Statistics for Economics · ECON 509, by Dr. M. Zainal Ch. 6-14. Ch. 6-15 The Standard Normal Table The Standard Normal table in the textbook gives the probability from the mean (zero)

Statistics for Economics

Dr. Mohammad Zainal

Chapter 6Sampling and

Sampling Distributions

ECON 509

Department of Economics

Chapter Goals

After completing this chapter, you should be able to:

Describe a simple random sample and why sampling is

important

Explain the difference between descriptive and

inferential statistics

Define the concept of a sampling distribution

Determine the mean and standard deviation for the

sampling distribution of the sample mean,

Describe the Central Limit Theorem and its importance

Determine the mean and standard deviation for the

sampling distribution of the sample proportion,

Describe sampling distributions of sample variances

p̂

X

ECON 509, by Dr. M. Zainal Chap 6-2

The Normal Distribution

Descriptive statistics

Collecting, presenting, and describing data

Inferential statistics

Drawing conclusions and/or making decisions

concerning a population based only on

sample data

6.0


Probability Distributions

Continuous

Probability

Distributions

Binomial

Hypergeometric

Poisson

Probability

Distributions

Discrete

Probability

Distributions

Normal

Uniform

Exponential

.

.

.

.


Ch. 6-5

Continuous Probability Distributions

A continuous random variable is a variable that

can assume any value on a continuum (can

assume an uncountable number of values)

thickness of an item

time required to complete a task

temperature of a solution

height, in inches

These can potentially take on any value,

depending only on the ability to measure

accurately.

ECON 509, by Dr. M. Zainal

Ch. 6-6

The Normal Distribution

‘Bell Shaped’

Symmetrical

Mean, Median and Modeare Equal

Location is determined by the mean, μ

Spread is determined by the standard deviation, σ

The random variable has an infinite theoretical range: + to

Mean

= Median

= Mode

x

f(x)

μ

σ


Ch. 6-7

The Normal Distribution Shape

x

f(x)

μ

σ

Changing μ shifts the

distribution left or right.

Changing σ increases

or decreases the

spread.


Ch. 6-8

The Normal Distribution Shape

By varying the parameters μ and σ, we obtain

different normal distributions


Ch. 6-9

Finding Normal Probabilities

a b x

f(x) P a x b( )

Probability is measured by the area

under the curve


ECON 509, by Dr. M. Zainal Ch. 6-10

f(x)

xμ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is

symmetric, so half is above the mean, half is below

1.0)xP(

0.5)xP(μ 0.5μ)xP(

Ch. 6-11

The Standard Normal Distribution

Also known as the “z” distribution

Mean is defined to be 0

Standard Deviation is 1

z

f(z)

0

1

Values above the mean have positive z-values,

values below the mean have negative z-values


Ch. 6-12

The Standard Normal

Any normal distribution (with any mean and standard deviation combination) can be transformed into the standard normaldistribution (z)

Need to transform x units into z units

σ

μxz


Ch. 6-13

Example

If x is distributed normally with mean of 100

and standard deviation of 50, the z value for

x = 250 is

This says that x = 250 is three standard

deviations (3 increments of 50 units) above

the mean of 100.

3.050

100250

σ

μxz


Comparing x and z units

z

100

3.00

250 x

Note that the distribution is the same, only the

scale has changed. We can express the problem in

original units (x) or in standardized units (z)

μ = 100

σ = 50


Ch. 6-15

The Standard Normal Table

The Standard Normal table in the textbook

gives the probability from the mean (zero)

up to a desired value for z

z0 2.00

.4772Example:

P(0 < z < 2.00) = .4772


The Standard Normal Table

The value within the

table gives the

probability from z = 0

up to the desired z

value

z 0.00 0.01 0.02 …

0.1

0.2

.4772

2.0P(0 < z < 2.00) = .4772

The row shows

the value of z

to the first

decimal point

The column gives the value of

z to the second decimal point

2.0

.

.

.

