EF 507 QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE FALL 2008 Chapter 9

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EF 507

QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE

FALL 2008

Chapter 9

Estimation: Additional Topics

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Estimation: Additional Topics

Chapter Topics

Population Means,

Independent Samples

Population Means,

Dependent Samples

Population Variance

Group 1 vs. independent

Group 2

Same group before vs. after

treatment

Variance of a normal distribution

Examples:

Population Proportions

Proportion 1 vs. Proportion 2

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1. Dependent Samples

Tests Means of 2 Related Populations Paired or matched samples Repeated measures (before/after) Use difference between paired values:

Eliminates Variation Among Subjects Assumptions:

Both Populations Are Normally Distributed

Dependent samples

di = xi - yi

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Mean Difference

The ith paired difference is di , where

di = xi - yi

The point estimate for the population mean paired difference is d : n

dd

n

1ii

n is the number of matched pairs in the sample

1n

)d(dS

n

1i

2i

d

The sample standard deviation is:

Dependent samples

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Confidence Interval forMean Difference

The confidence interval for difference between population means, μd , is

Where

n = the sample size (number of matched pairs in the paired sample)

n

Stdμ

n

Std d

α/21,ndd

α/21,n

Dependent samples

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The margin of error is

tn-1,/2 is the value from the Student’s t distribution with (n – 1) degrees of freedom for which

Confidence Interval forMean Difference

(continued)

2

α)tP(t α/21,n1n

n

stME d

α/21,n

Dependent samples

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Six people sign up for a weight loss program. You collect the following data:

Paired Samples Example

Weight: Person Before (x) After (y) Difference, di

1 136 125 11 2 205 195 10 3 157 150 7 4 138 140 - 2 5 175 165 10 6 166 160 6 42

d = di

n

4.82

1n

)d(dS

2i

d

= 7.0

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For a 95% confidence level, the appropriate t value is tn-1,/2 = t5,0.025 = 2.571

The 95% confidence interval for the difference between means, μd , is

12.06μ1.94

6

4.82(2.571)7μ

6

4.82(2.571)7

n

Stdμ

n

Std

d

d

dα/21,nd

dα/21,n

Paired Samples Example (continued)

Since this interval contains zero, we can not be 95% confident, given this limited data, that the weight loss program helps people lose weight

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2. Difference Between Two Means

Population means, independent

samples

Goal: Form a confidence interval for the difference between two population means, μx – μy

x – y

Different data sources Unrelated Independent

Sample selected from one population has no effect on the sample selected from the other population

The point estimate is the difference between the two sample means:

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Difference Between Two Means


samples

Confidence interval uses z/2

Confidence interval uses a value from the Student’s t distribution

σx2 and σy

2 assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

(continued)

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samples

2. σx2 and σy

2 Known

Assumptions:

Samples are randomly and independently drawn

both population distributions are normal

Population variances are known

*σx2 and σy

2 known

σx2 and σy

2 unknown

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samples

…and the random variable

has a standard normal distribution

When σx and σy are known and

both populations are normal, the variance of X – Y is

y

2y

x

2x2

YX n

σ

n

σσ

(continued)

*

Y

2y

X

2x

YX

n

σ

nσ

)μ(μ)yx(Z

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 Known

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samples

The confidence interval for

μx – μy is:

Confidence Interval, σx

2 and σy2 Known

*

y

2Y

x

2X

α/2YXy

2Y

x

2X

α/2 n

σ

n

σz)yx(μμ

n

σ

n

σz)yx(

σx2 and σy

2 known

σx2 and σy

2 unknown

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samples

3.a) σx2 and σy

2 Unknown, Assumed Equal

Assumptions:


Populations are normally distributed

Population variances are unknown but assumed equal*σx

2 and σy2

assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

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samples

(continued)

Forming interval estimates:

The population variances are assumed equal, so use the two sample standard deviations and pool them to estimate σ

use a t value with (nx + ny – 2) degrees of freedom

*σx2 and σy

2 assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

σx2 and σy2 Unknown,

Assumed Equal

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samples

The pooled variance is

(continued)

* 2nn

1)s(n1)s(ns

yx

2yy

2xx2

p

σx

2 and σy2

assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

σx2 and σy

2 Unknown,Assumed Equal

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μ1 – μ2 is:

