assignment #1 - zoology.ubc.camfscott/lectures/05_probability.pdf · assignment #2 chapter 3: 17,...
TRANSCRIPT
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Assignment #1
Chapter 1: 14, 17, 19, 20 Chapter 2: 25, 32 Due Tomorrow by 2 pm in your TA’s homework box
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Assignment #2
Chapter 3: 17, 21 Chapter 4: 18, 20 Due Next Friday, Oct. 2nd by 2pm in your TA’s homework box
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Reading
For Today: Chapter 5 For Tuesday: Chapter 6
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Chapter 4 Review
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Sampling Distribution
The probability distribution of all values for an estimate that we might obtain when we sample a population
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N
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The standard error of an estimate is the standard deviation of its sampling
distribution.
The standard error predicts the sampling error of the
estimate.
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Standard error of the mean
€
σY =σn
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Estimate of the standard error of the mean
€
SEY =sn
This gives us some knowledge of the likely difference between our sample mean and the true population mean.!
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95% Confidence Interval
If you took 1000 independent samples and calculated the 95% confidence
interval of your estimate from each, ~950 of them would contain the true population
parameter
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The 2SE rule-of-thumb
The interval from - 2 to + 2 provides a rough estimate of the 95% confidence interval for the mean. "
€
SEY
€
SEY
€
Y
€
Y
(Assuming normally distributed population and/or sufficiently large sample size.)!
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Pseudoreplication
The error that occurs when samples are not "independent, but they are treated as though they are."
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Probability
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The probability of an event is its true relative frequency; the proportion of times the event would occur if we repeated the
same process over and over again.""
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Two events are mutually exclusive if they cannot both be true. "
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Mutually exclusive
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Mutually exclusive
Heads, Tails Boy, Girl Ace, King Apple, Orange
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Mutually exclusive
Pr(A and B) = 0
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Not mutually exclusive
Pr(A and B) ≠ 0 Pr(purple AND square) ≠ 0
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Not mutually exclusive
Heads 1st flip, Tails 2nd flip Boy, Green eyes Ace, Hearts Apple, Red
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For example
Event A: First child is female Event B: Second child is female P(A) = 0.48 P(B) = 0.48 But P(A and B) ≠ 0, so these events are NOT mutually exclusive.
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Probability distribution
A probability distribution describes the true relative frequency (a.k.a. the probability) of all
possible values of a random variable.
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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 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 principle
P[1st or 2nd roll is 3] = 1/6 + 1/6 = 1/3"
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The probability of a range
Pr[Number of boys ≥ 6] = Pr[6] + Pr[7] + Pr[8]....
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The probabilities of all possibilities add to 1.
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Probability of Not
Pr[NOT rolling a 2] = 1 – Pr[Rolling a 2] = 5/6
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General Addition Principle
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General Addition Principle Pr[A OR B] = Pr[A] + Pr[B] - Pr[A AND B].
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General addition principle Pr[A OR B] = Pr[A] + Pr[B] - Pr[A AND B].
If two events A and B are mutually exclusive, then Pr[A AND B] = 0,
therefore:
Pr[A OR B] = Pr[A] + Pr[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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The multiplication principle
The multiplication principle: If two events A and B are independent, then
Pr[A AND B] = Pr[A] × Pr[B]
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The multiplication principle
P[1st and 2nd roll is 3] = 1/6 x 1/6 = 1/36"
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Offspring of two "carriers":
Pr[congenital nightblindness]=0.25
Pr[ (first child has nightblindness) AND
(second child has nightblindness)] = 0.25 × 0.25 = 0.0625. "
What is the probability that two kids from this family both have nightblindedness?"
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Probability trees
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Phenotypes in two-child family
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Phenotypes in two-child family
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The probability of A OR B involves addition. Pr(A or B) = Pr(A) + Pr(B) if the two are mutually
exclusive. The probability of A AND B involves multiplication
Pr(A and B) = Pr(A) Pr(B) if the two are independent "
Short summary
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Dependent events
Variables are not always independent.""The probability of one event may depend on the "outcome of another event"
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Washing hands
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Freq
uenc
y"
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Hand washing after using the restroom
• Pr[male] = 0.495
• Pr[male washes his hands] = 0.74
• Pr[female washes her hands] = 0.83
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Hand washing
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Are sex and hand washing independent? Pr(male) = 0.495 Pr(hand washing) = 0.366 + 0.419 = 0.785 Pr(male AND hand washing) = 0.366 ≠ Pr(male) × Pr(hand washing) = 0.495 × 0.785 = 0.389 So these two events are NOT independent.