(continued)


General Procedure for Finding Probabilities

Draw the normal curve for the problem in

terms of x

Translate x-values to z-values

Use the Standard Normal Table

To find P(a < x < b) when x is distributed

normally:


Z Table example

Suppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 8.6)

P(8 < x < 8.6)

= P(0 < z < 0.12)

Z0.120

x8.68

05

88

σ

μxz

0.125

88.6

σ

μxz

Calculate z-values:


Z Table example

Suppose x is normal with mean 8.0 and

standard deviation 5.0. Find P(8 < x < 8.6)

P(0 < z < 0.12)

z0.120x8.68

P(8 < x < 8.6)

= 8

= 5

= 0

= 1

(continued)



Z

0.12

z .00 .01

0.0 .0000 .0040 .0080

.0398 .0438

0.2 .0793 .0832 .0871

0.3 .1179 .1217 .1255

Solution: Finding P(0 < z < 0.12)

.0478.02

0.1 .0478

Standard Normal Probability

Table (Portion)

0.00

= P(0 < z < 0.12)

P(8 < x < 8.6)



Suppose x is normal with mean 8.0

and standard deviation 5.0.

Now Find P(x < 8.6)

Z

8.6

8.0

Chap 6-21





Now Find P(x < 8.6)

(continued)

Z

0.12

.0478

0.00

.5000P(x < 8.6)

= P(z < 0.12)

= P(z < 0) + P(0 < z < 0.12)

= .5 + .0478 = .5478

Chap 6-22


Upper Tail Probabilities



Now Find P(x > 8.6)

Z

8.6

8.0

Chap 6-23


Now Find P(x > 8.6)…

(continued)

Z

0.12

0Z

0.12

.0478

0

.5000 .50 - .0478

= .4522

P(x > 8.6) = P(z > 0.12) = P(z > 0) - P(0 < z < 0.12)

= .5 - .0478 = .4522

Upper Tail Probabilities

Chap 6-24


Lower Tail Probabilities



Now Find P(7.4 < x < 8)

Z

7.48.0


Lower Tail Probabilities

Now Find P(7.4 < x < 8)…

Z

7.48.0

The Normal distribution is

symmetric, so we use the

same table even if z-values

are negative:

P(7.4 < x < 8)

= P(-0.12 < z < 0)

= .0478

(continued)

.0478

Chap 6-26


Z Table example

Example: Find the following areas under the

standard normal curve.

a) P(0 < z < 5.65)

b) P( z < - 5.3)

Chap 6-27

Z Table example

Example: The lifetime of a calculator

manufactured by a company has a normal

distribution with a mean of 54 months and a

standard deviation of 8 months. The company

guarantees that any calculator that starts

malfunctioning within 36 months of the purchase

will be replaced by a new one. What percentage

of such calculators are expected to be

replaced?ECON 509, by Dr. M. Zainal Chap 6-28

Determining the z and x values

We reverse the procedure of finding the area

under the normal curve for a specific value of z

or x to finding a specific value of z or x for a

known area under the normal curve.

z 0.00 0.01 0.02 …

0.1

0.2

.47722.0

.

.

.


Determining the z and x values

Example: Find a point z such that the area

under the standard normal curve between 0 and

z is .4251 and the value of z is positive


Finding x for a normal dist.

To find an x value when an area under a normal

distribution curve is given, we do the following

▪ Find the z value corresponding to that x value

from the standard normal curve.

▪ Transform the z value to x by substituting the

values of , , and z in the following formula

x z


Finding x for a normal dist.

Example: Recall the calculators example, it is

known that the life of a calculator manufactured

by a factory has a normal distribution with a

mean of 54 months and a standard deviation of

8 months. What should the warranty period be

to replace a malfunctioning calculator if the

company does not want to replace more than

0.5 % of all the calculators sold?


Tools of Business Statistics

Descriptive statistics

Collecting, presenting, and describing data

Inferential statistics

Drawing conclusions and/or making decisions

concerning a population based only on

sample data

6.1


Populations and Samples

A Population is the set of all items or individuals of interest

Examples: All likely voters in the next election

All parts produced today

All sales receipts for November

A Sample is a subset of the population

Examples: 1000 voters selected at random for interview

A few parts selected for destructive testing

Random receipts selected for audit

Ch. 6-34ECON 509, by Dr. M. Zainal

Population vs. Sample

Ch. 6-35

a b c d

ef gh i jk l m n

o p q rs t u v w

x y z

Population Sample

b c

g i n

o r u

y


Why Sample?

Less time consuming than a census

Less costly to administer than a census

It is possible to obtain statistical results of a

sufficiently high precision based on samples.


