statistics quick overview
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Statistics Quick Overview. Class #2. A New Game. 1. 2. 3. Conditional Probability Examples. Given that a consumer has looked at a green shirt 5 times What is the probability that they will buy? What is the probability that they will buy if you give them 5% off? - PowerPoint PPT PresentationTRANSCRIPT
Copyright by Michael S. Watson, 2012
Statistics Quick Overview
Class #2
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 2
A New Game
1 2 3
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 3
Conditional Probability Examples
Given that a consumer has looked at a green shirt 5 times What is the probability that they will buy? What is the probability that they will buy if you give them 5% off? What is the probability that they will buy if you give them 10% off? What is the probability given that they are a new customer/existing
customer?
Given that a consumer has bought a flashlight What is the probability that they will buy batteries? What is the probability that they will buy a 2nd flashlight?
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 4
Basic Probability
A BA and B
π (π΅|π΄ )= π (π΄ππππ΅)π ( π΄)
Solution space = 1
π ( π΄ππ π΅ )=P ( A )+P (B )βP (Aβ§B)
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 5
Car Example- Neither Will Start?
A (90%) B (80%)A and B (75%)
Solution space = 1
= 90% + 80% - 75%=95% 1-95% = 5% chance neither will start
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 6
Neither Will Start?A Table Can Be Helpful
BMW
Acura
Starts
Doesnβt
Starts Doesnβt
75% 80%
90%
20%
10%
15%
5%
5%
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 7
Car Example- A Starting if B Starts?
A (90%) B (80%)A and B (75%)
= 75% / 80% = 93.75%
Solution space = 1
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 8
Conditional Probability:A Table Can Be Helpful
BMW
Acura
Starts
Doesnβt
Starts Doesnβt
75% 80%
90%
20%
10%
15%
5%
5%
Normalize this row:75% / 80%
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 9
Counting: What Do These Five Guys in Frontβ¦β¦
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 10
β¦.Have to Do with the Front Line in Hockeyβ¦.
Right Center Left
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 11
β¦or the Front Line In Quiditch?
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 12
Letβs Say A Coach (Maybe Mr. Brown?) Had to Pick 3 Players for Hockey and Then Quiditch
Letβs start with hockeyβ¦
Here, order matters The person on the left must stay on the left The person on the right must stay on the right
So, how many different potential line-ups does Mr. Brown have to consider? Choices are: Mr Blonde, Mr White, Mr Orange, Mr Pink, and Mr Blue
5 x 4 x 3 = 60
π !(πβπ)!
Where n is the number of choices, and k is the number picked. In Excel, this is PERMUT(n,k)
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 13
Letβs Carefully Write Out the Permutations
(Blnd, Blue, Orng) (Blue, Blnd, Orng) (Blnd, Orng, Blue) (Blue, Orng, Blnd) (Orng, Blnd, Blue) (Orng, Blue, Blnd)
(Blnd, Blue, Wht) (Blue, Blnd, Wht) (Blnd, Wht, Blue) (Blue, Wht, Blnd) (Wht, Blnd, Blue) (Wht, Blue, Blnd)
(Blnd, Blue, Pink) (Blue, Blnd, Pink) (Blnd, Pink, Blue) (Blue, Pink, Blnd) (Pink, Blnd, Blue) (Pink, Blue, Blnd)
(Blnd, Orng, Wht) (Blnd, Wht, Orng) (Orng, Blnd, Wht) (Orng, Wht, Blnd) (Wht, Blnd, Orng) (Wht, Orng, Blnd)
(Blnd, Orng, Pink) (Blnd, Pink, Orng) (Orng, Blnd, Pink) (Orng, Pink, Blnd) (Pink, Blnd, Orng) (Pink, Orng, Blnd)
(Blnd, Pink, Wht) (Blnd, Wht, Pink) (Pink, Blnd, Wht) (Pink, Wht, Blnd) (Wht, Blnd, Pink) (Wht, Pink, Blnd)
(Blue, Orng, Wht) (Blue, Wht, Orng) (Orng, Blue, Wht) (Orng, Wht, Blue) (Wht, Blue, Orng) (Wht, Orng, Blue)
(Blue, Orng, Pink) (Blue, Pink, Orng) (Orng, Blue, Pink) (Orng, Pink, Blue) (Pink, Blue, Orng) (Pink, Orng, Blue)