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Freq
uenc
y"
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Conditional probability
Pr[A|B]"
The conditional probability of an event is the probability of that event occurring given that a
condition is met.
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Pr(hand washing | male) = 0.74.
Pr(A | B) means the probability of A if B is true. It is read as "the probability of A given B." "
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Pr A[ ] = Pr A | B[ ]All valuesof B∑ Pr B[ ]
Law of total probability
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Freq
uenc
y"
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The probability of hand washing is Pr[hand washing] =
Pr(hand washing | male) Pr(male) + Pr(hand washing | female) Pr(female)
= 0.74 (0.495) + 0.83 (0.505) = 0.785
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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]
If two events A and B are independent, then Pr[B|A] = Pr[B], therefore:
Pr[A AND B] = Pr[A] × Pr[B]
![Page 56: Assignment #1 - zoology.ubc.camfscott/lectures/05_Probability.pdf · Assignment #2 Chapter 3: 17, 21 Chapter 4: 18, 20 Due Next Friday, Oct. 2nd by 2pm in your TA’s homework box](https://reader030.vdocument.in/reader030/viewer/2022040417/5d57e63a88c9936f1a8b9ddf/html5/thumbnails/56.jpg)
The general multiplication rule
Pr[A AND B] = Pr[A] Pr[B | A] Pr[A AND B] = Pr[B] Pr[A | B] Therefore Pr[B] Pr[A | B] = Pr[A] Pr[B | A]
![Page 57: Assignment #1 - zoology.ubc.camfscott/lectures/05_Probability.pdf · Assignment #2 Chapter 3: 17, 21 Chapter 4: 18, 20 Due Next Friday, Oct. 2nd by 2pm in your TA’s homework box](https://reader030.vdocument.in/reader030/viewer/2022040417/5d57e63a88c9936f1a8b9ddf/html5/thumbnails/57.jpg)
Bayes' theorem
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Pr A |B[ ] =Pr B |A[ ]Pr[A]
Pr[B]
![Page 58: Assignment #1 - zoology.ubc.camfscott/lectures/05_Probability.pdf · Assignment #2 Chapter 3: 17, 21 Chapter 4: 18, 20 Due Next Friday, Oct. 2nd by 2pm in your TA’s homework box](https://reader030.vdocument.in/reader030/viewer/2022040417/5d57e63a88c9936f1a8b9ddf/html5/thumbnails/58.jpg)
In class exercise Using data collected in 1975, the probability of women had cervical cancer was 0.0001. The probability that a biopsy would correctly 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 probability that a woman with a positive result actually has cancer?
![Page 59: Assignment #1 - zoology.ubc.camfscott/lectures/05_Probability.pdf · Assignment #2 Chapter 3: 17, 21 Chapter 4: 18, 20 Due Next Friday, Oct. 2nd by 2pm in your TA’s homework box](https://reader030.vdocument.in/reader030/viewer/2022040417/5d57e63a88c9936f1a8b9ddf/html5/thumbnails/59.jpg)
Pr[cancer | positive result] = ???
Using data collected in 1975, the probability of women had cervical cancer was 0.0001. The probability that a biopsy would correctly 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 probability that a woman with a positive result actually has cancer?
![Page 60: Assignment #1 - zoology.ubc.camfscott/lectures/05_Probability.pdf · Assignment #2 Chapter 3: 17, 21 Chapter 4: 18, 20 Due Next Friday, Oct. 2nd by 2pm in your TA’s homework box](https://reader030.vdocument.in/reader030/viewer/2022040417/5d57e63a88c9936f1a8b9ddf/html5/thumbnails/60.jpg)
Pr[cancer | positive result] = ?
Using data collected in 1975, the probability of women had cervical cancer was 0.0001. The probability that a biopsy would correctly 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 probability that a woman with a positive result actually has cancer?
€
Pr A |B[ ] =Pr B |A[ ]Pr[A]
Pr[B]