Simple Random Samples

Every object in the population has an equal chance of

being selected

Objects are selected independently

Samples can be obtained from a table of random

numbers or computer random number generators

A simple random sample is the ideal against which

other sample methods are compared


Inferential Statistics

Making statements about a population by

examining sample results

Sample statistics Population parameters

(known) Inference (unknown, but can

be estimated from

sample evidence)

Ch. 6-38

SamplePopulation


Inferential Statistics

Estimation

e.g., Estimate the population mean

weight using the sample mean

weight

Hypothesis Testing

e.g., Use sample evidence to test

the claim that the population mean

weight is 120 pounds

Ch. 6-39

Drawing conclusions and/or making decisions concerning a population based on sample results.


Sampling Distributions

A sampling distribution is a distribution of

all of the possible values of a statistic for

a given size sample selected from a

population

Ch. 6-40

6.2


Chapter Outline

Ch. 6-41

Sampling

Distributions

Sampling

Distribution of

Sample

Mean

Sampling

Distribution of

Sample

Proportion

Sampling

Distribution of

Sample

Variance


Sampling Distributions ofSample Means

Ch. 6-42

Sampling

Distributions

Sampling

Distribution of

Sample

Mean

Sampling

Distribution of

Sample

Proportion

Sampling

Distribution of

Sample

Variance


Developing a Sampling Distribution

Assume there is a population …

Population size N=4

Random variable, X,

is age of individuals

Values of X:

18, 20, 22, 24 (years)

Ch. 6-43

A B CD



Ch. 6-44

.25

018 20 22 24

A B C D

Uniform Distribution

P(x)

x

(continued)

Summary Measures for the Population Distribution:

214

24222018

N

Xμ i

2.236N

μ)(Xσ

2

i


Now consider all possible samples of size n = 2

Ch. 6-45

1st 2nd Observation Obs 18 20 22 24

18 18,18 18,20 18,22 18,24

20 20,18 20,20 20,22 20,24

22 22,18 22,20 22,22 22,24

24 24,18 24,20 24,22 24,24

16 possible samples

(sampling with

replacement)

1st 2nd Observation

Obs 18 20 22 24

18 18 19 20 21

20 19 20 21 22

22 20 21 22 23

24 21 22 23 24

(continued)


16 Sample

Means


Sampling Distribution of All Sample Means

Ch. 6-46

1st 2nd Observation

Obs 18 20 22 24

18 18 19 20 21

20 19 20 21 22

22 20 21 22 23

24 21 22 23 24

18 19 20 21 22 23 240

.1

.2

.3

P(X)

X

Sample Means

Distribution16 Sample Means

_


(continued)

(no longer uniform)

_


Summary Measures of this Sampling Distribution:

Ch. 6-47

Developing aSampling Distribution

(continued)

μ2116

24211918

N

X)XE( i

1.5816

21)-(2421)-(1921)-(18

N

μ)X(σ

222

2i

X


Comparing the Population with its Sampling Distribution

Ch. 6-48

18 19 20 21 22 23 240

.1

.2

.3 P(X)

X18 20 22 24

A B C D

0

.1

.2

.3

Population

N = 4

P(X)

X_

1.58σ 21μXX2.236σ 21μ

Sample Means Distributionn = 2

_


Expected Value of Sample Mean

Let X1, X2, . . . Xn represent a random sample from a

population

The sample mean value of these observations is

defined as

Ch. 6-49

n

1i

iXn

1X


Standard Error of the Mean

Different samples of the same size from the same

population will yield different sample means

A measure of the variability in the mean from sample to

sample is given by the Standard Error of the Mean:

Note that the standard error of the mean decreases as

the sample size increases

Ch. 6-50

n

σσ

X


If sample values are not independent

If the sample size n is not a small fraction of the

population size N, then individual sample members

are not distributed independently of one another

Thus, observations are not selected independently

A correction is made to account for this:

or

Ch. 6-51

(continued)

1N

nN

n

σσ

X

1N

nN

n

σ)XVar(

2


If the Population is Normal

If a population is normal with mean μ and

standard deviation σ, the sampling distribution

of is also normally distributed with

and

If the sample size n is not large relative to the population size N, then

and

Ch. 6-52

X

μμX

n

σσ

X

1N

nN

n

σσ

X

μμ

X


Z-value for Sampling Distributionof the Mean

Z-value for the sampling distribution of :

Ch. 6-53

where: = sample mean

= population mean

= standard error of the mean

Xμ

Xσ

μ)X(Z

X

xσ


Sampling Distribution Properties

(i.e. is unbiased )

Ch. 6-54

Normal Population

Distribution

Normal Sampling

Distribution

(has the same mean)

xx

x

μμx

μ

xμ


Sampling Distribution Properties

For sampling with replacement:

As n increases,

decreases

Ch. 6-55

Larger

sample size

Smaller

sample size

x

(continued)

xσ

μECON 509, by Dr. M. Zainal

If the Population is not Normal

We can apply the Central Limit Theorem:

Even if the population is not normal,

…sample means from the population will beapproximately normal as long as the sample size is large enough.

