WARNING! 175 slides! Print in draft mode, 6 to a page to conserve paper and ink!
TR 555 Statistics “Refresher”
Lecture 2: Distributions and Tests
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Binomial, Normal, Log Normal distributions
Chi Square and K.S. tests for goodness of fit
and independence
Poisson and negative exponential
Weibull distributions
Test Statistics, sample size and Confidence
Intervals
Hypothesis testing
WARNING!
175 slides! Print in draft mode, 6 to a page to conserve paper and ink!
1
Another good reference
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2
http://www.itl.nist.gov/div898/handbook/index.htm
Another good reference
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3
http://www.ruf.rice.edu/~lane/stat_sim/index.html
Bernoulli Trials
1
Only two possible outcomes on each trial
(one is arbitrarily labeled success, the other failure)
2
The probability of a success = P(S) = p is the same
for each trial
(equivalently, the probability of a failure = P(F) =
1-P(S) = 1- p is the same for each trial
3
4
The trials are independent
Binomial, A Probability Distribution
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n = a fixed number of Bernoulli trials
p = the probability of success in each trial
X = the number of successes in n trials
The random variable X is called a binomial
random variable. Its distribution is called a
binomial distribution
5
The binomial distribution with n trials and success
probability p is denoted by the equation
 n x
n x
f x   PX  x    p 1  p 
 x
or
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7
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The binomial distribution with n trials and success probability
p has

Mean =

Variance =

Standard deviation =
np
 2  np1  p 
   2  np 1  p 
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10
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Binomial Distribution with p=.2, n=5
0.4
n=5
0.3
0.2
0.1
0.0
0
5
10
C1
12
15
Binomial Distribution with p=.2, n=10
0.3
n=10
0.2
0.1
0.0
0
5
10
C1
13
15
Binomial Distribution with p=.2, n=30
n=30
0.2
0.1
0.0
0
5
10
C1
14
15
Binomial Distributions with p=.2
0.4
n=5
n=5
0.3
0.2
0.1
0.0
0
5
10
15
C1
0.3
n=10
n=10
0.2
0.1
0.0
0
5
10
15
C1
n=30
n=30
0.2
0.1
0.0
15
0
5
10
C1
15
Transportation Example
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16
The probability of making it safely from city A
to city B is.9997 (do we generally know this?)
Traffic per day is 10,000 trips
Assuming independence, what is the
probability that there will be more than 3
crashes in a day
What is the expected value of the number of
crashes?
Transportation Example
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Expected value = np = .0003*10000 = 3
P(X>3) = 1- [P (X=0) + P (X=1) + P (X=2) + P
(X=3)]
e.g.,P (x=3) = 10000!/(3!*9997!) *.0003^3 *
.9997^9997 = .224
don’t just hit 9997! On your calculator!
P(X>3) = 1- [.050 + .149 + .224 + .224] =
65%
Continuous probability
density functions
18
Continuous probability
density functions
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The curve describes probability of getting
any range of values, say P(X > 120),
P(X<100), P(110 < X < 120)
Area under the curve = probability
Area under whole curve = 1
Probability of getting specific number is 0,
e.g. P(X=120) = 0
Histogram
(Area of rectangle = probability)
IQ
(Intervals of size 20)
Density
0.02
0.01
0.00
55
75
95
IQ
20
115
135
Decrease interval size...
IQ
(Intervals of size 10)
Density
0.02
0.01
0.00
55
65
75
85
95
IQ
21
105
115
125
135
Decrease interval size more….
IQ
(Intervals of size 5)
0.03
Density
0.02
0.01
0.00
50
22
60
70
80
90
100
IQ
110 120
130 140
Normal: special kind of continuous
p.d.f
Bell-shaped curve
0.08
Mean = 70 SD = 5
0.07
Density
0.06
0.05
0.04
Mean = 70 SD = 10
0.03
0.02
0.01
0.00
40
50
60
70
Grades
23
80
90
100
Normal distribution
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Characteristics of
normal distribution
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Symmetric, bell-shaped curve.
Shape of curve depends on population mean
 and standard deviation .
Center of distribution is .
Spread is determined by .
Most values fall around the mean, but some
values are smaller and some are larger.
Probability = Area under curve
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Normal integral cannot be solved, so must be
numerically integrated - tables
We just need a table of probabilities for every
possible normal distribution.
But there are an infinite number of normal
distributions (one for each  and )!!
Solution is to “standardize.”
Standardizing
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Take value X and subtract its mean  from it,
and then divide by its standard deviation .
Call the resulting value Z.
That is, Z = (X- )/
Z is called the standard normal. Its mean 
is 0 and standard deviation  is 1.
Then, use probability table for Z.
Using Z Table
Standard Normal Curve
0.4
Density
0.3
0.2
Tail probability
P(Z > z)
0.1
0.0
-4
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-3
-2
-1
0
Z
1
2
3
4
Suppose we want to calculate
P X  b
where X ~ N (  ,  )
We can calculate
z
b

