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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
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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?
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Sampling distribution of the mean
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Sampling distribution of the mean
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Sampling distribution of the mean
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Sampling distribution of the mean
How confident can we be about this one estimate of the mean?
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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:
€
SEY
=s
n
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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
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![Page 10: 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:](https://reader036.vdocuments.mx/reader036/viewer/2022062314/56649d495503460f94a25b17/html5/thumbnails/10.jpg)
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= 5.14
Y = 5.26
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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
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Mean ± 95% C.I.
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Mean ± 95% C.I.
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What if we took larger samples? Say, n=20 instead of n=10?
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Probability
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
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The Birthday Challenge
![Page 20: 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:](https://reader036.vdocuments.mx/reader036/viewer/2022062314/56649d495503460f94a25b17/html5/thumbnails/20.jpg)
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
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Two events are mutually exclusive if they cannot both be
true.
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Two events are mutually exclusive if they cannot both be
true.
(cannot both occur simultaneously)
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A
B
Mutually exclusive
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A
B
Mutually exclusive
Venn diagram
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A
B
Mutually exclusive
Venn diagram
Sample space
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A
B
Mutually exclusive
Venn diagram
Sample space
Possible outcome
Pr[B] proportionalto area
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Mutually exclusive
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Mutually exclusive
Pr(A and B) = 0
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Mutually exclusive
Visual definition - areas do not overlap in Venn diagram
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Not mutually exclusive
Pr(A and B) 0
Pr(purple AND square) 0
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For example
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Probability distribution
A probability distribution describes the true relative
frequency of all possible values of a random variable.
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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
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Probability distribution for the
outcome of a roll of a die
Number rolled
Frequency
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Probability distribution for the sum of a roll of two
dice
Sum of two dice
Frequency
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The addition rule
The addition principle: If two events A and B are mutually
exclusive, then
Pr[A OR B] = Pr[A] + Pr[B]
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Addition Rule
Pr[1 or 2] = ?
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Addition Rule
Pr[1 or 2] = Pr[1]+Pr[2]
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Addition Rule
Pr[1 or 2] = Pr[1]+Pr[2]
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Addition Rule
Pr[1 or 2] = Pr[1]+Pr[2]
Sum of areas
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The probability of a range
For families of 8 children,
Pr[Number of boys ≥ 6] = ?
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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]
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The probabilities of all possibilities add
to 1.
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Addition Rule
Pr[1 or 2 or 3 or 4 or 5 or 6] = ?
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Addition Rule
Pr[1 or 2 or 3 or 4 or 5 or 6] = 1
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Probability of Not
QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.
Pr[NOT rolling a 2] = ?
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Probability of Not
QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.
Pr[NOT rolling a 2] = 1 - Pr[2] = 5/6
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Probability of Not
QuickTime™ and aTIFF (Uncompressed) decompressorare needed to see this picture.
Pr[NOT rolling a 2] = 1 - Pr[2] = 5/6
Pr[not A] = 1-Pr[A]
![Page 49: 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:](https://reader036.vdocuments.mx/reader036/viewer/2022062314/56649d495503460f94a25b17/html5/thumbnails/49.jpg)
The addition rule
The addition principle: If two events A and B are mutually
exclusive, then
Pr[A OR B] = Pr[A] + Pr[B]
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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?
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General Addition Rule
A
B
Pr[A or B] = ?
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General Addition Rule
A
B
Pr[A or B] = ?
AB
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General Addition Rule
A
B
Pr[A or B] = ?
AB
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General Addition Rule
A
B
Pr[A or B] = ?
AB
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General Addition Rule
A
B
AB
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General Addition Rule
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General Addition RulePr[Walks or flies] = ?
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General Addition RulePr[Walks or flies] = ?
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General Addition RulePr[Walks or Flies] = Pr[Walks] + Pr[Flies]
- Pr[Walks and Flies]
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General Addition Rule
Pr[A OR B] = Pr[A] + Pr[B] - Pr[A AND B].
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Independence
Two events are independent if the occurrence of one gives
no information about whether the second will occur.
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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
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Multiplication rule
If two events A and B are independent, then
Pr[A and B] = Pr[A] x Pr[B]
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Pr[boy]=0.512
Pr[ (first child is a boy) AND (second child is a boy)]
= 0.512 × 0.512 = 0.262.
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Pr[boy]=0.512
Pr[ (first child is a boy) AND (second child is a boy)]
= 0.512 × 0.512 = 0.262.
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General Addition Rule
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Multiplication rule
If two events A and B are independent, then
Pr[A and B] = Pr[A] x Pr[B]
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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
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Probability trees
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Sex of two children family
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Sex of two children family
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Dependent events
Variables are not always independent; in fact they are often not
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Fig wasps
QuickTime™ and aTIFF (Uncompressed) decompressor
are needed to see this picture.
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Fig wasps
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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]
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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.
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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.
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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.
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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.
![Page 80: 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:](https://reader036.vdocuments.mx/reader036/viewer/2022062314/56649d495503460f94a25b17/html5/thumbnails/80.jpg)
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.
≠
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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
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Conditional probability
Pr[X|Y]
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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:
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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
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The general multiplication rule
Pr[A AND B] = Pr[A] Pr[B | A].
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The general multiplication rule
Pr[A AND B] = Pr[A] Pr[B | A].
Does not require independence between A and B
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Bayes' theorem
€
Pr A| B[ ] =Pr B| A[ ]Pr[A]
Pr[B]
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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?
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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] = ???
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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] = ???
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Answer
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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
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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
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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