Statistics Quick Overview

Copyright by Michael S. Watson, 2012

Statistics Quick Overview

Class #2

Copyright by Michael S. Watson, 2012; Slides from Managerial Statistics book 2

Thought Exercise with Our Packaging ExampleOriginal Case (mean = 290, sd = 53)

Less Variability (m = 290, sd = 5) More Variability (m = 290, sd = 186)

If a store manager came to you and said, “what will my sales be?” how would you answer?

If CEO came to you and said, “what will average sales be?” how would you answer?


Thought Exercise II- We Doubled The Samples

If a store manager came to you and said, “what will my sales be?” how would you answer?

If CEO came to you and said, “what will average sales be?” how would you answer?

(mean = 290, sd = 53) (mean = 290, sd = 53)

What do you think of these questions now?


Sampling Distribution

is approximately normally distributed with a mean of µ and st dev of

Since we never know the actual σ, we approximate it with the sample standard deviation, s.

Xn

is commonly used in statistics

We call this term the standard error of the mean

Xssn

Let’s see how this applies to our examples


We Have 3 Measures for a Sample of Data

Mean (average)

Standard Deviation (sample standard deviation)

Standard Error of the Mean


Central Limit Theorem– General Idea

is approximately normally distributed with a mean of µ and st dev of

In other words, as you take various samples, the collection of these samples will be approximately normally distributed The larger the value of n, the closer to normally distributed

The population data does not have to be normally distributed

Xn


A New Game

1 2 3


Basic Probability

A BA and B

𝑃 (𝐵|𝐴 )= 𝑃 (𝐴𝑎𝑛𝑑𝐵)𝑃 ( 𝐴)

Solution space = 1

𝑃 ( 𝐴𝑜𝑟 𝐵 )=P ( A )+P (B )−P (A∧B)


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


Neither Will Start?A Table Can Be Helpful

BMW

Acura

Starts

Doesn’t

Starts Doesn’t

75% 80%

90%

20%

10%

15%

5%

5%


Car Example- A Starting if B Starts?

A (90%) B (80%)A and B (75%)

= 75% / 80% = 93.75%

Solution space = 1


Conditional Probability:A Table Can Be Helpful

BMW

Acura

Starts

Doesn’t

Starts Doesn’t

75% 80%

90%

20%

10%

15%

5%

5%

Normalize this row:75% / 80%


Counting: What Do These Five Guys in Front……


….Have to Do with the Front Line in Hockey….

Right Center Left


…or the Front Line In Quiditch?


Let’s Say A Coach (Maybe Mr. Brown?) Had to Pick 3 Players for Hockey and Then Quiditch

Let’s start with hockey…

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

𝑛 !(𝑛−𝑘)!

Where n is the number of choices, and k is the number picked. In Excel, this is PERMUT(n,k)


Let’s 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


Mr. Brown’s 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


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


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}


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

Let’s assume: s is the probability of success f is the probability of failure


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


Bayes’ 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%

𝑃 ( 𝐴1|𝐵 )=𝑃 ( 𝐴1𝑎𝑛𝑑𝐵)𝑃 (𝐵)

What we want….


Bayes Rule- Calculate

𝑃 (𝐵|𝐴1 )=𝑃 ( 𝐴1𝑎𝑛𝑑𝐵)𝑃 (𝐴1)

𝑃 ( 𝐴1|𝐵 )=𝑃 ( 𝐴1𝑎𝑛𝑑𝐵)𝑃 (𝐵)

𝑃 (𝐴 1)𝑃 (𝐵|𝐴1 )= 𝑃 (𝐴 1𝑎𝑛𝑑𝐵)1

P (𝐵 )=𝑃 ( 𝐴1𝑎𝑛𝑑𝐵 )+𝑃 ( 𝐴 2𝑎𝑛𝑑𝐵 )=𝑃 ( 𝐴1 ) 𝑃 (𝐵|𝐴1 )+𝑃 ( 𝐴2 ) 𝑃 (𝐵|𝐴2 )


Bayes Rule- Calculation

=𝑃 (𝐴 1)𝑃 (𝐵|𝐴1 )

𝑃 ( 𝐴1 ) 𝑃 (𝐵|𝐴 1 )+𝑃 ( 𝐴2 ) 𝑃 (𝐵|𝐴2 )

= 5%*(98%) / ((5%*98%)+(95%*2%)) =72%

Statistics Quick Overview

Documents