(A) and the results of repeated trials do not depend on results of earlier trials (they are independent). It states: If the process is repeated through many trials, the proportion of the trials in which event A occurs will be close to the probability P(A). The larger the number of trials, the closer the proportion should be to P(A).

22. Roulette Example

23. Expected Value The expected value of a variable is the weighted average of all its possible events. Because it is an average, we should expect to find the “expected value” only when there are a large number of events, so that the law of large numbers comes into play. Consider two events, each with its own value and probability. The expected value is: expected value = (value of event 1) * (probability of event 1) + (value of event 2) * (probability of event 2) This formula can be extended to any number of events by including more terms in the sum.

24. Example: Suppose you pay $2 to roll a loaded die and win $6 if it comes up a 1 or 6, but nothing otherwise. What is your expected value? NumProb 11/21 22/21 33/21 44/21 55/21 66/21 ev = (value of event 1) * (probability of event 1) + (value of event 2) * probability of event 2) = (NET win of $4)(prob of rolling 1 or 6) + (NET loss of $6) (prob of rolling 2,3,4,5) = ($4) (7/21) + (-$2) (14/21) = (28/21) – (28/21) = 0. Your expected value is 0.

25. Expected Value - example Question: You go to a social function and buy a $1 raffle ticket to win the one grand prize of $500. If 1200 tickets are sold what is your expected value? Answer: ev =(value of event 1) * (probability of event 1) + (value of event 2) * (probability of event 2) = (NET win of $499)(prob of winning) + (NET loss of $1) (prob of losing) = (499)(1/1200) + (-1)(1199/1200) = (499/1200) – (1199/1200) = -700/1200 = -7/12 = -.5833. If you did this 100 times, you should expect to lose $58.33.

26. Winning the Lottery A $1 lottery tickets have the following probabilities: 1 in 5 to win a free ticket (worth $1); 1 in 100 to win $5; 1 in 100,000 to win $1,000; and 1 in 10 million to win $1 million. What is the expected value of a lottery ticket? Thus, averaged over many tickets, you should expect to lose 64¢ for each lottery ticket that you buy. If you buy, say, 1,000 tickets, you should expect to lose about 1,000 × $0.64 = $640. EventValueProbabilityValue * ProbabilityResult Purchase ticket-$11-$1 x 1-$1.00 Win free ticket$11/5$1 x 1/5$0.20 Win $5$51/100$5 x $100$0.05 Win $1,000$1,0001/100,000$1,000 x 1/100,000$0.01 Win $1,000,000 $1,000,0001/10,000,000$1,000,000/10,000,000$.10 Expected Value =$-0.64

27. Accident Rates Travel risk is often expressed in terms of an accident rate or death rate. For example, suppose an annual accident rate is 750 accidents per 100,000 people. This means that, within a group of 100,000 people, on average 750 will have an accident over the period of a year. The statement is in essence an expected value, which means it also represents a probability: It tells us that the probability of a person being involved in an accident (in one year) is 750 in 100,000, or 0.0075.

28. Death Rates

29. Vital Statistics

30. Computing Probabilities in StatCrunch Create a Contingency table from data sets to compute answers to And, Or, and Conditional Probabilities problems you may find in The textbook, MSL homework or Quiz, and the final exam.

31. Computing Probabilities in StatCrunch Example 1: Find the probability of a house having exactly 3 bedrooms or 3 full bathrooms. This is an OR problem. P(of 3 bedrooms or 3 bathrooms) = P(3 bedrooms) + P(3 bathrooms) - P(3 bedrooms and 3 bathrooms) = (19/40) + (6/40) - (2/40) = 23 / 40 =.575 or.58 rounded to 2 decimal places n = 40

32. Questions?