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Transcript
```Chapter 10 Sampling and Sampling
Distributions
•
•
•
•
•
10.1 Random sampling
10.4 Stratified sampling
10.6 Sampling distribution
10.7 The standard error of the mean
10.8 The central limit theorem
10.1 Random sampling
• Example: At a parts depot the inventory
shows 1000 parts in stock. What percentage
of those are actually in stock?
– Let p = true %
– To estimate p, take a sample of n parts to check
out in the supply room.
– Then pˆ  sample proportion
– How do we choose the parts to check?
Random sample
• A random sample is an insurance
policy to protect against bias.
• A simple random sample gives each of the
N 
 n  possible sample choices the same
 
chance of being selected.
Random sampling
In other words, for a finite population of N
sample points:
A sample of size n is random if any n
sample points have a probability
1
N

n 

 
to be selected.
To generalize
For an infinite population, or a finite population but
sampling with replacement.
A value is observed according to a probability
distribution.
A random sample of size n consists of observed
values that are independent and have the same
distributions.
10.2 &10.3 Skip
10.4 Stratified sample
• A simple random sample is not always the
best option.
• Consider the following example. We want
to estimate the average number of vireos per
area.
Conifer
Desiduous
Then the population mean and sample mean
are
total red-eyed vireos

total area
total observed red-eyed vireos
x
total observed area
Question: if we use simple random sampling, is x
always a good estimator of  ?
• Reasoning: A simple random sample of
n=10 locations might all end up Conifer. We
would be better off putting n1 samples in
conifer and n2 samples in Desiduous.
Stratified sampling
• A stratified random sample breaks the
population into strata and samples randomly
within each stratum.
• In our previous example:
– Stratum 1 = Deciduous
– Stratum 2 = Conifer
Notations
N1=number of units in stratum 1
N2=number of units in stratum 2
N = N1+N2 = number of units in entire population
n1=number of units sampled from stratum 1
n2=number of units sampled from stratum 2
n = n1+n2 = number of sampled units
Optimal allocation
• How do we decide n1 and n2 (when n is fixed)?
N1=# of possible sampling locations in conifer
N2=# in desiduous
σ1=standard deviation of bird counts in conifer
σ2=standard deviation of bird counts in desiduous
Optimal allocation: (Problem 10.29)
nN1 1
n1 
N1 1  N 2 2
nN 2 2
n2 
N1 1  N 2 2
Sample more in strata, with
--more units (area)
--higher variability
Proportional allocation
• Proportional allocation doesn’t consider σ
and makes the sample number from each
stratum proportional to the size of the
stratum.
N1
n1  n
N1  N 2
N2
n2  n
N1  N 2
Example of proportional allocation
If
N1=100
Then
N2=300
N1
100 1


N1  N 2 400 4
1
n1  40   10
4
3
n2  40   30
4
n=40
A comparison between simple random
sampling and stratified sampling
Example: Population: Weights of rocks
4, 6, 10, 12. So N=4 and =8.
Sample n=2.
Possible results for simple random samples
Sample
4
6
4
10
4
12
6
10
6
12
10 12
Probability
1/6
1/6
1/6
1/6
1/6
1/6
sample mean
5
7
8
8
9
11
Probability
1/6
1/6
2/6
1/6
1/6
Example Continued
• Possible results for a stratified sample. Choose
1 from the smaller rocks and choose 1 from the
larger rocks.
Sample
4 10
4 12
6 10
6 12
Probability
1/4
1/4
1/4
1/4
sample mean
7
8
8
9
Probability
1/4
1/2
1/4
A comparison between simple random
sampling and stratified sampling
2/5
3/5
1/3
2/5
1/5
Series2
Series2
1/5
0
0
0
5
6
7
8
SRS
9
10 11
7
8
9
Stratified
The stratified sample is more likely to be close to
the true population value, 8 here.
10.6 Sampling Distributions
A statistic (e.g. sample mean) from a random
sample or randomized experiment is a random
variable and its probability distribution is a
sampling distribution.
The population distribution of a variable is the
distribution of its values for all members of the
population. The population distribution is also
the probability distribution of the variable when
choosing one subject at random from the
population.