And then use the fact that
P[ X  b]  PZ  z 
We can find PZ  z  from our Z table
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Probability below 65?
0.08
0.07
Density
0.06
0.05
0.04
0.03
0.02
P(X < 65)
0.01
0.00
55
65
75
Grades
31
85
Suppose we wanted to calculate
P Z  z 
The using the law of complements, we have
P Z  z  1  P Z  z
This is the area under the curve to the right of z.
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Probability above 75?
Probability student scores higher than 75?
0.08
0.07
Density
0.06
0.05
P(X > 75)
0.04
0.03
0.02
0.01
0.00
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55
60
65
70
Grades
75
80
85
Now suppose we want to calculate
P a  Z  b
This is the area under the curve between a and b.
We calculate this by first calculating the area to
the left of b then subtracting the area to the left
of a.
P a  Z  b  P Z  b  P Z  a
Key Formula!
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Probability between 65 and 70?
0.08
0.07
Density
0.06
0.05
P(65 < X < 70)
0.04
0.03
0.02
0.01
0.00
35
55
60
65
70
Grades
75
80
85
Transportation Example
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36
Average speeds are thought to be normally
distributed
Sample speeds are taken, with X = 74.3 and sigma =
6.9
What is the speed likely to be exceeded only 5% of
the time?
Z95 = 1.64 (one tail) = (x-74.3)/6.9
x = 85.6
What % are obeying the 75mph speed limit within a
5MPH grace?
Assessing Normality
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the normal distribution requires that the
mean is approximately equal to the median,
bell shaped, and has the possibility of
negative values
Histograms
Box plots
Normal probability plots
Chi Square or KS test of goodness of fit
Transforms:
Log Normal
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If data are not normal, log
of data may be
If so, …
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Example of Lognormal
transform
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Example of Lognormal
transform
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Chi Square Test
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AKA cross-classification
Non-parametric test Use for nominal scale data (or convert your
data to nominal scale/categories)
Test for normality (or in general, goodness of fit)
Test for independence(can also use Cramer’s coefficient for
independence or Kendall’s tau for ratio, interval or ordinal
data)
if used it is important to recognize that it formally applies only to
discrete data, the bin intervals chosen influence the outcome,
and exact methods (Mehta) provide more reliable results
particularly for small sample size
Chi Square Test
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Tests for goodness of fit
Assumptions
–
–
–
–
–
–
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The sample is a random sample.
The measurement scale is at least nominal
Each cell contains at least 5 observations
N observations
Break data into c categories
H0 observations follow some f(x)
Chi Square Test
44

Expected number of observations in any cell
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The test statistic

Reject (not from the distribution of interest) if chi
square exceeds table value at 1-α (c-1-w degrees of
freedom, where w is the number of parameters to be
estimated)
Chi Square Test
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Tests independence of 2 variables
Assumptions
–
–
–
–
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N observations
R categories for one variable
C categories for the other variable
At least 5 observations in each cell
Prepare an r x c contingency table
H0 the two variables are independent
Chi Square Test
46
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Expected number of observations in any cell