(Blue, Pink, Wht) (Blue, Wht, Pink) (Pink, Blue, Wht) (Pink, Wht, Blue) (Wht, Blue, Pink) (Wht, Pink, Blue)
(Orng, Pink, Wht) (Orng, Wht, Pink) (Pink, Orng, Wht) (Pink, Wht, Orng) (Wht, Orng, Pink) (Wht, Pink, Orng)
Note: Each column is a unique combination of players
Note: The entries within each row are different permutations of the players. This is our same problem again where n= 3 and k = 3
==6
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 14
Mr. Brownβs Choices for a Quiditch Front Line
Here, order does not matter He just needs a front line
All that matters is the number of unique combinations
What observation from the permutation table helps us determine the unique combinations
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 15
Figuring out the Combinations
(Blnd, Blue, Orng) (Blue, Blnd, Orng) (Blnd, Orng, Blue) (Blue, Orng, Blnd) (Orng, Blnd, Blue) (Orng, Blue, Blnd)
(Blnd, Blue, Wht) (Blue, Blnd, Wht) (Blnd, Wht, Blue) (Blue, Wht, Blnd) (Wht, Blnd, Blue) (Wht, Blue, Blnd)
(Blnd, Blue, Pink) (Blue, Blnd, Pink) (Blnd, Pink, Blue) (Blue, Pink, Blnd) (Pink, Blnd, Blue) (Pink, Blue, Blnd)
(Blnd, Orng, Wht) (Blnd, Wht, Orng) (Orng, Blnd, Wht) (Orng, Wht, Blnd) (Wht, Blnd, Orng) (Wht, Orng, Blnd)
(Blnd, Orng, Pink) (Blnd, Pink, Orng) (Orng, Blnd, Pink) (Orng, Pink, Blnd) (Pink, Blnd, Orng) (Pink, Orng, Blnd)
(Blnd, Pink, Wht) (Blnd, Wht, Pink) (Pink, Blnd, Wht) (Pink, Wht, Blnd) (Wht, Blnd, Pink) (Wht, Pink, Blnd)
(Blue, Orng, Wht) (Blue, Wht, Orng) (Orng, Blue, Wht) (Orng, Wht, Blue) (Wht, Blue, Orng) (Wht, Orng, Blue)
(Blue, Orng, Pink) (Blue, Pink, Orng) (Orng, Blue, Pink) (Orng, Pink, Blue) (Pink, Blue, Orng) (Pink, Orng, Blue)
(Blue, Pink, Wht) (Blue, Wht, Pink) (Pink, Blue, Wht) (Pink, Wht, Blue) (Wht, Blue, Pink) (Wht, Pink, Blue)
(Orng, Pink, Wht) (Orng, Wht, Pink) (Pink, Orng, Wht) (Pink, Wht, Orng) (Wht, Orng, Pink) (Wht, Pink, Orng)
When calculating the permutations, we naturally determined the unique combinations (the columns) and then ran the permutations for each combination. If we divide out that last step, we will have just the combinations:
= 60 / 6=10
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 16
By Convention, we write these with brackets {}
{Blnd, Blue, Orng}
{Blnd, Blue, Wht}
{Blnd, Blue, Pink}
{Blnd, Orng, Wht}
{Blnd, Orng, Pink}
{Blnd, Pink, Wht}
{Blue, Orng, Wht}
{Blue, Orng, Pink}
{Blue, Pink, Wht}
{Orng, Pink, Wht}
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 17
Binomial Distribution
Sample Size of 10
Case #1: Assume that the lot is good with 5% defectives When will you reject because you find 3 or more defectives
Case #2: Assume that the lot has 40% defectives When will you accept because you find 2 or less defectives
Letβs assume: s is the probability of success f is the probability of failure
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 18
Case #1 (only 5% of the lot is defective)
Example of getting 3 Failures fssfsssssf Probability of this is (5%)3(95%)7
Example of getting 4 Failures fssfsfsssf Probability of this is (5%)4(95%)6
What are we missing?
The number of combinations
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 19
Bayesβ Rule
A1 uses drugs P(A1) = 5% A2 does not use drugs P(A2) = 95% B tests shows drug use P(B | A1) = 98% P(B | A2) = 2%
π ( π΄1|π΅ )=π ( π΄1ππππ΅)π (π΅)
What we wantβ¦.
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 20
Bayes Rule- Calculate
π (π΅|π΄1 )=π ( π΄1ππππ΅)π (π΄1)
π ( π΄1|π΅ )=π ( π΄1ππππ΅)π (π΅)
π (π΄ 1)π (π΅|π΄1 )= π (π΄ 1ππππ΅)1
P (π΅ )=π ( π΄1ππππ΅ )+π ( π΄ 2ππππ΅ )=π ( π΄1 ) π (π΅|π΄1 )+π ( π΄2 ) π (π΅|π΄2 )
Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 21
Bayes Rule- Calculation
=π (π΄ 1)π (π΅|π΄1 )
π ( π΄1 ) π (π΅|π΄ 1 )+π ( π΄2 ) π (π΅|π΄2 )
= 5%*(98%) / ((5%*98%)+(95%*2%)) =72%