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• Today: Quizz 4
• Tomorrow: Lab 3 – SN 4117
• Wed: A3 due
• Friday: Lab 3 due
• Mon Oct 1: Exam I  this room, 12 pm
• Mon Oct 1: No grad seminar
Key concepts so far
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Quantity
Measurement scale
Dimensions & Units
Equations
Data Equations
– Sums of squared residuals quantify
improvement in fit, compare models
• Quantify uncertainty through frequency
distributions
– Empirical
– Theoretical
– 4 forms, 4 uses
Today
Selected examples from:
Read lecture notes
Logic of Hypothesis Testing
Reject JUST LUCK Hypothesis
A
B
C
D
Skill!
Just
Luck!!
Izaak
Walton
Reject JUST LUCK
• Compared observed outcome to all possible
outcomes  more tractable to restrict to all possible
outcomes given that JUST LUCK hyp is true
Arrangements of 8 fish such that IW catches 7?
Reject JUST LUCK
Arrangements of 8 fish such that IW catches 7?
Assign probabilities to each outcome, assuming that the
H0 ‘JUST LUCK’ is true
For each fish, there is a 1 in 5 chance that IW will catch it
IW=8
IW=7
IW=7
IW=7
IW=7
A=1
B=1
C=1
D=1
(1/5)8
(1/5)7
(1/5)7
(1/5)7
(1/5)7
0.00000256
0.0000128
0.0000128
0.0000128
0.0000128
p=0.00005376, i.e. 5 times in 10,000
Hypothesis Testing
• Set of rules for making decisions in the face of
uncertainty
• Logic is inductive: from specific to general
• Structure is binary
3 styles of statistical inference
• Likelihood, frequentist and Bayesian inference
• All based on the principle of maximum likelihood
Definition: a model that makes the data more probable
(best predicts the observed data) is said to be more
likely to have generated the data
3 styles of statistical inference
Likelihood inference
∑ res2 = 0.1171
∑ res2 = 0.0204
Reduction in squared deviance
∑ res2 = 0.0966
Which model is more likely to have generated the data?
Frequentist inference
Use expected distribution of outcomes to calculate a probability
3 styles of statistical inference
Bayesian inference
Find the probability that a hypothesis is true, given the
observed data
Contrast to: finding the probability of observing the data I observed
(or more extreme data), assuming that the null hypothesis is true
Integrates prior knowledge we have on the system with new
observations to make an informed decision
3 styles of statistical inference
Bayesian inference
e.g.: coin flip. Hypothesis: the coin is biased
Observe flips: HTHHHTTHHHH
Frequentist approach
Null Hypothesis H0
• H0  just chance
• Research hypothesis (what we really care about) is
stated as HA
• So, why work with H0 and not HA?
– Easier to work out probabilities
– Permits yes/no decision
• Working with H0 is not intuitive.
Logic is backwards because we want to reject H0,
not explain how the world functions through H0
Choice of HA
• Start with research hyp, then challenge it with H0
• HA/H0 defined with respect to population, not sample
• HA/H0 must be defined prior to analysis
• Choice of HA/H0 determines how we calculate p-value
• HA/H0 pair must be exhaustive
• HA/H0 must be mutually exclusive
Choice of HA
How do we choose it?
Often HA=effect, H0= no effect
BUT, more informative choices are available:
G: growth rate of plants. c:Control, t: treated with fertilizer
1..
2..
3..
‘tails’
‘scale’
Type I & Type II error
• Type I (α): reject H0 when it is true
‘false positive’
e.g. in a trial, accused is innocent but goes to jail
H0:
• Type II (β): not rejecting H0 when it is false
‘false negative’
e.g. in a trial, accused is guilty but is set free
H0:
Type I & Type II error
• Type I (α): reject H0 when it is true
‘false positive’
• Type II (β): not rejecting H0 when it is false
‘false negative’
H0 True
Not rejecting H0
Reject H0
H0 False
Type I & Type II error
True H0
Reject H0 when it is true
Type I & Type II error
Draw not rejecting H0 when it is false, i.e. β
Tradeoff between α and β
Draw rejecting H0 when H0 is false, i.e. power
True HA
Selected examples from:
Will present 2 examples (if time allows)
More examples in lecture notes
Table 7.1 Generic recipe for decision making with statistics
1.
