copyright (c) bani k. mallick1 stat 651 lecture #15
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Copyright (c) Bani K. Mallick 1
STAT 651
Lecture #15
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Topics in Lecture #15 Some basic probability
The binomial distribution
Inference about a single population proportions
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Book Sections Covered in Lecture #15
Chapters 4.7-4.8
Chapter 10.2
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Lecture 14 Review: Nonparametric Methods
Replace each observation by its rank in the pooled data
Do the usual ANOVA F-test
Kruskal-Wallis
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Lecture 14 Review: Nonparametric Methods
Once you have decided that the populations are different in their means, there is no version of a LSD
You simply have to do each comparison in turn
This is a bit of a pain in SPSS, because you physically must do each 2-population comparison, defining the groups as you go
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Categorical Data
Not all experiments are based on numerical outcomes
We will deal with categorical outcomes, i.e., outcomes that for each individual is a category
The simplest categorical variable is binary:
Success or failure
Male of female
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Categorical Data
For example, consider flipping a fair coin, and let
X = 0 means “tails”
X = 1 means “heads”
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Categorical Data
The fraction of the population who are “successes” will be denoted by the Greek symbol
Note that because it is a Greek symbol, it represents something to do with a population
For coin flipping, if you flipped all the fair coins in the world (the population), the fraction of the times they turn up heads equals
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Categorical Data
The fraction of the population who are “successes” will be denoted by the Greek symbol
The fraction of the sample of size n who are “successes” is going to be denoted by
We want to relate to
Let X = number of successes in the sample. The fraction = (# successes)/n = X / n
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Categorical Data
Suppose you flip a coin 10 times, and get 6 heads.
The proportion of heads = 0.60
The percentage of heads = 60%
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Categorical Data
The number of success X in n experiments each with probability of success is called a binomial random variable
There is a formula for this:
Pr(X = k) =
0! = 1, 1! = 1, 2! = 2 x 1 = 2, 3! = 3 x 2 x 1 = 6, 4! = 4 x 3 x 2 x 1 = 24, etc.
k n kn!Pr( k/ n) (1 )ˆ
k! (n-k)!
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Categorical Data
0! = 1, 1! = 1, 2! = 2 x 1 = 2, 3! = 3 x 2 x 1 = 6, 4! = 4 x 3 x 2 x 1 = 24, etc.
The idea is to relate the sample fraction to the population fraction using this formula
Key Point: if we knew , then we could entirely characterize the fraction of experiments that have k successes
k n kn!Pr(X k) Pr( k/ n) (1 )ˆ
k! (n-k)!
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Categorical Data
The probability that the coin lands on heads will be denoted by the Greek symbol
Suppose you flip a coin 2 times, and count the number of heads.
So here, X = number of heads that arise when you flip a coin 2 times
X takes on the values 0, 1 and 2
takes on the values 0/2, ½, 2/2
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Categorical Data: What the binomial formula does
The experiment results in 4 equally likely outcomes: each occurs ¼ of the time
Tails on toss #1
Heads on toss #1
Tails of toss #2
¼ ¼
Heads on Toss #2
¼ ¼
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Categorical Data
Heads = “success”:
Tails on toss #1
Heads on toss #1
Tails on toss #2
¼ ¼
Heads on Toss #2
¼ ¼
Pr(X 0) Pr( 0/ 2) 1/ 4ˆ Pr(X 1) Pr( 1/ 2) 1/ 2ˆ
Pr(X 2) Pr( 2/ 2) 1/ 4ˆ The binomial formula can be used to give these results without thinking
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Categorical Data
0! = 1, 1! = 1, 2! = 2 x 1 = 2, 3! = 3 x 2 x 1 = 6, 4! = 4 x 3 x 2 x 1 = 24, etc.
n=2, k=1, k! = 1, n! = 2, (n-k)! = 1
The binomial formula gives the answer ½, which we know to be correct
k n kn!Pr(X k) Pr( k/ n) (1 )ˆ
k! (n-k)!
k n k.5, and(1 ) .5
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Categorical Data
Roll a fair dice
1 2 3 4 5 6
First Dice
Every combination is equally likely, so what are the probabilities?
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Categorical Data
Roll a fair dice
1 2 3 4 5 6
1/6 1/6 1/6 1/6 1/6 1/6
First Dice
Every combination is equally likely, so what are the probabilities?
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Copyright (c) Bani K. Mallick 19
Categorical Data
Roll a fair dice
1 2 3 4 5 6
1/6 1/6 1/6 1/6 1/6 1/6
First Dice
Every combination is equally likely, so what are the probabilities?
What is the chance of rolling a 1 or a 2?
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Categorical Data
Roll a fair dice
1 2 3 4 5 6
1/6 1/6 1/6 1/6 1/6 1/6
First Dice
Every combination is equally likely, so what are the probabilities?