Properties of the sampling distribution:

and

Ch. 6-56

μμ x n

σσ x


Central Limit Theorem

Ch. 6-57

n↑As the

sample

size gets

large

enough…

the sampling

distribution

becomes

almost normal

regardless of

shape of

population

xECON 509, by Dr. M. Zainal

If the Population is not Normal

Ch. 6-58

Population Distribution

Sampling Distribution

(becomes normal as n increases)

Central Tendency

Variation

x

x

Larger

sample

size

Smaller

sample size

(continued)

Sampling distribution

properties:

μμ x

n

σσ x

xμ

μ


How Large is Large Enough?

For most distributions, n > 25 will give a

sampling distribution that is nearly normal

For normal population distributions, the

sampling distribution of the mean is always

normally distributed


Example

Suppose a large population has mean μ = 8

and standard deviation σ = 3. Suppose a

random sample of size n = 36 is selected.

What is the probability that the sample mean is

between 7.8 and 8.2?


Example

Solution:

Even if the population is not normally

distributed, the central limit theorem can be

used (n > 25)

… so the sampling distribution of is

approximately normal

… with mean = 8

…and standard deviation

Ch. 6-61

(continued)

x

xμ

0.536

3

n

σσ x


Example

Solution (continued):

Ch. 6-62

(continued)

0.38300.5)ZP(-0.5

363

8-8.2

nσ

μ- μ

363

8-7.8P 8.2) μ P(7.8

X

X

Z7.8 8.2-0.5 0.5

Sampling

Distribution

Standard Normal

Distribution .1915

+.1915

Population

Distribution

??

??

????

????

Sample Standardize

8μ 8μX 0μz xX


Acceptance Intervals

Goal: determine a range within which sample means are

likely to occur, given a population mean and variance

By the Central Limit Theorem, we know that the distribution of X

is approximately normal if n is large enough, with mean μ and

standard deviation

Let zα/2 be the z-value that leaves area α/2 in the upper tail of the

normal distribution (i.e., the interval - zα/2 to zα/2 encloses

probability 1 – α)

Then

is the interval that includes X with probability 1 – α

Ch. 6-63

Xσ

X/2σzμ


Sampling

Distributions

Sampling

Distribution of

Sample

Mean

Sampling

Distribution of

Sample

Proportion

Sampling

Distribution of

Sample

Variance

Sampling Distributions of Sample Proportions

Ch. 6-64

6.3


Sampling Distributions of Sample Proportions

P = the proportion of the population having some characteristic

Sample proportion ( ) provides an estimateof P:

0 ≤ ≤ 1

has a binomial distribution, but can be approximated by a normal distribution when nP(1 – P) > 5

Ch. 6-65

size sample

interest ofstic characteri the having sample the in itemsofnumber

n

Xp ˆ

p̂

p̂

p̂


Sampling Distribution of p

Normal approximation:

Properties:

and

Ch. 6-66

(where P = population proportion)

Sampling Distribution

.3

.2

.1

00 . 2 .4 .6 8 1

P)pE( ˆn

P)P(1

n

XVarσ2

p

ˆ

^

)PP( ˆ

P̂


Z-Value for Proportions

Ch. 6-67

n

P)P(1

Pp

σ

PpZ

p

ˆˆ

ˆ

Standardize to a Z value with the formula:p̂


Example

If the true proportion of voters who support

Proposition A is P = .4, what is the probability

that a sample of size 200 yields a sample

proportion between .40 and .45?

Ch. 6-68

▪ i.e.: if P = .4 and n = 200, what is

P(.40 ≤ ≤ .45) ?p̂


Example

if P = .4 and n = 200, what is

P(.40 ≤ ≤ .45) ?

Ch. 6-69

(continued)

.03464200

.4).4(1

n

P)P(1σ

p

ˆ

1.44)ZP(0

.03464

.40.45Z

.03464

.40.40P.45)pP(.40

ˆ

Find :

Convert to

standard

normal:

pσ ˆ

p̂


Example

if P = .4 and n = 200, what is

P(.40 ≤ ≤ .45) ?