Sampling Variability
Approximating the Sampling Distribution
Results of many random samples of size n=100, from a population where it
is known that 60% of the people hate to shop for clothes, population
proportion (parameter) p = .6) and sample proportion p̂ (statistic).
Most samples of size n = 100 gave estimates close to .6,
but some were far off. On average, they centered on .6,
they are variable, but unbiased.
Sampling Variability (cont.)
Approximating the Sampling Distribution
Results of many random samples of size n=2500, from the same
population, with population proportion (parameter) p = .6) and sample
proportion p̂ (statistic).
Larger samples are more precise (have less variability)
AND are unbiased.
Sample size
What advantage is there of taking a larger sample? Larger n?
Taking a larger sample decreases the potential deviation of
away from .
Let  x be the standard deviation of the sampling
distribution of x , then the larger the sample size is,
the smaller  x is.
x
Unbiased estimators
• P.77 “Estimators having the desirable property that
their values will on average equal the quantity they
are supposed to estimate are said to be unbiased.”
• If  x   then x is an unbiased estimator of .
• Another example of unbiased estimator is s2 for 2.
• Choosing non-random samples can introduce bias.
10.7 The Standard Error of the Mean
If x is the mean of a sample of size n from a
population having mean  and standard deviation
. The mean and standard deviation of x are:
x  
 x 

n

(standard error of the mean)
N n
for finite populations
or  x 
n N 1
So, as n increases,  x decreases. If the sample size is
multiplied by 4, the “standard deviation” (standard error of
the mean) will be divided in half for infinite populations.
Standard Deviation
vs.
Standard Error of the Mean
• The standard error of the mean is smaller than the
standard deviation by a factor of the square root of the
sample size.
• The standard deviation describes the variability of
individuals.
• The standard error of the mean describes the
accuracy of means of a given size or the potential error
in sample mean as a guess at .
Examples
• Population size N=4, sample size n=2,
110
150
112
152
  131
  20.025
What is the standard error of the mean?
20.025 4  2
x 
 11.56
4 1
2
Examples
• Infinite Population
P(3)=P(5)=
P(7)=P(9)=
1/4
 6
 5
2
What is the standard error of the mean?
x 
5
 1.58
2
Example
• Governor’s Poll
Estimate
guess
1, 1, 1
0, 1, 0
0, …
x  observed fraction of 1' s  p̂
What is the standard error of the mean?
  fraction of 1s  p
 2  (0   ) 2 P(0)  (1   ) 2 P(1)
 (0  p ) 2 (1  p )  (1  p) 2 p
 p 2 (1  p)[ p  1  p]
 p (1  p )
 x   pˆ 
Related to variance of a binomial x
p(1  p)
n
If a population has the N(,) distribution, then the sample mean x
 distribution.
of n independent observations has the N ( ,
)
n
10.8 Central Limit Theorem
For ANY population with mean  and standard deviation  the
sample mean x of a random sample of size n is approximately
N ( ,

n
) when n is LARGE.
Central Limit Theorem
If the population is normal or sample size is large,
sample mean x follows a normal distribution
N ( ,
and
z

n
)
x  x
x

x

n
follows a standard normal distribution.
• The closer x’s distribution is to a normal
distribution, the smaller n can be and have
the sample mean nearly normal.
• If x is normal, then the sample mean is
normal for any n, even n=1.
Central Limit Theorem At Work
n=1
n=2
n=10
n=25
• Usually n=30 is big enough so that the sample
mean is approximately normal unless the
distribution of x is very asymmetrical.
• If x is not normal, there are often better procedures
than using a normal approximation, but we won’t
study those options.
Example
• X=ball bearing diameter
• X is normal distributed with =1.0cm and =0.02cm
• x =mean diameter of n=25
• Find out what is the probability that x will be off by
less than 0.01 from the true population mean.
 x  1.0

0.02
x 

 0.004
n
25
0.99  1.0
1.01  1.0
P(0.99  x  1.01)  P(
z
)
0.004
0.004
 P(2.5  z  2.5)  98.76%
Exercise
The mean of a random sample of size
n=100 is going to used to estimate the mean
daily milk production of a very large herd of
dairy cows. Given that the standard
deviation of the population to be sampled is
s=3.6 quarts, what can we assert about he
probabilities that the error of this estimate
will be more then 0.72 quart?
```
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