Sampling Distributions - PowerPoint Presentation

Slide1

Sampling DistributionsSlide2

Parameter & Statistic

Parameter

Summary measure about populationSample StatisticSummary measure about sample

P

in Population & ParameterS in Sample & StatisticSlide3

Common Statistics & Parameters

Sample Statistic

Population Parameter

Variance

S

22

Standard

Deviation

S

MeanXBinomial Proportionp

p

^Slide4

Theoretical

probability distribution

Random variable is sample statistic Sample mean, sample proportion, etc.

Results from drawing

all possible samples of a fixed size

List of all possible [x, p(x)] pairsSampling distribution of the sample meanSampling DistributionSlide5

Sampling from

Normal PopulationsSlide6

定理

1

平均數

變異數Slide7

定理

2Slide8

定理：

Let Y

1,Y2

,…,Y

n be a random sample of size n from a normal distribution with mean μand varianceσ

2. Thenis normally distribution with mean And variance Slide9

Proof:Slide10
Slide11
Slide12

Properties of the Sampling Distribution of

xSlide13

3.

Formula (sampling with replacement)

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 MeanSlide14

Central Tendency

Dispersion

Sampling with replacement

m

= 50

s

= 10

X

n =16





X

= 2.5

n = 4X = 5

m

X = 50

-

X

Sampling Distribution

Population Distribution

Sampling from Normal PopulationsSlide15

Standardizing the Sampling Distribution of

x

Standardized Normal Distribution

m

= 0

s

= 1

Z

Sampling

Distribution

X

m

X

s

XSlide16

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 ChallengeSlide17

Sampling

Distribution

8

s

`

X

= .4

7.8

8.2

`

X

0

s

= 1

–.50

Z

.50

.3830

Standardized Normal Distribution

.1915

.1915

Sampling Distribution Solution*Slide18

Sampling from

Non-Normal PopulationsSlide19

Population size,

N

= 4

Random variable,

x

Values of x: 1, 2, 3, 4Uniform distribution© 1984-1994 T/Maker Co.

Suppose There’s a Population ...

Developing Sampling DistributionsSlide20

Population Distribution

Summary Measures

.0

.1

.2

.3

1

2

3

4

P(

x

)

x

Population CharacteristicsSlide21

Sample with replacement

1.0

1.5

2.0

2.5

1.52.02.5

3.0

2.0

2.5

3.03.52.53.03.54.016 Samples1st

Obs

1,1

1,21,3

1,42,12,22,3

2,43,13,2

3,33,4

4,14,2

4,34,42nd Observation

12

3

4

1

2

3

4

2nd Observation

1

2

34

12

34

1st

Obs

16 Sample Means

All Possible Samples of Size

n = 2Slide22

1.0

1.5

2.0

2.5

1.5

2.02.53.02.0

2.5

3.0

3.5

2.53.03.54.02nd Observation123

4

1

2

34

1st

Obs

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

4.0P(x)x

Sampling Distribution of All Sample MeansSlide23

Summary Measures of All Sample MeansSlide24

Population

Sampling Distribution

.0

.1

.2

.3

1

2

3

4

.0

.1

.2

.3

1.0

1.5

2.0

2.5

3.0

3.5

4.0

P(

x

)

x

P(

x

)xComparisonSlide25

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.

25x

1

2

3

456P(x)1/6

1/6

1/6

1/6

1/61/6Sampling Distribution of the Mean…Slide26

Sampling Distribution of Two Dice

A sampling distribution is created by looking at

all 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.

26Slide27

9.

27

9.

27

P( )

6/36

5/36

4/36

3/36

2/36

1/36

P( )

Sampling Distribution of Two Dice…Slide28

9.

28

Compare…

Compare the distribution of X…

…with the sampling distribution of .

As well, note that:Slide29

Sampling from

Non-Normal PopulationsSlide30

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.Slide31
Slide32

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.Slide33

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.Slide34
Slide35
Slide36

Central Tendency

Dispersion

Sampling with replacement

Population Distribution

Sampling Distribution

n =30X = 1.8n = 4X = 5

m

= 50

s = 10

X

m

X

= 50

-

X

Sampling from Non-Normal PopulationsSlide37

X

As sample size gets

large

enough

(n



30) ...

sampling distribution becomes almost normal.

Central Limit TheoremSlide38

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?

SODASlide39

Sampling

Distribution

12

s

`

X

= .03

11.95

`

X

0

s

= 1

–1.77

Z

.0384

Standardized Normal Distribution

.4616

Shaded area exaggerated

Central Limit Theorem Solution*Slide40

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. Slide41

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 deviation

9.

41ExampleSlide42

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. Slide43

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) = .95

Substituting the formula for the sampling distributionWith a little algebraSlide44

9.

44

Using the Sampling Distribution for Inference

Returning to the chapter-opening example where µ = 800,

σ

= 100, and n = 25, we computeorThis 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.Slide45

9.

45

Using the Sampling Distribution for Inference

For example, with µ = 800,

σ

= 100, n = 25 and α= .01, we produceSlide46

Sampling Distributions The Proportion

The proportion of the population having some characteristic is denoted

π

.Slide47

Standard error for the proportion:

Z value for the proportion:

Sampling Distributions The ProportionSlide48

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 isP(.40 ≤ p ≤ .45) ?Sampling Distributions The Proportion: ExampleSlide49

Sampling Distributions The Proportion: Example

Find :

Convert to standardized normal: Slide50

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

0

p