The test statistic

Reject (not independent) if chi square exceeds table
value at 1-α distribution with (r - 1)(c - 1) degrees of
freedom
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Transportation Example
Number of crashes during a year
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Transportation Example
49
Transportation Example
Adapted from Ang and Tang, 1975
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K.S. Test for goodness of fit
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Kolmogorov-Smirnov
Non-parametric test
Use for ratio, interval or ordinal scale data
Compare experimental or observed data to a
theoretical distribution (CDF)
Need to compile a CDF of your data (called
an EDF where E means empirical)
OK for small samples
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Poisson Distribution
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the Poisson distribution
requires that the mean
be approximately equal
to the variance
Discrete events, whole
numbers with small
values
Positive values
e.g., number of crashes
or vehicles during a
given time
Transportation Example #1
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On average, 3 crashes per day are
experienced on a particular road segment
What is the probability that there will be more
than 3 crashes in a day
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P(X>3) = 1- [P (X=0) + P (X=1) + P (X=2) + P
(X=3)]
e.g.,P (x=3) =
= .224
P(X>3) = 1- [.050 + .149 + .224 + .224] = 65%
(recognize this number???)
56
Transportation Example #2
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Negative Binomial Distribution
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An “over-dispersed” Poisson
Mean > variance
Also used for crashes, other count data, especially
when combinations of poisson distributed data
Recall binomial:
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Negative binomial:
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(Negative) Exponential Distribution
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Good for inter-arrival time (e.g., time
between arrivals or crashes, gaps)
Assumes Poisson counts
P(no occurrence in time t) =
Transportation Example
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In our turn bay design example, what is the
probability that no car will arrive in 1 minute?
(19%)
How many 7 second gaps are expected in
one minute??? 82% chance that any 7 sec.
Period has no car … 60/7*82%=7/minute
Weibull Distribution
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Very flexible empirical model
Sampling
Distributions
63
Sampling Distributions
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Some Definitions
Some Common Sense Things
An Example
A Simulation
Sampling Distributions
Central Limit Theorem
Definitions
• Parameter: A number describing a population
• Statistic: A number describing a sample
• Random Sample: every unit in the population has an
equal probability of being included in the sample
• Sampling Distribution: the probability distribution of a
statistic
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Common Sense Thing #1
A random sample should represent the
population well, so sample statistics
from a random sample should provide
reasonable estimates of population
parameters
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Common Sense Thing #2
All sample statistics have some error in
estimating population parameters
67
Common Sense Thing #3
If repeated samples are taken from a
population and the same statistic (e.g.
mean) is calculated from each sample,
the statistics will vary, that is, they will
have a distribution
68
Common Sense Thing #4
A larger sample provides more
information than a smaller sample so a
statistic from a large sample should
have less error than a statistic from a
small sample
69
Distribution of X when sampling
from a normal distribution
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X
has a normal distribution with
mean =  x  
and

standard deviation =  x 
n
70
Central Limit Theorem
If the sample size (n) is large enough,
has a normal distribution with
mean =  x  
and

