Homework Assignment
• Chapter 1, Problems 6, 15• Chapter 2, Problems 6, 8, 9, 12• Chapter 3, Problems 4, 6, 15• Chapter 4, Problem 16• Due a week from Friday:• Sept. 22, 12 noon.• Your TA will tell you where to hand these in
Random Sampling - what did we learn?
• It’s difficult to do properly• Why not just point?• Computers and random numbers• Can you tell if your numbers were random?
Sampling distribution of the mean
Sampling distribution of the mean
Sampling distribution of the mean
Sampling distribution of the mean
How confident can we be about this one estimate of the mean?
Estimating error of the mean
• Hard method: take a few MORE random samples, and get more estimates for the mean
• Easy method: use the formula:
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SEY
=s
n
Confidence interval
• Confidence interval– a range of values surrounding the sample estimate that is likely to contain the population parameter
• We are 95% confident that the true mean lies in this interval
= 5.14
Y = 5.26
What if we calculate 95% confidence
intervals?• Approximately ± 2 S.E.• Expect that 95% of the intervals from the class will contain the true population mean, 5.14
• 70 invervals * 5% = 3.5• Expect that 3 or 4 will not contain the mean, and the rest will
Mean ± 95% C.I.
Mean ± 95% C.I.
What if we took larger samples? Say, n=20 instead of n=10?
Probability
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The Birthday Challenge
Probability
• The proportion of times the event occurs if we repeat a random trial over and over again under the same conditions
• Pr[A] – The probability of event A
Two events are mutually exclusive if they cannot both be
true.
Two events are mutually exclusive if they cannot both be
true.
(cannot both occur simultaneously)
A
B
Mutually exclusive
A
B
Mutually exclusive
Venn diagram
A
B
Mutually exclusive
Venn diagram
Sample space
A
B
Mutually exclusive
Venn diagram
Sample space
Possible outcome
Pr[B] proportionalto area
Mutually exclusive
Mutually exclusive
Pr(A and B) = 0
Mutually exclusive
Visual definition - areas do not overlap in Venn diagram
Not mutually exclusive
Pr(A and B) 0
Pr(purple AND square) 0
For example
Probability distribution
A probability distribution describes the true relative
frequency of all possible values of a random variable.
Probability distribution
A probability distribution describes the true relative
frequency of all possible values of a random variable.
Random variable - a measurement that changes from one observation to the next because of chance
Probability distribution for the
outcome of a roll of a die
Number rolled
Frequency
Probability distribution for the sum of a roll of two
dice
Sum of two dice
Frequency
The addition rule
The addition principle: If two events A and B are mutually
exclusive, then
Pr[A OR B] = Pr[A] + Pr[B]
Addition Rule
Pr[1 or 2] = ?
Addition Rule
Pr[1 or 2] = Pr[1]+Pr[2]
Addition Rule
Pr[1 or 2] = Pr[1]+Pr[2]
Addition Rule
Pr[1 or 2] = Pr[1]+Pr[2]
Sum of areas
The probability of a range
For families of 8 children,
Pr[Number of boys ≥ 6] = ?
The probability of a range
For families of 8 children,
Pr[Number of boys ≥ 6] = Pr[6 or 7 or 8]
= Pr[6]+Pr[7]+Pr[8]
The probabilities of all possibilities add
to 1.
Addition Rule
Pr[1 or 2 or 3 or 4 or 5 or 6] = ?
Addition Rule
Pr[1 or 2 or 3 or 4 or 5 or 6] = 1
Probability of Not
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Pr[NOT rolling a 2] = ?
Probability of Not
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Pr[NOT rolling a 2] = 1 - Pr[2] = 5/6
Probability of Not
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Pr[NOT rolling a 2] = 1 - Pr[2] = 5/6
Pr[not A] = 1-Pr[A]
The addition rule
The addition principle: If two events A and B are mutually
exclusive, then
Pr[A OR B] = Pr[A] + Pr[B]
The addition rule
The addition principle: If two events A and B are mutually
exclusive, then
Pr[A OR B] = Pr[A] + Pr[B]
What if they are not mutually exclusive?
General Addition Rule
A
B
Pr[A or B] = ?
General Addition Rule
A
B
Pr[A or B] = ?
AB
General Addition Rule
A
B
Pr[A or B] = ?
AB
General Addition Rule
A
B
Pr[A or B] = ?
AB
General Addition Rule
A
B
AB
General Addition Rule
General Addition RulePr[Walks or flies] = ?
General Addition RulePr[Walks or flies] = ?
General Addition RulePr[Walks or Flies] = Pr[Walks] + Pr[Flies]
- Pr[Walks and Flies]
General Addition Rule
Pr[A OR B] = Pr[A] + Pr[B] - Pr[A AND B].
Independence
Two events are independent if the occurrence of one gives
no information about whether the second will occur.
Independence
Two events are independent if the occurrence of one gives
no information about whether the second will occur.
Equivalent definition: The occurrence of one does notchange the probability that the second will occur
Multiplication rule
If two events A and B are independent, then
Pr[A and B] = Pr[A] x Pr[B]
Pr[boy]=0.512
Pr[ (first child is a boy) AND (second child is a boy)]
= 0.512 × 0.512 = 0.262.
Pr[boy]=0.512
Pr[ (first child is a boy) AND (second child is a boy)]
= 0.512 × 0.512 = 0.262.
