Date post: | 29-Mar-2015 |
Category: |
Documents |
Upload: | tony-pember |
View: | 213 times |
Download: | 0 times |
Sampling Distributions
Parameter & Statistic
Parameter• Summary measure
about population
Sample Statistic• Summary measure
about sample
• P in Population & Parameter
• S in Sample & Statistic
Common Statistics & Parameters
Sample Statistic Population Parameter
Variance S2 2
StandardDeviation S
Mean X
Binomial Proportion
pp̂
1. Theoretical probability distribution
2. Random variable is sample statistic• Sample mean, sample proportion,
etc.
3. Results from drawing all possible samples of a fixed size
4. List of all possible [x, p(x)] pairs•Sampling distribution of the sample
mean
Sampling Distribution
Sampling from Normal Populations
定理 1
平均數
變異數
),(~ , , ),(~ 222 bbaNYthenbxaYNxif 若
( ) ( ) ( ) ( ) ( )E Y E a bx E a E bx a bE x a b
2 2 2( ) ( ) ( ) ( ) 0 ( )V Y V a bx V a V bx b V x b
定理 2
2 2 X ~ ( , ) , ~ ( , ) , X and Y are independentx x Y Yif N Y N
2 2 2 2 , then ~ ( , )X Y X Yif w aX bY w N a b a b
定理: Let Y1,Y2,…,Yn be a random sample of size n from a normal distribution with mean μand varianceσ2. Then
is normally distribution with mean And variance
n
iiYn
Y1
1
Y 2 2 /Y n
Proof:
niYVYE ii ,...,2,1for )( and )( 2
1 21
1 1 1 1( ) ( ) ( )
n
i ni
Y Y Y Y Yn n n n
1 1 2 2
where 1/ , 1, 2,...,n n
i
a Y a Y a Y
a n i n
1 2
1 1 1( ) ( ) ( ) ( )
1 1 1 ( ) ( ) ( )
nE Y E Y Y Yn n n
n n n
1 2
2 2 22 2 2
1 1 1( ) ( ) ( ) ( )
1 1 1 ( ) ( ) ( )
nV Y V Y Y Yn n n
n n n
22
2
1( )n
n n
Properties of the Sampling Distribution of x
xn
3. Formula (sampling with replacement)
2. Less than population standard deviation
1. Standard deviation of all possible sample means, x
● Measures scatter in all sample means, x
Standard Error of the Mean
Central Tendency
Dispersion
Sampling with replacement
m = 50
s = 10
X
n =16
X = 2.5
n = 4
X = 5
mX = 50- X
Sampling Distribution
Population Distributionx
xn
Sampling from Normal Populations
Standardizing the Sampling Distribution of x
Standardized Normal Distribution
m = 0
s = 1
Z
x
x
X XZ
n
Sampling Distribution
XmX
sX
You’re an operations analyst for AT&T. Long-distance telephone calls are normally distribution with = 8 min. and = 2 min. If you select random samples of 25 calls, what percentage of the sample means would be between 7.8 & 8.2 minutes?
© 1984-1994 T/Maker Co.
Thinking Challenge
Sampling Distribution
8
s `X = .4
7.8 8.2 `X 0
s = 1
–.50 Z.50
.3830
Standardized Normal Distribution
.1915.1915
7.8 8.50
225
8.2 8.50
225
XZ
n
XZ
n
Sampling Distribution Solution*
Sampling from Non-Normal Populations
Population size, N = 4
Random variable, x
Values of x: 1, 2, 3, 4
Uniform distribution
© 1984-1994 T/Maker Co.
Suppose There’s a Population ...
Developing Sampling Distributions
1 2.5
N
ii
X
N
2
1 1.12
N
ii
X
N
Population Distribution
Summary Measures
.0
.1
.2
.3
1 2 3 4
P(x)
x
Population Characteristics
Sample with replacement
1.0 1.5 2.0 2.5
1.5 2.0 2.5 3.0
2.0 2.5 3.0 3.5
2.5 3.0 3.5 4.0
16 Samples
1stObs
1,1 1,2 1,3 1,4
2,1 2,2 2,3 2,4
3,1 3,2 3,3 3,4
4,1 4,2 4,3 4,4
2nd Observation1 2 3 4
1
2
3
4
2nd Observation1 2 3 4
1
2
3
4
1stObs
16 Sample Means
All Possible Samples of Size n = 2
1.0 1.5 2.0 2.5
1.5 2.0 2.5 3.0
2.0 2.5 3.0 3.5
2.5 3.0 3.5 4.0
2nd Observation1 2 3 4
1
2
3
4
1stObs
16 Sample Means Sampling Distribution of the
Sample Mean
.0
.1
.2
.3
1.0 1.5 2.0 2.5 3.0 3.5 4.0
P(x)
x
Sampling Distribution of All Sample Means
1 1.0 1.5 ... 4.02.5
16
N
ii
X
X
N
2
1
N
i Xi
X
X
N
2 2 2(1.0 2.5) (1.5 2.5) ... (4.0 2.5).79
16
Summary Measures of All Sample Means
Population Sampling Distribution
2.5x .79x
.0
.1
.2
.3
1 2 3 4
2.5 1.12
.0
.1
.2
.3
1.0 1.5 2.0 2.5 3.0 3.5 4.0
P(x)
x
P(x)
x
Comparison
A fair die is thrown infinitely many times,with the random variable X = # of spots on any throw.
