Business Statistics CONTENTS Data as a random sample Probability - - PowerPoint PPT Presentation

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BASIC PROBABILITY Business Statistics CONTENTS Data as a random sample Probability theory Be careful with probability Old exam question Further study DATA AS A RANDOM SAMPLE In business and economics, the data represents a small sample of


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BASIC PROBABILITY

Business Statistics

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Data as a random sample Probability theory Be careful with probability Old exam question Further study CONTENTS

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In business and economics, the data represents a small sample of the phenomenon of interest

▪ poll with random 100 telephone calls ▪ interviews with random 25 customers ▪ batch of random 50 meals for food quality ▪ car accidents on 5 random days in a year ▪ etc.

All are supposed to represent a bigger population

▪ the entire population of the US (or only those that have a telephone?) ▪ all of your customers (or all of your customers that have a phone, e- mail, or that visit your shop) ▪ all meals produced in some factory or some restaurant ▪ all accidents in a particular year, particular country, ... ▪ etc.

DATA AS A RANDOM SAMPLE

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Task: to infer properties of the population from the sample

▪ Inferential statistics ▪ example: 12 out of 100 random clients liked our new product – what proportion of the total population will like it?

In order to get an understanding: study the variation of samples

▪ Probability theory ▪ example: if 15% of the population likes our new product, what is the probability that we find 12 in a random sample of 100?

DATA AS A RANDOM SAMPLE

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Sample space 𝑇: all possible outcomes of a random experiment

▪ coin: 𝑇 = heads, tails ▪ die: 𝑇 = 1,2,3,4,5,6 or ▪ number of accidents on a day: 𝑇 ⊂ ℕ0 ▪ body height: 𝑇 ⊂ ℝ+ ▪ two dice: 𝑇 = 1,1 , 1,2 , … , 6,6

  • r

PROBABILITY THEORY

⊂ means “is a subset of”

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Probability 𝑄: likelihood of a particular outcome

▪ coin: 𝑄 heads = 1

2

▪ die: 𝑄 3 = 1

6

▪ number of accidents on a day: 𝑄 1 = 0.19 (for example) ▪ body height: 𝑄 1.8304678 = 0 (why?)

Types of probabilty

▪ a priori (classical theory) ▪ 𝑄 head = 1

2 or 𝑄 event = number of outcomes in event number of possible outcomes

▪ empirical ▪ 𝑄 event = number of occurrences of event

number of observations

▪ subjective ▪ 𝑄 Italy will win next World Cup = ⋯

PROBABILITY THEORY

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Probability function 𝑄 event

▪ for every event 𝐵 ⊂ 𝑇: 𝑄 𝐵 ≥ 0 ▪ for entire sample space 𝑇: 𝑄 𝑇 = 1 ▪ for 𝐵 = ∅: 𝑄 𝐵 = 0

Events:

▪ elementary (e.g. for a die, outcome = 3) ▪ compound (e.g. for a die, outcome = even)

Example: die with 𝑇 = 1,2,3,4,5,6

▪ 𝑄 3 =

1 6

▪ 𝑄 even = 𝑄 2,4,6 =

3 6 = 1 2

▪ 𝑄 even or odd = 𝑄 𝑇 = 1 ▪ 𝑄 −1 = 𝑄 7 = 𝑄 2.43 = 𝑄 ∅ = 0

PROBABILITY THEORY

∅ is the empty set

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The complement of an event 𝐵 is denoted by 𝐵′ and consists

  • f everything in the sample space 𝑇 except event 𝐵

Since 𝐵 and 𝐵′ together comprise the entire sample space 𝑇, 𝑄 𝐵 + 𝑄 𝐵′ = 1 or 𝑄 𝐵′ = 1 − 𝑄 𝐵 PROBABILITY THEORY

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The union of two events consists of all outcomes in the sample space 𝑇 that are contained either in compound event 𝐵 (e.g., “≤ 3”) or in compound event 𝐶 (e.g., “even”) or both

▪ denoted by 𝐵 ∪ 𝐶 ▪ pronounced “𝐵 or 𝐶”

PROBABILITY THEORY

“or” means “and/or”, so not the exclusive or (as in “either 𝐵 or 𝐶”) 1 3 2 4 6

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The intersection of two events 𝐵 and 𝐶 is the event consisting

  • f all outcomes in the sample space 𝑇 that are contained in

both event 𝐵 and event 𝐶

▪ denoted by 𝐵 ∩ 𝐶 ▪ pronounced “𝐵 and 𝐶” ▪ also known as the joint probability

PROBABILITY THEORY

1 3 2 4 6

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Given are two dies, their random outcomes are 𝑌 and 𝑍.

