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Using Simulations and Samples to Estimate Probability in the Real World Chapter 5.5 – Probability Distributions and Predictions Mathematics of Data Management (Nelson) MDM 4U Authors: Gary Greer (with K. Myers) Simulating the Stanley Cup Final suppose that the Leafs have a 70% chance of winning a single Stanley Cup game the team needs to win 4 games each game can be considered a trial, in which the probability of success is p = 0.7 and the probability of failure is q = 0.3 each game may be considered a Bernoulli trial Simulating the Stanley Cup Final however, this is not a binomial experiment because the random variable is the number of games that need to be completed rather than the number of successes in 7 trials it is called a waiting time problem we can simulate the problem using graphing calculators, generating numbers between 1 and 10 numbers between 1 and 7 are a win for the leafs, between 8 and 10 are considered a loss what is the expected number of games? The Simulation we will do 10 trials each on the TI83+ Math → PRB 5: rand int ( 1, 10, 7) Enter (then hit enter 9 more times) record whether each game was a win or a loss for the leafs using a table like the one on the next slide The Table of Raw Data Simulation Winning Team 1 M M L L L 2 L L L L 3 L M M L 4 L L L L 5 L L M L M 6 M L L L L 7 L L M M M 8 M L L L L 9 L L L L 10 M M L L # Games L 6 4 M L L 7 4 L 6 5 M 6 5 4 M ex: 8 4 3 1 5 gives M L L L L L L 7 What does the Frequency Table Look Like? Number of Games 4 5 6 7 Frequency 3 2 3 2 0.3 0.2 0.3 0.2 Experimental Probability is this a sufficient number of trials? we can simulate more trials using the TI83+ calculators Student Simulations each pair of students can generate 100 sets of 7 random numbers seed MATH → PRB 5: randInt( 1, 10, 7) Enter by hitting enter again, another set will appear a large scale simulation can be done using a Java program What would the outcome be if we performed a large number of trials? a Java program running 100 000 000 trials found the following probabilities: P(4 Games) = 0.24824748 P(5 Games) = 0.31078485 P(6 Games) = 0.25578749 P(7 Games) = 0.18518018 So the expected number of games is E(x) 4(0.248) + 5(0.311) + 6(0.256) + 7(0.185) = 5.378 Simulations Based on Sample Results imagine for years students have used a literacy technique which improves reading accuracy 70% of the time a new technique is tested on 200 people and 160 of the students improve their accuracy is the new technique really more effective? performing a straight calculation, it appears that the new technique works in 80% of cases however, this success could have occurred by chance! (perhaps this group was unusual) So how do we proceed? in statistics, we often look to see how probable an event is if the event is quite improbable, we assume that it did not occur by chance generally, if something has an occurrence probability of less than 5% it is not considered likely to occur by chance if something is not likely to occur by chance, it is likely that the effect we observe is significant So how can we test this? using the TI83+ we can generate random numbers between 1 and 10 if a number is 7 or less it models the 70% success rate of the original treatment by carrying out a number of trials it is possible to establish a distribution of outcomes on the TI83+ we can use the following: sum(randInt(1,10,100)≤7) Finally… once we generate the outcomes we can determine the likelihood of each different outcome for example we can find the likelihood of a 75% success, 76% success, etc. if the sum of the probabilities greater than 80% is less than 5% of the total, the effect likely did not occur by chance and therefore is significant to test this, we turned to a Java program… What does the distribution look like? P(61 Successes) = 0.0090 P(62 Successes) = 0.011 P(63 Successes) = 0.026 P(64 Successes) = 0.042 P(65 Successes) = 0.048 P(66 Successes) = 0.054 P(67 Successes) = 0.057 P(68 Successes) = 0.084 P(69 Successes) = 0.094 P(70 Successes) = 0.097 P(71 Successes) = 0.096 P(72 Successes) = 0.075 P(73 Successes) = 0.075 P(74 Successes) = 0.06 P(75 Successes) = 0.052 P(76 Successes) = 0.033 P(77 Successes) = 0.025 P(78 Successes) = 0.018 P(79 Successes) = 0.0050 P(80 Successes) = 0.012 P(81 Successes) = 0.0020 P(82 Successes) = 0.0010 P(83 Successes) = 0.0020 P(84 Successes) = 0.0030 P(85 Successes) = 0.0 P(86 Successes) = 0.0 P(87 Successes) = 0.0 P(88 Successes) = 0.0 P(89 Successes) = 0.0 P(90 Successes) = 0.0 Our Conclusion the probability of greater than say 80% success is 2% in this simulation of 1000 repetitions of 100 trials so it is extremely unlikely that this is a chance result! therefore we would be confident that the new technique is more effective Exercises / Homework do Page 321 # 6, 7, 8, 9