Transcript Chapter5.5 - Robinson Math

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
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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
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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
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we will do 10 trials each
on the TI83+
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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
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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
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is this a sufficient number of trials?
we can simulate more trials using the TI83+
calculators
Student Simulations
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each pair of students can generate 100 sets
of 7 random numbers
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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?
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a Java program running 100 000 000 trials
found the following probabilities:
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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
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E(x) 4(0.248) + 5(0.311) + 6(0.256) + 7(0.185)
= 5.378
Simulations Based on Sample Results
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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?
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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?
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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:
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sum(randInt(1,10,100)≤7)
Finally…
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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?
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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
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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
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do Page 321 # 6, 7, 8, 9