Where

*


2 and σy2 Unknown, Equal

σx2 and σy

2 assumed equal

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

y

2p

x

2p

α/22,nnYXy

2p

x

2p

α/22,nn n

s

n

st)yx(μμ

n

s

n

st)yx(

yxyx

2nn

1)s(n1)s(ns

yx

2yy

2xx2

p

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Pooled Variance Example

You are testing two computer processors for speed. Form a confidence interval for the difference in CPU speed. You collect the following speed data (in Mhz):

CPUx CPUy

Number Tested 17 14Sample mean 3,004 2,538Sample std dev 74 56

Assume both populations are normal with equal variances, and use 95% confidence

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Calculating the Pooled Variance

4427.03

1)141)-(17

5611474117

1)n(n

S1nS1nS

22

y

2yy

2xx2

p

(()1x

The pooled variance is:

The t value for a 95% confidence interval is:

2.045tt 0.025 , 29α/2 , 2nn yx

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Calculating the Confidence Limits

The 95% confidence interval is

y

2p

x

2p

α/22,nnYXy

2p

x

2p

α/22,nn n

s

n

st)yx(μμ

n

s

n

st)yx(

yxyx

14

4427.03

17

4427.03(2.054)2538)(3004μμ

14

4427.03

17

4427.03(2.054)2538)(3004 YX

515.31μμ416.69 YX

We are 95% confident that the mean difference in CPU speed is between 416.69 and 515.31 Mhz.

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samples

3.b) σx2 and σy

2 Unknown, Assumed Unequal

Assumptions:


Populations are normally distributed

Population variances are unknown and assumed unequal

*

σx2 and σy

2 assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

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samples

σx2 and σy

2 Unknown,Assumed Unequal

(continued)

Forming interval estimates:

The population variances are assumed unequal, so a pooled variance is not appropriate

use a t value with degrees of freedom, where

σx2 and σy

2 known

σx2 and σy

2 unknown

*

σx2 and σy

2 assumed equal

σx2 and σy

2 assumed unequal

1)/(nn

s1)/(n

ns

)n

s()

ns

(

y

2

y

2y

x

2

x

2x

2

y

2y

x

2x

v

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μ1 – μ2 is:

*


2 and σy2 Unknown, Unequal

σx2 and σy

2 assumed equal

σx2 and σy

2 unknown

σx2 and σy

2 assumed unequal

y

2y

x

2x

α/2,YXy

2y

x

2x

α/2, n

s

n

st)yx(μμ

n

s

n

st)yx(

1)/(nn

s1)/(n

ns

)n

s()

ns

(

y

2

y

2y

x

2

x

2x

2

y

2y

x

2x

vWhere

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Two Population Proportions

Goal: Form a confidence interval for the difference between two population proportions, Px – Py

The point estimate for the difference is

Population proportions

Assumptions:

Both sample sizes are large (generally at least 40 observations in each sample)

yx pp ˆˆ

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Two Population Proportions


(continued)

The random variable

is approximately normally distributed

y

yy

x

xx

yxyx

n

)p(1p

n)p(1p

)p(p)pp(Z

ˆˆˆˆ

ˆˆ

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Confidence Interval forTwo Population Proportions


The confidence limits for

Px – Py are:

y

yy

x

xxyx n

)p(1p

n

)p(1pZ )pp(

ˆˆˆˆˆˆ

2/

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Example: Two Population Proportions

Form a 90% confidence interval for the difference between the proportion of men and the proportion of women who have college degrees.

In a random sample, 26 of 50 men and 28 of 40 women had an earned college degree

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Men:

Women:

0.101240

0.70(0.30)

50

0.52(0.48)

n

)p(1p

n

)p(1p

y

yy

x

xx

ˆˆˆˆ

0.5250

26px ˆ

0.7040

28py ˆ

(continued)

For 90% confidence, Z/2 = 1.645

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The confidence limits are:

so the confidence interval is

-0.3465 < Px – Py < -0.0135

Since this interval does not contain zero we are 90% confident that the two proportions are not equal

(continued)

y yx xx y α/2

x y

ˆ ˆp (1 p )ˆ ˆp (1 p )ˆ ˆ(p p ) Z

n n

(0.52 0.70) 1.645(0.1012)