standard deviation =  x 
X
n
regardless of the population distribution
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n  30
72
Does X have a normal distribution?
Is the population normal?
Yes
No
Is n  30 ?
X is normal
Yes
X is considered to be
normal
No
X may or may not be
considered normal
(We need more info)
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Situation
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Different samples produce different results.
Value of a statistic, like mean or proportion,
depends on the particular sample obtained.
But some values may be more likely than
others.
The probability distribution of a statistic
(“sampling distribution”) indicates the
likelihood of getting certain values.
Transportation Example
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75
Speed is normally distributed with mean 45
MPH and standard deviation 6 MPH.
Take random samples of n = 4.
Then, sample means are normally distributed
with mean 45 MPH and standard error 3
MPH [from 6/sqrt(4) = 6/2].
Using empirical rule...
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68% of samples of n=4 will have an average
speed between 42 and 48 MPH.
95% of samples of n=4 will have an average
speed between 39 and 51 MPH.
99% of samples of n=4 will have an average
speed between 36 and 54 MPH.
What happens if we take larger
samples?
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Speed is normally distributed with mean 45
MPH and standard deviation 6 MPH.
Take random samples of n = 36 .
Then, sample means are normally distributed
with mean 45 MPH and standard error 1 MPH
[from 6/sqrt(36) = 6/6].
Again, using empirical rule...
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78
68% of samples of n=36 will have an
average speed between 44 and 46 MPH.
95% of samples of n=36 will have an
average speed between 43 and 47 MPH.
99% of samples of n=36 will have an
average speed between 42 and 48 MPH.
So … the larger the sample, the less the
sample averages vary.
Sampling Distributions for
Proportions
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Thought questions
Basic rules
ESP example
Taste test example
Rule for Sample
Proportions
80
Proportion “heads” in 50 tosses
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Bell curve for possible proportions
Curve centered at true proportion (0.50)
SD of curve = Square root of [p(1-p)/n]
SD = sqrt [0.5(1-0.5)/50] = 0.07
By empirical rule, 68% chance that a
proportion will be between 0.43 and 0.57
ESP example
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Five cards are randomly shuffled
A card is picked by the researcher
Participant guesses which card
This is repeated n = 80 times
Many people participate
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Researcher tests hundreds of people
Each person does n = 80 trials
The proportion correct is calculated for each
person
Who has ESP?
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What sample proportions go beyond luck?
What proportions are within the normal
guessing range?
Possible results of ESP
experiment
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1 in 5 chance of correct guess
If guessing, true p = 0.20
Typical guesser gets p = 0.20
SD of test = Sqrt [0.2(1-0.2)/80] = 0.035
Description of possible
proportions
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Bell curve
Centered at 0.2
SD = 0.035
99% within 0.095 and 0.305 (+/- 3SD)
If hundreds of tests, may find several (does it
mean they have ESP?)
Transportation Example
87
Concepts of Confidence
Intervals
88
Confidence Interval
A range of reasonable guesses at a
population value, a mean for instance
Confidence level = chance that range of
guesses captures the the population
value
Most common confidence level is 95%
89
General Format of a Confidence
Interval
estimate +/-
90
margin of error
Transportation Example:
Accuracy of a mean
A sample of n=36 has mean speed = 75.3.
The SD = 8 .
How well does this sample mean estimate
the population mean ?
91
Standard Error of Mean
SEM = SD of sample / square root of n
SEM = 8 / square root ( 36) = 8 / 6 = 1.33
Margin of error of mean = 2 x SEM
Margin of Error = 2.66 , about 2.7
92
Interpretation
95% chance the sample mean is within 2.7
MPH of the population mean. (q. what is
implication on enforcement of type I
error? Type II?)
A 95% confidence interval for the
population mean
sample mean +/- margin of error
75.3 +/-2.7 ; 72.6 to 78.0
93
For Large Population
Could the mean speed be 72 MPH ?
Maybe, but our interval doesn't include
72.
It's likely that population mean is above
72.
94
C.I. for mean speed at another
location
n=49
sample mean=70.3 MPH, SD = 8
SEM = 8 / square root(49) = 1.1
margin of error=2 x 1.1 = 2.2
Interval is 70.3 +/- 2.2
68.1 to 72.5
95
Do locations 1 and 2 differ in mean
speed?
C.I. for location 1 is 72.6 to 78.0
C.I. for location 2 is 68.1 to 72.5
No overlap between intervals
Looks safe to say that population means
differ
96
Thought Question
Study compares speed reduction due to
enforcement vs. education
95% confidence intervals for mean speed
reduction
•
•
97
Cop on side of road :
Speed monitor only :
13.4 to 18.0
6.4 to 11.2
Part A
Do you think this means that 95% of
locations with cop present will lower
speed between 13.4 and 18.0 MPH?
Answer : No. The interval is a range of
guesses at the population mean.
This interval doesn't describe individual
variation.
98
Part B
Can we conclude that there's a difference
between mean speed reduction of the two
programs ?
This is a reasonable conclusion. The two
confidence intervals don't overlap.
It seems the population means are
different.
99
Direct look at the difference
For cop present, mean speed reduction =
15.8 MPH
For sign only, mean speed reduction = 8.8
MPH
Difference = 7 MPH more reduction by
enforcement method
10
0
Confidence Interval for Difference
95% confidence interval for difference in
mean speed reduction is 3.5 to 10.5 MPH.
•
Don't worry about the calculations.
This interval is entirely above 0.
This rules out "no difference" ; 0
difference would mean no difference.
10
1
Confidence Interval for a Mean
when you have a
“small” sample...
10
2
As long as you have a
“large” sample….
A confidence interval for a population mean is:
 s 
x  Z

 n
10
3
where the average, standard deviation, and n depend
on the sample, and Z depends on the confidence level.
Transportation Example
Random sample of 59 similar locations
produces an average crash rate of 273.2.
Sample standard deviation was 94.40.
 94.4 
273.20  1.96
  273.20  24.09
 59 
10
4
We can be 95% confident that the average crash rate
was between 249.11 and 297.29
What happens if you can only take
a “small” sample?


10
5
Random sample of 15 similar location crash
rates had an average of 6.4 with standard
deviation of 1.
What is the average crash rate at all similar
locations?
If you have a “small” sample...
Replace the Z value with a t value to get:
 s 
x  t

 n
where “t” comes from Student’s t distribution,
and depends on the sample size through the
degrees of freedom “n-1”
10
6
Can also use the tau test for very small samples
Student’s t distribution versus
Normal Z distribution
T-distribution and Standard Normal Z distribution
0.4
Z distribution
0.3
0.2
T with 5 d.f.
0.1
0.0
10
7
-5
0
Value
5
T distribution



