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Binomial model (cont’d)
Introduction to Statistics
Lecture 16
„
For n random Bernoulli trials, X is the
number of successes. If the probability of
success is p, q=1-p is the probability of
failure,
⎛n⎞
P ( X = k ) = ⎜ ⎟ p k q n −k
⎝k ⎠
"n choose k"
Reminder: Quiz 3 next week.
⎛n⎞
n!
⎜ ⎟=
!(
k
k
n
− k )!
⎝ ⎠
5! = 5 × 4 × 3 × 2 × 1
Mean: np
Standard deviation: np(1 − p )
0! = 1
Example
„
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Normal approximation of binomial
distribution
A large pool of candies. 30% red.
We randomly sampled 10, what is the
probability 5 of them are red?
Binom(100, 0.3)
Binom(10, 0.3)
Binom(100, 0.3)
Binom(10, 0.3)
25
8
20
„
What is the probability that fewer than 5 of
them are red?
Percent of Total
Percent of Total
6
15
10
4
2
5
0
0
0
2
4
6
rbinom(50000, 10, 0.3)
8
10
0
20
40
60
80
100
rbinom(50000, 100, 0.3)
When n is large, so that (np>10) and (nq>10), the binomial
distribution Binom(n, p) can be approximated by N (np, np(1 − p ))
Calculation steps
Normal approximation of Binomial model
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Check Bernoulli trials?
‰ Success/failure
‰ Probability of success
‰ Independence
Find n—number of trials
Find p—probability of success for each trial
Write down the distribution Binom(n, p)
Check Normal approximation rules:
‰ np>10
‰ n(1-p)>10
Find mean and standard deviation of Binom(n, p)
Use normal distribution with the same mean and standard
deviation to carry out the calculation.
What if p is too small that np<10 even
when n is large
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Poisson distribution can be used to
approximate the binomial distribution when
‰
‰
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n is large
np is small
where λ=np.
P( X = k ) = e−λ
λk
k!
1
Poisson approximation of the
binomial distribution
Binom(10000, 0.0001)
Poisson with rate 1
Why study binomial distribution and its
approximations
„
Consider the following scenario:
0.3
0.3
‰
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In a large population, 60% of the people support
the current mayor.
You random sample one individual
0.2
0.2
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0.1
A survey of 100 people, X is the number of people
that supports the mayor in this survey.
0.0
0.0
0.1
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‰
Success: he/she supports the mayor
Failure: he/she doesn’t support the mayor
Probability of success = 60% ?
0 1 2 3 4 5
0 1 2 3 4 5
Different samples out of a population
Why probability models are
important? Or why binomial
distribution is important?
We are starting chapters on statistical
inference NOW.
„
„
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Given a simple random sample with n observations
of a Bernoulli trial
X: the number of success
p: probability of success
Sample proportion:
X
pˆ =
Sample
2
(a,b,c,d)
1
Green
Red
2
Red
Green
3
Green
Green
4
Red
Green
c
d
2
3
b
4
a
Sampling distribution models
of the sample proportion
Sample proportion
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Sample
1
(1,2,3,4)
1
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‰
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n
Sample proportion is a random variable since X
is a random variable that has Binomial model.
Sampling variation: the p-hat calculated based on
different random samples from the sample
population differs due to chance.
Parameters:
„
The sample size, n
The probability of success: long-term relative frequency
(proportion) of success in the population—population
proportion, p
We can study the variation of sample proportion
using some probability model under some
assumptions—sampling distribution model.
2
By the normal approximation …
X is approximately normally distributed with
General situation …
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mean np and standard deviation np (1 − p )
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Then,
„
pˆ =
X
is approximately normally distributed with
n
mean p and standard deviation
p (1- p )
.
n
N: population size
n: sample size
To use normal approximation of binomial
model to study the random behavior of the
sample proportion, we need
n < N/10
np > 10 AND nq=n(1-p) > 10
‰
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Sampling distribution model of sample proportion!
Example
We just discussed about
proportion, what about mean,
the average?
As historically studied,15% of faculty
members leave campus during spring break.
For a (simple) random sample of 100 faculty
members, what is the chance that more than
20% of them left campus during the past
spring break?
100
-1
0
1
0
50
0
-2
2
-0.5
0.0
0.5
-0.2
100
6
8
10
1
2
3
4
5
1.4
4
2.0
2.2
2.4
2.6
-2
0
2
sample mean of 4 obs
4
1.9
2.0
2.1
2.2
150
100
300
50
200
0
100
0
-4
1.8
sample mean of 1000 obs
400
300
200
100
2
1.8
sample mean of 100 obs
0
0
population
1.6
sample mean of 10 obs
0.12
-2
0.2
0
0
0
population
-4
0.1
50
50
50 100
0
4
0.0
100 150 200 250
200
150
0.4
0.3
2
-0.1
sample mean of 1000 obs
200
sample mean of 100 obs
300
0.5
sample mean of 10 obs
0.2
density curve
50 100 150 200 250 300
150
300
200
4
0.1
0
density curve
2
2
0.0
1 n
1
X = ∑ X i so µ X = ( µ + ... + µ ) = µ
n i =1
n
σ2
⎛1⎞
σ X2 = ⎜ ⎟ (σ 2 + ... + σ 2 ) =
n
⎝n⎠
0
population
0.11
1
-2
0.10
1
50 100
-4
independent, and have the same probability distribution
µ X = µ σ X2 = σ 2
0.15
0
X 1 ,… , X n
n observations:
0.10
0.05
density curve
0.20
Central limit theorem
0.09
Mean and Variance
of Sample mean
Let’s see an online demo first.
0.08
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-4
-2
0
2
sample mean of 2 obs
4
-3
-2
-1
0
1
2
3
sample mean of 10 obs
3
Sampling distribution model of
Sample mean
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If the population distribution is
X ∼ N (µ ,
„
n
N (µ , σ )
σ
n
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)
If the population distribution is not normal and with
mean µ and standard deviation σ
X ∼ N (µ ,
„
σ
In practice
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Standard error: estimated standard deviation
of a sampling distribution.
Distinguish the sampling distribution and the
distribution of a sample.
) approximately when n is large.
How large is large?
Reading
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Chapter 18
4