  • a. Find 𝑄

𝑌 = 2 ∩ 𝑍 = 3 ∪ 𝑌 = 3 ∩ 𝑍 = 2

  • b. Find 𝑄 𝑌 + 𝑍 = 5

EXERCISE 1

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The general law of addition states that 𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 + 𝑄 𝐶 − 𝑄 𝐵 ∩ 𝐶

▪ when you add the 𝑄(𝐵) and 𝑄(𝐶) together, you count the 𝑄(𝐵 ∩ 𝐶) twice ▪ so, you have to subtract 𝑄(𝐵 ∩ 𝐶) to avoid over-stating the probability ▪ often, the right-hand side is easier to find

PROBABILITY THEORY

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Example: standard deck of cards

▪ 52 cards ▪ 4 queens: 𝑄 𝑅 =

4 52

▪ 26 red cards: 𝑄 𝑆 =

26 52

▪ 2 red queens: 𝑄 𝑅 ∩ 𝑆 =

2 52

▪ the probability that a random card is red or a queen: 𝑄 𝑅 ∪ 𝑆 = 𝑄 𝑅 + 𝑄 𝑆 − 𝑄 𝑅 ∩ 𝑆 =

4+26−2 52

=

28 52

▪ so 53.85%

PROBABILITY THEORY

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Events 𝐵 and 𝐶 are mutually exclusive (or disjoint) if their intersection is the null set (∅) that contains no elements

▪ if 𝐵 ∩ 𝐶 = ∅, then 𝑄 𝐵 ∩ 𝐶 = 0

In the case of mutually exclusive events, the addition law reduces to:

▪ 𝑄 𝐵 ∪ 𝐶 = 𝑄 𝐵 + 𝑄 𝐶

Example:

▪ 𝑄 𝑅 ∪ 𝐿 = 𝑄 𝑅 + 𝑄 𝐿 because 𝑅 ∩ 𝐿 = ∅ ▪ more general: 𝑄 𝑅 ∪ 𝑅′ = 𝑄 𝑅 + 𝑄 𝑅′ because 𝑅 ∩ 𝑅′ = ∅

PROBABILITY THEORY

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Events are collectively exhaustive if their union is the entire sample space 𝑇 Two mutually exclusive, collectively exhaustive events are dichotomous (or binary) events

▪ for example, a car repair is either covered by the warranty (𝐵) or not (𝐵′)

Mutually exclusive, collectively exhaustive events are a partition of the sample space PROBABILITY THEORY

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The probability of event 𝐵 given that event 𝐶 has occurred

▪ written as 𝑄 𝐵 𝐶 ▪ pronounced as “probability of 𝐵 given 𝐶” ▪ for example, the probability of passing the statistics exam, given that you have passed the mathematics exam

𝑄 𝐵 𝐶 = 𝑄 𝐵 ∩ 𝐶 𝑄 𝐶

▪ only defined when 𝑄 𝐶 > 0 ▪ this is a conditional probability

PROBABILITY THEORY

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Example: highs school drop-outs Facts about population aged 16-21 and not in college:

▪ unemployed (U): 12% ▪ high school drop-out (D): 30% ▪ unemployed high school drop-out: 4%

What is the conditional probability that a member of this population is unemployed, given that the person is a high school dropout?

▪ 𝑄 𝑉 𝐸 =

𝑄 𝑉∩𝐸 𝑄 𝐸

=

0.04 0.30 = 0.133

Compare to 𝑄 𝑉 = 0.12

▪ so, being a high school dropout is related to being unemployed

PROBABILITY THEORY

Here is a first example of learning from data, although it is not yet inferential statistics (why not?)

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Given are two dies, their random outcomes are 𝑌 and 𝑍.