10
8
Very similar to standard normal distribution,
except:
t depends on the degrees of freedom “n-1”
more likely to get extreme t values than
extreme Z values
10
9
11
0
Let’s compare t and Z values
Confidence t value with Z value
level
5 d.f
2.015
1.65
90%
2.571
1.96
95%
4.032
2.58
99%
For small samples, T value is larger than Z value.
11
1
So,T interval is made to be longer than Z interval.
OK, enough theorizing!
Let’s get back to our example!
Sample of 15 locations crash rate of 6.4 with
standard deviation of 1.
Need t with n-1 = 15-1 = 14 d.f.
For 95% confidence, t14 = 2.145
 s 
 1 
x  t
  6.4  2.145
  6.4  0.55
 n
 15 
11
2
We can be 95% confident that average crash
rate is between 5.85 and 6.95.
What happens as
sample gets larger?
T-distribution and Standard Normal Z distribution
0.4
Z distribution
density
0.3
T with 60 d.f.
0.2
0.1
11
3
0.0
-5
0
Value
5
What happens to CI as
sample gets larger?
11
4

x  Z


x  t

s 

n
s 

n
For large samples:
Z and t values
become almost
identical, so CIs are
almost identical.
One not-so-small problem!


11
5
It is only OK to use the t interval for small
samples if your original measurements are
normally distributed.
We’ll learn how to check for normality in a
minute.
Strategy for
deciding how to analyze



11
6
If you have a large sample of, say, 60 or
more measurements, then don’t worry about
normality, and use the z-interval.
If you have a small sample and your data are
normally distributed, then use the t-interval.
If you have a small sample and your data are
not normally distributed, then do not use the
t-interval, and stay tuned for what to do.
Hypothesis tests

Test should begin with a set of specific, testable
hypotheses that can be tested using data:
Not a meaningful hypothesis – Was safety improved by
improvements to roadway
– Meaningful hypothesis – Were speeds reduced when traffic
calming was introduced.
–
Usually to demonstrate evidence that there is a
difference in measurable quantities
 Hypothesis testing is a decision-making tool.

11
7
Hypothesis Step 1
one working hypothesis – the null hypothesis
– and an alternative
 The null or nil hypothesis convention is generally that
nothing happened
 Provide
–
Example
were not reduced after traffic calming – Null Hypothesis
 Speed were reduced after traffic calming – Alternative Hypothesis
 speeds
 When
stating the hypothesis, the analyst must think
of the impact of the potential error.
11
8
Step 2, select appropriate
statistical test
 The
analyst may wish to test
– Changes
in the mean of events
– Changes in the variation of events
– Changes in the distribution of events
11
9
Step 3, Formulate decision rules
and set levels for the probability of
error
Accept Reject
Area where we
incorrectly accept
Type II error, referred
to as b
12
0
Area where we
incorrectly reject
Type I error,
referred to as a
significance level)
Type I and II errors
 lies in
 lies in
acceptance interval rejection interval
Accept the
Claim
Reject the
claim
12
1
No error
Type I error
Type II error
No error
Levels of a and b
Often b is not considered in the development of the
test.
 There is a trade-off between a and b
 Over emphasis is placed on the level of significance
of the test.
 The level of a should be appropriate for decision
being made.