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Confidence Intervals for the Population Variance

Population Variance

Goal: Form a confidence interval for the population variance, σ2

The confidence interval is based on the sample variance, s2

Assumed: the population is normally distributed

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Population Variance

The random variable

2

22

1n σ

1)s(n

follows a chi-square distribution with (n – 1) degrees of freedom

(continued)

The chi-square value denotes the number for which2

, 1n

α)P( 2α , 1n

21n χχ

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Population Variance

The (1 - )% confidence interval for the population variance is

2/2 - 1 , 1n

22

2/2 , 1n

2 1)s(nσ

1)s(n

αα χχ

(continued)

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Example

You are testing the speed of a computer processor. You collect the following data (in Mhz):

CPUx

Sample size 17Sample mean 3,004Sample std dev 74

Assume the population is normal. Determine the 95% confidence interval for σx

2

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Finding the Chi-square Values

n = 17 so the chi-square distribution has (n – 1) = 16 degrees of freedom

= 0.05, so use the the chi-square values with area 0.025 in each tail:

probability α/2 = 0.025

216

216

= 28.85

6.91

28.85

20.975 , 16

2/2 - 1 , 1n

20.025 , 16

2/2 , 1n

χχ

χχ

α

α

216 = 6.91

probability α/2 =0 .025

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Calculating the Confidence Limits

The 95% confidence interval is

Converting to standard deviation, we are 95% confident that the population standard deviation of

CPU speed is between 55.1 and 112.6 Mhz

2/2 - 1 , 1n

22

2/2 , 1n

2 1)s(nσ

1)s(n

αα χχ

6.91

1)(74)(17σ

28.85

1)(74)(17 22

2

23,037 σ 12,683

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Sample PHStat Output

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Sample PHStat Output

Input

Output

(continued)

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Sample Size Determination

For the Mean

DeterminingSample Size

For theProportion

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Margin of Error

The required sample size can be found to reach a desired margin of error (ME) with a specified level of confidence (1 - )

The margin of error is also called sampling error the amount of imprecision in the estimate of the

population parameter the amount added and subtracted to the point estimate

to form the confidence interval

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For the Mean


n

σzx α/2

n

σzME α/2

Margin of Error (sampling error)


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For the Mean


n

σzME α/2

(continued)

2

22α/2

ME

σzn Now solve

for n to get


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To determine the required sample size for the mean, you must know:

The desired level of confidence (1 - ), which determines the z/2 value

The acceptable margin of error (sampling error), ME

The standard deviation, σ

(continued)

2

22α/2

ME

σzn

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Required Sample Size Example

If = 45, what sample size is needed to estimate the mean within ± 5 with 90% confidence?

(Always round up)

219.195

(45)(1.645)

ME

σzn

2

22

2

22α/2

So the required sample size is n = 220

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n

)p(1pzp α/2

ˆˆˆ

n

)p(1pzME α/2

ˆˆ


For theProportion

Margin of Error (sampling error)


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For theProportion

2

2α/2

ME

z 0.25n

Substitute 0.25 for and solve for n to get

(continued)


n

)p(1pzME α/2

ˆˆ

cannot be larger than 0.25, when = 0.5

)p(1p ˆˆ

p̂)p(1p ˆˆ

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The sample and population proportions, and P, are generally not known (since no sample has been taken yet)

P(1 – P) = 0.25 generates the largest possible margin of error (so guarantees that the resulting sample size will meet the desired level of confidence)

To determine the required sample size for the proportion, you must know:

The desired level of confidence (1 - ), which determines the critical z/2 value

The acceptable sampling error (margin of error), ME Estimate P(1 – P) = 0.25

(continued)


p̂

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How large a sample would be necessary to estimate the true proportion defective in a large population within ±3%, with 95% confidence?

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Solution:

For 95% confidence, use z0.025 = 1.96

ME = 0.03

Estimate P(1 – P) = 0.25

So use n = 1,068

(continued)

2 2α/2

2 2

0.25 z (0.25)(1.96)n 1,067.11

ME (0.03)

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PHStat Sample Size Options

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EF 507 QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE FALL 2008 Chapter 9