2.
3.
4.
5.
6.
State population, conditions for taking sample
State the model or measure of pattern……………………………
State null hypothesis about population……………………………
State alternative hypothesis…………………………………………
State tolerance for Type I error………………………………………
State frequency distribution that gives probability of outcomes when
the Null Hypothesis is true. Choices:
a) Permutations: distributions of all possible outcomes
b) Empirical distribution obtained by random sampling of all possible
outcomes when H0 is true
c) Cumulative distribution function (cdf) that applies when H0 is true
State assumptions when using a cdf such as Normal, F, t or chisquare
7. Calculate the statistic. This is the observed outcome
8. Calculate p-value for observed outcome relative to distribution of
outcomes when H0 is true
9. If p less than α then reject H0 in favour of HA
If greater than α then not reject H0
10.Report statistic, p-value, sample size
Declare decision
Example: jackal bones
Length of bones from 10 female and 10 male jackals
(Manly 1991)
Male
Female
L = length of mandible
120
110
(L=mm) of Golden jackals
107
111
110
107
116
108
114
110
111
105
113
107
117
106
114
111
112
111
113.4
108.6 mean
13.82
5.16 var
Example: jackal bones
1. Population:
All possible measurements on these bones
All jackals in the world? Need to know if sample
representative
2. Measure of pattern: ST = D0 =
3. H0:
4. HA:
5. α=
6. Theoretical dist of D0? Unknown
Solution: construct empirical freq dist of D0 when H0 is true by
randomization….
Example: jackal bones
2. D0 = mean(Lmale)-mean(Lfem) 3.H0: D0<=0
4.HA:D0>0
5. α=5%
6. Empirical FD. Randomization
a) Assign bones randomly to 2 groups (forget M/F)
b) Compute mean(gr1) and mean(gr2)
c) D0,res= mean(gr1) - mean(gr2)
d) Repeat many times (the more the better, continued later)
e) Assemble random differences into a FD
7. Statistic. Do= 113.4 – 108.6 = 4.8 mm
Example: jackal bones
2. D0 = mean(Lmale)-mean(Lfem) 3.H0: D0<=0
4.HA:D0>0
8. Compute p-value:
100,000 values of D0,res
360 values exceed 4.8
p = 360/100000
p = 0.0036
9. p =0.0036< α=0.05
 reject H0
in favour of HA (D0>0)
10.D0 = 4.8 mm
n=
p=
male jackal mandible bones
significantly longer than those of females
5. α=5%
Example: jackal bones
This was laborious
Can be made easier
by using theoretical
frequency distributions
Trade off: must make
assumptions
Example: jackal bones
6d) repeat many times
100,000 repetitions
Example: jackal bones
6d) repeat many times
10,000 repetitions
Example: jackal bones
6d) repeat many times
1,000 repetitions
Example: Oat Yield data
Yield of oats in 2 groups
1. Control
2. Chemical seed treatment
1 common mean
1 mean per group
Is the improvement better than random?
Example: Oat Yield data
1. Sample: 8 measurements
Population: all possible measurements taken with a stated
procedure
2. Measure of pattern: ST = SSmodel
3. H0: E(SSmodel) = 0
4. HA:E(SSmodel) > 0
5. α=5%
6. Theoretical dist of SSmodel? Unknown
Solution: construct empirical freq dist of SSmodel when H0 is
true by randomization….
Example: Oat Yield data
6. Empirical FD
a) Assign yields to 2 groups (forget treatment/control)
b) Fit common mean model
c) Fit 2 means model
d) Calculate SSmodel
e) Repeat many times (1000)
f) Assemble random differences into a FD
7. Statistic. SSmodel=192.08
Example: Oat Yield data
8. Compute p-value:
1,000 values of SSmodel
161 values exceed 192.08
p = 161/1000
p = 0.161
9. p = 0.161 > 0.05  do not reject H0
The improvement is not better than random
10.SSmodel = 192.08
n=8
p = 0.161
we can not reject the JUST LUCK hypothesis
QUIZZ 4
Good luck!