Small values for decisions where errors cannot be tolerated
and b errors are less likely
– Larger values where type I errors can be more easily tolerated
–
12
2
Step 4 Check statistical
assumption
 Draw
new samples to check answer
 Check the following assumption
– Are
data continuous or discrete
– Plot data
– Inspect to make sure that data meets assumptions
 For
example, the normal distribution assumes that mean =
median
– Inspect
12
3
results for reasonableness
Step 5 Make decision
 Typical
misconceptions
– Alpha
is the most important error
– Hypothesis tests are unconditional
 It
does not provide evidence that the working hypothesis is
true
– Hypothesis
test conclusion are correct
 Assume
–
12
4
300 independent tests
– 100 rejection of work hypothesis
 a = 0.05 and b = 0.10
– Thus 0.05 x 100 = 5 Type I errors
– And 0.1 x 200 = 20 Type II errors
– 25 time out of 300 the test results were wrong
Transportation Example
The crash rates below, in 100 million vehicle miles, were calculated for 50,
20 mile long segment of interstate highway during 2002
.34
.31
.20
.43
.26
.22
.43
.20
.45
.17
.40
.28
.36
.38
.30
.25
.33
.48
.54
.40
.31
.23
.36
.39
.16
.34
.40
.30
.55
.32
.26
.39
.27
.25
.34
.55
.38
.42
.35
.46
Frequency of Crash Rates by Sec tions
x = 0.35
s2 = 0.0090
s = 0.095
25
0.3 - 0.4
20
15
0.2 - 0.3
0.4 - 0.5
10
12
5
5
0.5-0.6
0.1 - 0.2
0
Crash Rates
.43
.21
.27
.39
.37
.34
.43
.28
.43
.33
Example continued
Crash rates are collected from non-interstate system
highways built to slightly lower design standards. Similarly
and average crash rate is calculated and it is greater (0.53).
Also assume that both means have the same standard
deviation (0.095). The question is do we arrive at the same
accident rate with both facilities. Our hypothesis is that both
have the same means f = nf Can we accept or reject our
hypothesis?
12
6
Example continued
Is this part of the crash rate distribution
for interstate highways
Accept Reject
Area where we
incorrectly accept
Type II error
Area where we
incorrectly reject
Type I error
12
7
0.34
0.53
Example continued
 Lets
set the probability of a Type I error at 5%
set (upper boundary - 0.35)/ 0.095 =1.645 (one tail
Z (cum) for 95%)
– Upper boundary = 0.51
– Therefore, we reject the hypothesis
– What’s the probability of a Type II error?
– (0.51 – 0.53)/0.095 = -0.21
– 41.7%
– There is a 41.7% chance of what?
–
12
8
The P value … example: Grade
inflation
H0: μ = 2.7
HA: μ > 2.7
Random sample
of students
Data
n = 36
s = 0.6
and
Decision Rule
Set
significance
level
α
=
0.05.
12
9 If p-value 0.05, reject null hypothesis.
X
Example: Grade inflation?
Population of
5 million college
students
Sample of
100 college students
13
0
Is the average
GPA 2.7?
How likely is it that
100 students would
have an average
GPA as large as 2.9
if the population
average was 2.7?
The p-value illustrated
How likely is it that 100 students would have an average
13
1 GPA as large as 2.9 if the population average was 2.7?
Determining the p-value
H0: μ = average population GPA = 2.7
HA: μ = average population GPA > 2.7
If 100 students have average GPA of 2.9 with standard
deviation of 0.6, the P-value is:
13
2
P( X  2.9)  P[Z  (2.9  2.7) /(0.6 / 100)]
 P[Z  3.33]  0.0004
Making the decision


13
3
The p-value is “small.” It is unlikely that we
would get a sample as large as 2.9 if the
average GPA of the population was 2.7.
Reject H0. There is sufficient evidence to
conclude that the average GPA is greater
than 2.7.
Terminology



13
4
H0: μ = 2.7 versus HA: μ > 2.7 is called a
“right-tailed” or a “one-sided” hypothesis
test, since the p-value is in the right tail.
Z = 3.33 is called the “test statistic”.
If we think our p-value small if it is less than
0.05, then the probability that we make a
Type I error is 0.05. This is called the
“significance level” of the test. We say,
α=0.05, where α is “alpha”.
Alternative Decision Rule


13
5
“Reject if p-value  0.05” is equivalent to
“reject if the sample average, X-bar, is larger
than 2.865”
X-bar > 2.865 is called “rejection region.”
Minimize chance of
Type I error...




13
6
… by making significance level a small.
Common values are a = 0.01, 0.05, or 0.10.
“How small” depends on seriousness of Type
I error.
Decision is not a statistical one but a
practical one (set alpha small for safety
analysis, larger for traffic congestion, say)
Type II Error and Power




13
7
“Power” of a test is the probability of
rejecting null when alternative is true.
“Power” = 1 - P(Type II error)
To minimize the P(Type II error), we
equivalently want to maximize power.
But power depends on the value under the
alternative hypothesis ...
Type II Error and Power
13
8
(Alternative is true)
Factors affecting power...




13
9
Difference between value under the null and
the actual value
P(Type I error) = a
Standard deviation
Sample size
Strategy for designing a
good hypothesis test





14
0
Use pilot study to estimate std. deviation.
Specify a. Typically 0.01 to 0.10.
Decide what a meaningful difference would
be between the mean in the null and the
actual mean.
Decide power. Typically 0.80 to 0.99.
Simple to use software to determine sample
size …
How to determine sample size
Depends on experiment
Basically, use the formulas and let sample size
be the factor you want to determine
Vary the confidence interval, alpha and beta
http://www.ruf.rice.edu/~lane/stat_sim/conf_inte
rval/index.html
14
1
If sample is too small ...

