STA 291 Lecture 9
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Transcript of STA 291 Lecture 9
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STA 291 - Lecture 9 1
STA 291Lecture 9
• Probability Review:
multiplication rule and independence
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Independent events
• If we know events A and B are independent, then all 3 hold
• P(A|B) = P(A)
• P(B|A) = P(B)
STA 291 - Lecture 9 2
( ) ( ) ( ).P A B P A P B
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• On the other hand, if we want to check that events A , B are independent, then we may check any one of the three equalities.
STA 291 - Lecture 9 3
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How to use independence?
• We might give you an RxC probability table, and ask you to check if events A, B are independent. [ you need to verify one of the three identities]
• We might tell you that A, B are independent events, and ask you to compute the probability of a related event. [in the computation, you may use any of the three identities]
STA 291 - Lecture 9 4
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STA 291 - Lecture 9 5
Example: Smoking and Lung Disease I
Lung Disease
Not Lung Disease
Marginal (smoke status)
Smoker .12 .19 .31
Nonsmoker .03 .66 .69
Marginal (disease status)
.15 .85 1.00
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STA 291 - Lecture 9 6
• The red probabilities are the joint probabilities
• The green ones are the two marginal probabilities.
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STA 291 - Lecture 9 7
Is smoking independent of lung disease?
• Check one of the three equations
P(smoker) = 0.31
P(lung disease) = 0.15
joint probability P(“smoker and lung disease”) = 0.12
Since
Therefore the two events are not independent
0.12 0.31 0.15 0.0465
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STA 291 - Lecture 9 8
• We may also check P( lung disease | smoker ) = 0.12 / 0.31 =0.387
This is different from P(lung disease) = 0.15(in fact this conditional probability increased a
lot compared to unconditional)
Therefore “smoker”, “lung disease” are not independent
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Example II
• Given that events A, B are independent, P(A) = 0.3, P(B) = 0.6. Find P(A or B) = ?
• = 0.72
STA 291 - Lecture 9 9
( ) ?P A B
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STA 291 - Lecture 9 10
Example of independent
• If the table look like the next one, then the two were independent.
• There, the proportion of disease/nondisease are the same across smokers and nonsmokers.
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STA 291 - Lecture 9 11
Example: Smoking and Lung Disease II independent case
Lung Disease
Not Lung Disease
Marginal (smoke status)
Smoker 0.0465 0.2635 0.31
Nonsmoker 0.1035 0.5865 0.69
Marginal (disease status)
0.15 0.85 1.00
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STA 291 - Lecture 9 12
• How did I come up with the red probabilities? (2 way street)
• Check for yourself that all three equations for independence are valid here.
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STA 291 - Lecture 9 13
Review: Joint, marginal and Conditional Probabilities
• P(A∩B) Joint probability of A and B
(of the intersection of A and B)• P(A|B) Conditional probability of A given B
“the probability that A occurs given that B has occurred.”
• P(A) (Marginal, or unconditional) probability of A
( )( | ) , provided ( ) 0
( )
P A BP A B P B
P B
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STA 291 - Lecture 9 14
( )( | ) , provided ( ) 0
( )
P B AP B A P A
P A
( ) ( )P A B P B A
A B B A
( ) ( ) ( | ) P A B P A P B A
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“Hot-hand theory” in basketball
• How can we verify or disprove the hot hand theory?
• Obtain a table of many two consecutive free throws and construct a 2x2 table. Verify one of the three equations.
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• May be it only apply to 3-point shoots?
• How do you define consecutive? Within 2 min?
STA 291 - Lecture 9 16
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• Disjoint events
• Independent events
STA 291 - Lecture 9 17
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Chap. 6 Display and describe quantitative data
• Center and spread
• Average (mean) and median
• Range, interquartile range and standard deviation
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examples
• Lexington Median family income: $59,193
• Lexington median home price $155,500
• Median age 33.9
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• For symmetric variables, mean value.
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STA 291 - Lecture 9 21
Attendance Survey Question
• On a 4”x6” index card– Please write down your name and
section number
– Today’s Question:
– Two most common measurements of central tendency of data are
– Mean and M______