  • a. Find 𝑄 𝑌 + 𝑍 > 4 𝑌 < 2
  • b. Find 𝑄 𝑌 = 3 𝑍 ≤ 2

EXERCISE 2

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Events 𝐵 and 𝐶 are independent if the 𝑄 𝐵 𝐶 = 𝑄 𝐵 (for 𝑄 𝐶 > 0)

▪ 𝑄 𝑉 𝐸 = 0.133 ≠ 𝑄 𝑉 = 0.12 ▪ so 𝑉 and 𝐸 are not independent ▪ that is, they are dependent

Another way to check for independence: 𝑄 𝐵 ∩ 𝐶 = 𝑄 𝐵 𝑄 𝐶 ⇔ ⇔ 𝑄 𝐵 =

𝑄 𝐵 ×𝑄 𝐶 𝑄 𝐶

=

𝑄 𝐵∩𝐶 𝑄 𝐶

= 𝑄 𝐵 𝐶 , so 𝑄 𝐵 ∩ 𝐶 = 𝑄 𝐵 𝑄 𝐶 ⇔ 𝐵 and 𝐶 are independent PROBABILITY THEORY

even when 𝑄 𝐵 = 0 or 𝑄 𝐶 = 0

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Contingency tables (see also summarizing data)

▪ from frequencies ... ▪ to proportions

PROBABILITY THEORY

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Table contains

▪ joint probabilities, like 𝑄 Christ ∩ GATT = 0.36 ▪ marginal (simple) probabilities, like 𝑄 Christ = 0.49

Table can be used to find

▪ conditional probabilities, like 𝑄 Christ GATT =

0.36 0.78 = 0.46

▪ dependence: 𝑄 Christ GATT ≠ 𝑄 Christ ▪ later, we will develop a statistical test for assessing this for inference from the sample to the population

PROBABILITY THEORY

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Independence

▪ flipping a fair coin 4 times, what is the probability of obtaining 4 heads? ▪ you are given that the outcomes are independent ▪ 𝐼1 means “heads” in experiment 1, 𝑈

1 “tails”, etc;

▪ 𝑄 4 heads = 𝑄 𝐼1𝐼2𝐼3𝐼4 = 𝑄 𝐼1 × 𝑄 𝐼2 × 𝑄 𝐼3 × 𝑄 𝐼4 =

1 2 × 1 2 × 1 2 × 1 2 = 1 16 ≈ 6%

Independence is often not realistic in business problems

▪ fashion: 𝑄 I buy a red shirt depends on what others do ▪ supply: 𝑄 I buy a red shirt depends on what is available in the shops ▪ combi: 𝑄 I buy a red shirt depends on if I also buy a red coat ▪ etc.

PROBABILITY THEORY

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  • Problem: aggregation
  • Context: which of two medicines (A and B) is best?
  • Facts:

▪ medicine A is better than B for treating cold ▪ medicine A is better than B for treating flue ▪ but altogether B looks better

BE CAREFUL WITH PROBABILITY

The most effective medicine Recovery Treatment A Treatment B Cold 93% 87% Flu 73% 69% Both 78% 83%

Treatment A Treatment B Yes No Tot Yes No Tot Cold 81 6 87 234 36 270 Flu 192 71 263 55 25 80 273 350 289 350

81 87 = 0.93

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  • Problem: aggregation
  • Context: which of two football players (A and B) should

take a penalty?

  • Facts:

▪ in 2012 A scored 8/12=0.67 and B scored 3/4=0.75 ▪ in 2013 A scored 1/6=0.17 and B scored 1/5=0.20 ▪ so B was best in both years separately ▪ but in 2012-2013 A scored 9/18=0.50 and B scored 4/9=0.44 ▪ so A was best in the two combined years

BE CAREFUL WITH PROBABILITY

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  • Problem: difference in point of view
  • Context: what is the class size of an elementary school?
  • Facts:

▪ there are 3 classes with 30 pupils and 3 classes with 10 pupils ▪ the director tells that average class size is

3×30+3×10 3+3

= 20 ▪ but pupils tell that average class size is

90×30+30×10 90+30

= 25

  • What do we mean by “mean”?

BE CAREFUL WITH PROBABILITY

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  • Problem: unrepresentative means
  • Context: if events are unlikely, the statistics are

unreliable

  • Facts:

▪ before July 2000, Concorde was one of the safest aircraft ▪ after July 2000, it was one of the unsafest ever

BE CAREFUL WITH PROBABILITY

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30 June 2014, Q1a-c OLD EXAM QUESTION

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Doane & Seward 5/E 5.1-5.5, 5.8 Tutorial exercises week 1 events FURTHER STUDY