… the power can be too low to identify even
large meaningful differences between the null
and alternative values.
–
–
14
2
Determine sample size in advance of conducting
study.
Don’t believe the “fail-to-reject-results” of a study
based on a small sample.
If sample is really large ...

… the power can be extremely high for
identifying even meaningless differences
between the null and alternative values.
–
–
14
3
In addition to performing hypothesis tests, use a
confidence interval to estimate the actual
population value.
If a study reports a “reject result,” ask how much
different?
The moral of the story
as researcher


14
4
Always determine how many measurements
you need to take in order to have high
enough power to achieve your study goals.
If you don’t know how to determine sample
size, ask a statistical consultant to help
you.
Important “Boohoo!” Point



14
5

Neither decision entails proving the null
hypothesis or the alternative hypothesis.
We merely state there is enough evidence to
behave one way or the other.
This is also always true in statistics! No
matter what decision we make, there is
always a chance we made an error.
Boohoo!
Comparing the Means
of Two Dependent Populations
The Paired T-test ….
14
6
Assumptions: 2-Sample T-Test



14
7
Data in each group follow a normal
distribution.
For pooled test, the variances for each
group are equal.
The samples are independent. That is, who
is in the second sample doesn’t depend on
who is in the first sample (and vice versa).
What happens if samples aren’t
independent?
That is, they are
“dependent” or
“correlated”?
14
8
Do signals with all-red clearance
phases have lower numbers of
crashes than those without?
All-Red
60
32
80
50
Sample Average: 55.5
No All-Red
32
44
22
40
34.5
Real question is whether intersections with similar traffic
volumes have different numbers of crashes. Better then to
14 compare the difference in crashes in “pairs” of intersections
9 with and without all-red clearance phases.
Now, a Paired Study
Traffic
Low
Medium
Med-high
High
Averages
St. Dev
Crashes
No all-red All-red
22
20
29
28
35
32
80
78
41.5
39.5
26.1
26.1
Difference
2.0
1.0
3.0
2.0
2.0
0.816
P-value = How likely is it that a paired sample would have a
15
difference as large as 2 if the true difference were 0? (Ho = no diff.)
0
- Problem reduces to a One-Sample T-test on differences!!!!
The Paired-T Test Statistic
If:
• there are n pairs
• and the differences are normally distributed
Then:
The test statistic, which follows a t-distribution with n-1
degrees of freedom, gives us our p-value:
td 
15
1
sample difference  hypothesized difference
standard error of differences
d μ
 s d
d
n
The Paired-T Confidence Interval
If:
• there are n pairs
• and the differences are normally distributed
Then:
The confidence interval, with t following t-distribution
with n-1 d.f. estimates the actual population difference:
15
2
d  t








s
d
n








Data analyzed as 2-Sample T
Two sample T for No all-red vs All-red
No
All
N
4
4
Mean
41.5
39.5
StDev
26.2
26.1
SE Mean
13
13
95% CI for mu No - mu All: ( -43, 47)
T-Test mu No = mu All (vs not =): T = 0.11
P = 0.92 DF = 6
Both use Pooled StDev = 26.2
15
3
P = 0.92. Do not reject null. Insufficient evidence to
conclude that there is a real difference.
Data analyzed as Paired T
Paired T for No all-red vs All-red
No
All
Difference
N
4
4
4
Mean
41.5
39.5
2.000
StDev
26.2
26.1
0.816
SE Mean
13.1
13.1
0.408
95% CI for mean difference: (0.701, 3.299)
T-Test of mean difference = 0 (vs not = 0):
T-Value = 4.90 P-Value = 0.016
15 P = 0.016. Reject null. Sufficient evidence to conclude that
4 there IS a difference.
What happened?



15
5
P-value from two-sample t-test is just plain
wrong. (Assumptions not met.)
We removed or “blocked out” the extra
variability in the data due to differences in
traffic, thereby focusing directly on the
differences in crashes.
The paired t-test is more “powerful” because
the paired design reduces the variability in
the data.
Ways Pairing Can Occur



15
6
When subjects in one group are “matched”
with a similar subject in the second group.
When subjects serve as their own control by
receiving both of two different treatments.
When, in “before and after” studies, the same
subjects are measured twice.
If variances of the measurements
of the two groups are not equal...
Estimate the standard error of the difference as:
s12 s22
n1  n 2
Then the sampling distribution is an approximate t
distribution with a complicated formula for d.f.
15
7
If variances of the measurements
of the two groups are equal...
Estimate the standard error of the difference using the
common pooled variance:










