sampling distributions. parameter & statistic parameter summary measure about population sample...

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