The probability distribution of X is:
…and the mean and variance are calculated as well:
9.25
x 1 2 3 4 5 6P(x
)1/6 1/6 1/6 1/6 1/6 1/6
Sampling Distribution of the Mean…
Sampling Distribution of Two Dice
A sampling distribution is created by looking atall samples of size n=2 (i.e. two dice) and their means…
While there are 36 possible samples of size 2, there are only 11 values for , and some (e.g. =3.5) occur more frequently than others (e.g. =1). 9.26
9.279.27
1.0 1/361.5 2/362.0 3/362.5 4/363.0 5/363.5 6/364.0 5/364.5 4/365.0 3/365.5 2/366.0 1/36
P( )
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
6/36
5/36
4/36
3/36
2/36
1/36
P(
)
Sampling Distribution of Two Dice…
9.28
Compare…
Compare the distribution of X…
…with the sampling distribution of .
As well, note that:
1 2 3 4 5 6 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0
Sampling from Non-Normal Populations
Law of Large Numbers
The law of large numbers states that, under general conditions, will be near with very high probability when n is large.
The conditions for the law of large numbers are Yi , i=1, …, n, are i.i.d. The variance of Yi , , is finite.
9.32
Central Limit Theorem…
The sampling distribution of the mean of a random sample drawn from any population is approximately normal for a sufficiently large sample size.
The larger the sample size, the more closely the sampling distribution of X will resemble a normal distribution.
9.33
Central Limit Theorem…
If the population is normal, then X is normally distributed for all values of n.
If the population is non-normal, then X is approximately normal only for larger values of n.
In most practical situations, a sample size of 30 may be sufficiently large to allow us to use the normal distribution as an approximation for the sampling distribution of X.
Central Tendency
Dispersion
Sampling with replacement
Population Distribution
Sampling Distributionn =30
X = 1.8n = 4
X = 5
m = 50
s = 10
X
mX = 50- X
x
xn
Sampling from Non-Normal Populations
X
As sample size gets large enough (n 30) ...
sampling distribution becomes almost normal.
x
xn
Central Limit Theorem
Central Limit Theorem Example
The amount of soda in cans of a particular brand has a mean of 12 oz and a standard deviation of .2 oz. If you select random samples of 50 cans, what percentage of the sample means would be less than 11.95 oz?
SODA
Sampling Distribution
12
s `X = .03
11.95 `X 0
s = 1
–1.77 Z
.0384
Standardized Normal
Distribution
.4616
11.95 121.77
.250
XZ
n
Shaded area exaggerated
Central Limit Theorem Solution*
9.40
Example
One survey interviewed 25 people who graduated one year ago and determines their weekly salary.
The sample mean to be $750.
To interpret the finding one needs to calculate the probability that a sample of 25 graduates would have a mean of $750 or less when the population mean is $800 and the standard deviation is $100.
After calculating the probability, he needs to draw some conclusions.
We want to find the probability that the sample mean is less than $750. Thus, we seek
The distribution of X, the weekly income, is likely to be positively skewed, but not sufficiently so to make the distribution of nonnormal. As a result, we may assume that is normal with mean
and standard deviation9.41
Example
)750X(P
X
800x
2025/100n/x
X
9.42
Example
Thus,
The probability of observing a sample mean as low as $750 when the population mean is $800 is extremely small. Because this event is quite unlikely, we would have to conclude that the dean's claim is not justified.
0062.
4938.5.
)5.2Z(P
20
800750XP
)750X(P
x
x
9.43
Using the Sampling Distribution for Inference
Here’s another way of expressing the probability calculated from a sampling distribution.
P(-1.96 < Z < 1.96) = .95Substituting the formula for the sampling distribution
With a little algebra
95.)96.1n/
X96.1(P
95.)n
96.1Xn
96.1(P
9.44
Using the Sampling Distribution for Inference
Returning to the chapter-opening example where µ = 800, σ = 100, and n = 25, we compute
or
This tells us that there is a 95% probability that a sample mean will fall between 760.8 and 839.2. Because the sample mean was computed to be $750, we would have to conclude that the dean's claim is not supported by the statistic.
95.)25
10096.1800X
25
10096.1800(P
95.)2.839X8.760(P
9.45
Using the Sampling Distribution for Inference
For example, with µ = 800, σ = 100, n = 25 and α= .01, we produce
01.1)n
zXn
z(P 005.005.
99.)25
100575.2800X
25
100575.2800(P
99.)5.851X5.748(P
Sampling Distributions The Proportion
size sample
interest of sticcharacteri thehaving sample in the ofnumber
n
X itemsp
The proportion of the population having some characteristic is denoted π.
Standard error for the proportion:
Z value for the proportion:
Sampling Distributions The Proportion
n
)(1σp
n
)(1σZ
p
pp
If the true proportion of voters who support Proposition A is π = .4, what is the probability that a sample of size 200 yields a sample proportion between .40 and .45?
In other words, if π = .4 and n = 200, what is
P(.40 ≤ p ≤ .45) ?
Sampling Distributions The Proportion: Example
Sampling Distributions The Proportion: Example
.03464200
.4).4(1
n
)(1σ
p
1.44)ZP(0
.03464
.40.45Z
.03464
.40.40P.45)P(.40
p
Find :
Convert to standardized normal:
pσ
Sampling Distributions The Proportion: Example
Use cumulative normal table:
P(0 ≤ Z ≤ 1.44) = P(Z ≤ 1.44) – 0.5 = .4251
Z.45
1.44
.4251
Standardize
Sampling Distribution
Standardized Normal
Distribution
.40
0p