Ch. 6-70

Z.45 1.44

.4251

Standardize

Sampling DistributionStandardized

Normal Distribution

(continued)

Use standard normal table: P(0 ≤ Z ≤ 1.44) = .4251

.40 0p̂

p̂


Sampling Distributions ofSample Variance

Ch. 6-71

Sampling

Distributions

Sampling

Distribution of

Sample

Mean

Sampling

Distribution of

Sample

Proportion

Sampling

Distribution of

Sample

Variance

6.4


Sample Variance

Let x1, x2, . . . , xn be a random sample from a

population. The sample variance is

the square root of the sample variance is called

the sample standard deviation

the sample variance is different for different

random samples from the same population

Ch. 6-72

n

1i

2

i

2 )x(x1n

1s


Sampling Distribution ofSample Variances

The sampling distribution of s2 has mean σ2

If the population distribution is normal, then

If the population distribution is normal then

has a 2 distribution with n – 1 degrees of freedom

Ch. 6-73

22 σ)E(s

1n

2σ)Var(s

42

2

2

σ

1)s-(n


The Chi-square Distribution

The chi-square distribution is a family of

distributions, depending on degrees of freedom

(Like the t distribution).

The chi-square distribution curve starts at the

origin and lies entirely to the right of the vertical

axis.

The chi-square distribution assumes

nonnegative values only, and these are denoted

by the symbol 2 (read as “chi-square”).


The Chi-square Distribution

The shape of a specific chi-square distribution

depends on the number of degrees of freedom.

peak of a 2 distribution curve with 1 or 2

degrees of freedom occurs at zero and for a

curve with 3 or more degrees of freedom at

(df−2).

0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28 0 4 8 12 16 20 24 28

d.f. = 1 d.f. = 5 d.f. = 15

2 22


Degrees of Freedom (df)

Idea: Number of observations that are free to varyafter sample mean has been calculated

Example: Suppose the mean of 3 numbers is 8.0

Let X1 = 7

Let X2 = 8

What is X3?

Ch. 6-76

If the mean of these three

values is 8.0,

then X3 must be 9

(i.e., X3 is not free to vary)

Here, n = 3, so degrees of freedom = n – 1 = 3 – 1 = 2

(2 values can be any numbers, but the third is not free to vary

for a given mean)ECON 509, by Dr. M. Zainal

Finding the Critical Value

The critical value, , is found from the

chi-square table

2

2

2



Example: Find the value of 2 for 7 degrees of freedom and an

area of .10 in the right tail of the chi-square distribution curve.



Example: Find the value of 2 for 9 degrees of freedom and an

area of .05 in the left tail of the chi-square distribution curve.


Chi-square Example

A commercial freezer must hold a selected

temperature with little variation. Specifications call

for a standard deviation of no more than 4 degrees

(a variance of 16 degrees2).

Ch. 6-81

▪ A sample of 14 freezers is to be

tested

▪ What is the upper limit (K) for the

sample variance such that the

probability of exceeding this limit,

given that the population standard

deviation is 4, is less than 0.05?


Finding the Chi-square Value

Use the chi-square distribution with area 0.05 in the upper tail:

Ch. 6-82

probability

α = .05

213

2

213

= 22.36

= 22.36 (α = .05 and 14 – 1 = 13 d.f.)

2

22

σ

1)s(nχ

Is chi-square distributed with (n – 1) = 13

degrees of freedom


Chi-square Example

Ch. 6-83

0.0516

1)s(nPK)P(s 213

22

χSo:

(continued)

213 = 22.36 (α = .05 and 14 – 1 = 13 d.f.)

22.3616

1)K(n

(where n = 14)

so 27.521)(14

)(22.36)(16K

If s2 from the sample of size n = 14 is greater than 27.52, there is

strong evidence to suggest the population variance exceeds 16.

or


Chapter Summary

Introduced sampling distributions

Described the sampling distribution of sample means

For normal populations

Using the Central Limit Theorem

Described the sampling distribution of sample

proportions

Introduced the chi-square distribution

Examined sampling distributions for sample variances

Calculated probabilities using sampling distributions


Copyright

The materials of this presentation were mostly

taken from the PowerPoint files accompanied

Business Statistics: A Decision-Making

Approach, 7e © 2008 Prentice-Hall, Inc.


Documents

Statistics for Economics · ECON 509, by Dr. M. Zainal Ch. 6-14. Ch. 6-15 The Standard Normal Table The Standard Normal table in the textbook gives the probability from the mean (zero)