What is the chance of rolling a 1 or 2? 2/6 = 1/3
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Categorical Data
Now roll two fair dice
1 2 3 4 5 6
1
2
3
4
5
6
Second Dice
First Dice
Every combination is equally likely, so what are the probabilities?
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Categorical Data
Roll two fair dice
1 2 3 4 5 6
1 1/36 1/36 1/36 1/36 1/36 1/36
2 1/36 1/36 1/36 1/36 1/36 1/36
3 1/36 1/36 1/36 1/36 1/36 1/36
4 1/36 1/36 1/36 1/36 1/36 1/36
5 1/36 1/36 1/36 1/36 1/36 1/36
6 1/36 1/36 1/36 1/36 1/36 1/36
Second Dice
First Dice
Every combination is equally likely, so what are the probabilities?
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Copyright (c) Bani K. Mallick 23
Categorical Data
Roll two fair dice
1 2 3 4 5 6
1 1/36 1/36 1/36 1/36 1/36 1/36
2 1/36 1/36 1/36 1/36 1/36 1/36
3 1/36 1/36 1/36 1/36 1/36 1/36
4 1/36 1/36 1/36 1/36 1/36 1/36
5 1/36 1/36 1/36 1/36 1/36 1/36
6 1/36 1/36 1/36 1/36 1/36 1/36
Second Dice
First Dice
Define a success as rolling a 1 or a 2. What is the chance of two successes?
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Copyright (c) Bani K. Mallick 24
Categorical Data
Roll two fair dice
1 2 3 4 5 6
1 1/36 1/36 1/36 1/36 1/36 1/36
2 1/36 1/36 1/36 1/36 1/36 1/36
3 1/36 1/36 1/36 1/36 1/36 1/36
4 1/36 1/36 1/36 1/36 1/36 1/36
5 1/36 1/36 1/36 1/36 1/36 1/36
6 1/36 1/36 1/36 1/36 1/36 1/36
Second Dice
First Dice
Define a success as rolling a 1 or a 2. What is the chance of two successes? 4/36 = 1/9
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Copyright (c) Bani K. Mallick 25
Categorical Data
Roll two fair dice
1 2 3 4 5 6
1 1/36 1/36 1/36 1/36 1/36 1/36
2 1/36 1/36 1/36 1/36 1/36 1/36
3 1/36 1/36 1/36 1/36 1/36 1/36
4 1/36 1/36 1/36 1/36 1/36 1/36
5 1/36 1/36 1/36 1/36 1/36 1/36
6 1/36 1/36 1/36 1/36 1/36 1/36
Second Dice
First Dice
Define a success as rolling a 1 or a 2. What is the chance of two failures? 16/36 = 4/9
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Categorical Data
So, a success occurs when you roll a 1 or a 2
Pr(success on a single die) = 2/6 = 1/3 =
Pr(2 successes) = 1/3 x 1/3 = 1/9
Use the binomial formula: pr(X=k) when k=2
k!=2, n!=2, (n-k)!=1,
k n k1/ 9,and(1 ) 1
k n kn!Pr(X k) Pr( k/ n) (1 ) 1/ 9ˆ
k! (n-k)!
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Categorical Data
In other words, the binomial formula works in these simple cases, where we can draw nice tables
Now think of rolling 4 dice, and ask the chance the 3 of the 4 times you get a 1 or a 2
Too big a table: need a formula
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Categorical Data
Does it matter what you call as “success” and hat you call a “failure”?
No, as long as you keep track
For example, in a class experiment many years ago, men were asked whether they preferred to wear boxers or briefs
This is binary, because there are only 2 outcomes
“success” = ?????
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Categorical Data
Binary experiments have sampling variability, just like sample means, etc.
Experiment: “success” = being under 5’10” in height
First 6 men with SSN < 5
First 6 men with SSN > 5
Note how the number of “successes” was not the same! (I might have to do this a few times)
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Categorical Data
The sample fraction is a random variable
This means that if I do the experiment over and over, I will get different values.
These different values have a standard deviation.
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Categorical Data
The sample fraction has a standard error
Its standard error is
Note how if you have a bigger sample, the standard error decreases
The standard error is biggest when = 0.50.
ˆ
(1 )n
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Categorical Data
The sample fraction has a standard error
Its standard error is
The estimated standard error based on the sample is
ˆ
(1 )n
ˆ
(1 )ˆ ˆˆ
n
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Categorical Data
It is possible to make confidence intervals for the population fraction if the number of successes > 5, and the number of failures > 5
If this is not satisfied, consult a statistician
Under these conditions, the Central Limit Theorem says that the sample fraction is approximately normally distributed (in repeated experiments)
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Categorical Data
(1100% CI for the population fraction
is by looking up 1 in Table 1
/ 2 ˆzˆ ˆ
ˆ
(1 )ˆ ˆˆ
n
/ 2z
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Categorical Data
Often, you will only know the sample proportion/percentage and the sample size
Computing the confidence interval for the population proportion: two ways By hand
By SPSS (this is a pain if you do not have the data entered already)
Because you may need to do this by hand, I will make you do this.