General Addition Rule
Multiplication rule
If two events A and B are independent, then
Pr[A and B] = Pr[A] x Pr[B]
OR versus AND
• OR statements:– Involve addition– It matters if the events are mutually exclusive
• AND statements:– Involve multiplication– It matters if the events are independent
Probability trees
Sex of two children family
Sex of two children family
Dependent events
Variables are not always independent; in fact they are often not
Fig wasps
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Fig wasps
Testing independence
• Are the previous state of the fig and the sex of an egg laid independent?
• Test the multiplication rule:Pr[A and B] ?=? Pr[A] x Pr[B]Pr[fig already has eggs and male] ?=?P[fig already has eggs] x Pr[male]
Are the previous state of the fig and the sex of an egg laidindependent?
Pr(male) = 0.18 + 0.04 = 0.22
Pr(fig already has eggs) = 0.2
Pr(male AND fig already has eggs) = 0.18 ≠
Pr(male) x Pr(fig already has eggs) = 0.22 x 0.2 = 0.044
So these two events are NOT independent.
Are the previous state of the fig and the sex of an egg laidindependent?
Pr(male) = 0.18 + 0.04 = 0.22
Pr(fig already has eggs) = 0.2
Pr(male AND fig already has eggs) = 0.18 ≠
Pr(male) x Pr(fig already has eggs) = 0.22 x 0.2 = 0.044
So these two events are NOT independent.
Are the previous state of the fig and the sex of an egg laidindependent?
Pr(male) = 0.18 + 0.04 = 0.22
Pr(fig already has eggs) = 0.2
Pr(male AND fig already has eggs) = 0.18 ≠
Pr(male) x Pr(fig already has eggs) = 0.22 x 0.2 = 0.044
So these two events are NOT independent.
Are the previous state of the fig and the sex of an egg laidindependent?
Pr(male) = 0.18 + 0.04 = 0.22
Pr(fig already has eggs) = 0.2
Pr(male AND fig already has eggs) = 0.18 ≠
Pr(male) x Pr(fig already has eggs) = 0.22 x 0.2 = 0.044
So these two events are NOT independent.
Are the previous state of the fig and the sex of an egg laidindependent?
Pr(male) = 0.18 + 0.04 = 0.22
Pr(fig already has eggs) = 0.2
Pr(male AND fig already has eggs) = 0.18 ≠
Pr(male) x Pr(fig already has eggs) = 0.22 x 0.2 = 0.044
So these two events are NOT independent.
≠
Short summary
The probability of A OR B involves addition.
P(A or B) = P(A) + P(B) if the two are mutually exclusive.
The probability of A AND B involves multiplication
P(A and B) = P(A) P(B) if the two are independent
Conditional probability
Pr[X|Y]
P(X | Y) means the probability of X if Y is true.
It is read as "the probability of X given Y."
P(female lays a male egg | fig has eggs already) = 0.9.
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Pr X[ ] = Pr Y[ ]Pr X |Y[ ]All valuesof Y
∑
Law of total probability:
The probability of a male egg is
Pr[male]=
Pr(male egg | fig has no eggs) Pr(fig has no eggs) + Pr(male egg |fig already has eggs) Pr(fig already has eggs)
= 0.05*0.8 + 0.9 (0.2) = 0.22
The general multiplication rule
Pr[A AND B] = Pr[A] Pr[B | A].
The general multiplication rule
Pr[A AND B] = Pr[A] Pr[B | A].
Does not require independence between A and B
Bayes' theorem
€
Pr A| B[ ] =Pr B| A[ ]Pr[A]
Pr[B]
In class exercise
Using data collected in 1975, the probabilit y of women being given
a biopsy having cervical cancer was 0.0001. The probabili ty that a
biopsy would corr ectly identify these women as having cancer was
0.90. The probabilities of a “false positive” (the test saying there
was cancer when there was not) was 0.001. What is the probabilit y
that a woman with a positive result actually has cancer?
Answer
Using data collected in 1975, the probability of women being given a biopsy havingcervical cancer was 0.0001. The probability that a biopsy would correctly identify thesewomen as having cancer was 0.90. The probabilities of a "false positi ve" (the test sayingthere was cancer when there was not) was 0.001. What is the probabilit y that a womanwith a positive result actually has cancer?
Pr[cancer | positive result] = ???
AnswerPr[cancer | positive result] = Pr[positive result | cancer] Pr[cancer]
Pr[positive result]
Pr[cancer] = 0.0001
Pr[no cancer] = 1-0.0001 = 0.9999
Pr[positive result | cancer]=0.9
Pr[positive result |no cancer] = 0.001
Pr[positive result] = ???
Answer
AnswerPr[cancer | positive result] = Pr[positive result | cancer] Pr[cancer]
Pr[positive result]
Pr[cancer] = 0.0001
Pr[no cancer] = 1-0.0001 = 0.9999
Pr[positive result | cancer]=0.9
Pr[positive result |no cancer] = 0.001
Pr[positive result] = 0.0010899
Answer
Pr[cancer | positive result] = (0.9)(0.0001) = 0.0826 0.0010899
Pr[cancer] = 0.0001
Pr[no cancer] = 1-0.0001 = 0.9999
Pr[positive result | cancer]=0.9
Pr[positive result |no cancer] = 0.001
Pr[positive result] = 0.0010899
Homework Assignment
• Chapter 1, Problems 6, 15• Chapter 2, Problems 6, 8, 9, 12• Chapter 3, Problems 4, 6, 15• Chapter 4, Problem 16• Due a week from Friday:• Sept. 22, 12 noon.• Your TA will tell you where to hand these in