2 
s2p n1  n1
1
where
2  (n 1)s2
(n

1)s
1
2
2
s2p  1
n1  n 2  2
Then the sampling distribution is a t distribution
with n1+n2-2 degrees of freedom.
Assume variances are equal only if neither sample standard deviation
15
8is more than twice that of the other sample standard deviation.
Assumptions for correct P-values



15
9
Data in each group follow a normal
distribution.
If use pooled t-test, the variances for each
group are equal.
The samples are independent. That is, who
is in the second sample doesn’t depend on
who is in the first sample (and vice versa).
Interpreting a confidence interval
for the difference in two means…
If the confidence
interval contains…
zero
16
0
only positive
numbers
only negative
numbers
then, we conclude …
the two means may
not differ
first mean is larger
than second mean
first mean is smaller
than second mean
Difference in variance






16
1
Use F distribution test
Compute F = s1^2/s2^2
Largest sample variance on
top
Look up in F table with n1
and n2 DOF
Reject that the variance is
the same if f>F
If used to test if a model is
the same (same coefficients)
during 2 periods, it is called
the Chow test
16
2
16
3
16
4
Experiments and pitfalls
 Types
of safety experiments
– Before
and after tests
– Cross sectional tests
 Control
sample
 Modified sample
 Similar or the same condition must take place for both
samples
16
5
Regression to the mean
 This
problem plagues before and after tests
– Before
and after tests require less data and
therefore are more popular
 Because
safety improvements are driven
abnormally high crash rates, crash rates are
likely to go down whether or not an
improvement is made.
16
6
San Francisco intersection crash
data
16
7
N o . o f A c c id e n ts / A c c id e n t/ A c c id e n ts / A c c id e n tsA c c id e n ts / % Ch a n g e
I n te rse c t io nIsn te rse c t io n Y e a r/
Y e a r in in 1 9 7 7
f Ionrte rse c t io n
W ith G iv e nI n 1 9 7 4 - 7I6n te rse c t io n1 9 7 6 – 7 6 G ro u p
In 1977
N o. of
in
1 9 7 4 -7 6
Fo r G ro u p
A c c id e n ts in
(ro u n d e d )
1974 - 76
256
0
0
0
64
0.25
L a rg e
I n c re a s e
218
1
0.33
72
120
0.55
67%
173
2
0.67
116
121
0.7
S ma ll
I n c re a s e
121
3
1
121
126
1.04
S ma ll
I n c re a s e
97
4
1.33
129
105
1.08
-1 9 %
70
5
1.67
117
93
1.33
-2 0 %
54
6
2
108
84
1.56
-2 2 %
32
7
2.33
75
72
2.25
-3 %
29
8
2.67
77
47
1.62
-3 9 %
Spill over and migration impacts
Improvements, particularly those that are expect to modify
behavior should be expected to spillover impacts at other
locations. For example, suppose that red light running
cameras were installed at several locations throughout a city.
Base on the video evidence, this jurisdiction has the ability to
ticket violators and, therefore, less red light running is
suspected throughout the system – leading to spill over
impacts. A second data base is available for a similar control
set of intersections that are believed not to be impacted by
spill over. The result are listed below:
Crashes at
16
8
Before
After
Comparison
Crashes at
Sites (no spill
Treated Sites
over)
100
140
64
112
Spill over impact
Assuming that spill over has occurred at the treated sites, the
reduction in accidents that would have occurred after
naturally is 100 x 112/140 = 80 had the intersection
remained untreated. therefore, the reduction is really from
80 crashes to 64 crashes (before vs after) or 20% rather than
100 crashes to 64 crashes or 36%.
16
9
Spurious correlations
17
0
During the 1980s and early 1990s the Japanese economy was
growing at a much greater rate than the U.S. economy. A professor
on loan to the federal reserved wrote a paper on the Japanese
economy and correlated the growth in the rate of Japanese economy
and their investment in transportation infrastructure and found a
strong correlation. At the same time, the U.S. invests a much lower
percentage of GDP in infrastructure and our GNP was growing at a
much lower rate. The resulting conclusion was that if we wanted to
grow the economy we would invest like the Japanese in public
infrastructure. The Association of Road and Transportation Builders
of America (ARTBA) loved his findings and at the 1992 annual
meeting of the TRB the economist from Bates college professor won
an award. In 1992 the Japanese economy when in the tank, the
U.S. economy started its longest economic boom.
Spurious Correlation Cont.
What is going on here?
What is the nature of the relationship
between transportation investment and
economic growth?
17
1