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Categorical Data
(1100% CI for the population fraction
95% CI, = 1.96
n = 25, = 0.30
/ 2 ˆzˆ ˆ
ˆ
(1 ) .3(1 .3)ˆ ˆ 0.09165ˆn 25
/ 2z
/ 2 ˆz 0.30 1.96x0.09165ˆ ˆ
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Categorical Data
(1100% CI for the population fraction
Interpretation?
/ 2 ˆz 0.30 1.96x0.09165ˆ ˆ
0.30 0.18 [0.12,0.48]
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Categorical Data
(1100% CI for the population fraction
Interpretation? The proportion of successes in the population is from 0.12 to 0.48 (12% to 48%) with 95% confidence
/ 2 ˆz 0.30 1.96x0.09165ˆ ˆ
0.30 0.18 [0.12,0.48]
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Categorical Data
You can use SPSS as long as the number of successes and the number of failures both exceed 5
To get the confidence intervals, you first have to define a numeric version of your variable that classifies whether an observation is a success or failure.
You then compute the 1-sample confidence interval from “descriptives” “Explore”: Demo
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Categorical Data
If you set up your data in SPSS, the “mean” will be the proportion/fraction/percentage of 1’s
Data = 0 1 1 1 0 0 0 1 0 0
n = 10
Mean = 4/10 = .40
= .40
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Boxers versus briefs for males
Case Processing Summary
188 100.0% 0 .0% 188 100.0%Boxers or BriefsPerference
N Percent N Percent N Percent
Valid Missing Total
Cases
In this output, boxers = 1 and briefs = 0
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Boxers versus briefs for males: what % prefer boxers? In the
sample, 46.81%. In the population???
Descriptives
.4681 3.649E-02
.3961
.5401
.4645
.0000
.250
.5003
.00
1.00
1.00
1.0000
.129 .177
-2.005 .353
MeanLower Bound
Upper Bound
95% ConfidenceInterval for Mean
5% Trimmed Mean
Median
Variance
Std. Deviation
Minimum
Maximum
Range
Interquartile Range
Skewness
Kurtosis
Boxers or BriefsPerference
Statistic Std. Error
In this output, boxers = 1 and briefs = 0. The proportionof 1’s is the mean
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Boxers versus briefs for males: what % prefer boxers? Between
39.61% and 54.01%
Descriptives
.4681 3.649E-02.3961
.5401
.4645
.0000
.250.5003
.00
1.001.00
1.0000.129 .177
-2.005 .353
Mean
Lower BoundUpper Bound
95% ConfidenceInterval for Mean
5% Trimmed Mean
MedianVariance
Std. DeviationMinimum
MaximumRange
Interquartile Range
SkewnessKurtosis
GenderMaleNumeric Boxers: 0
= Briefs, 1 = Boxers
Statistic Std. Error
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Boxers versus briefs
In the sample, 46.81% of the men preferred boxers to briefs: 53.19% preferred briefs.
Between 39.61% and 54.01% men prefer boxers to briefs (95% CI)
Is there enough evidence to conclude that men generally prefer briefs?
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Boxers versus briefs
In the sample, 46.81% of the men preferred boxers to briefs: 53.19% preferred briefs.
Between 39.61% and 54.01% men prefer boxers to briefs (95% CI)
Is there enough evidence to conclude that men generally prefer briefs?
No: since 50% is in the CI! This means that it is possible (95%CI) that 50% prefer boxers, 50% prefer briefs, = 0.50.
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Sample Size Calculations
The standard error of the sample fraction is
If you want an (1100% CI interval to be
you should set
ˆ
(1 )n
E
/ 2
(1 )E z
n
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Sample Size Calculations
This means that
/ 2
(1 )E z
n
2/ 2 2
(1 )n z
E
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Sample Size Calculations
The small problem is that you do not know . You have two choices: Make a guess for
Set = 0.50 and calculate (most conservative, since it results in largest sample size)
Most polling operations make the latter choice, since it is most conservative
2/ 2 2
(1 )n z
E
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Copyright (c) Bani K. Mallick 49
Sample Size Calculations: Examples
Set E = 0.04, 95% CI, you guess that = 0.30
You have no good guess:
2/ 2 2
(1 )n z
E
22
.3(1 .3)n 1.96 504
.04
22
.5(1 .5)n 1.96 601
.04