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Chapter 8 8-1

Statistics for Business and Economics, 7/eCopyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall

Chapter 8

Estimation: Additional Topics

ECON 509, by Dr. M. Zainal Ch. 8-1

Statistics for Business and Economics

7th Edition

Chapter Goals

After completing this chapter, you should be able to:

Form confidence intervals for the difference between two means from dependent samples

Form confidence intervals for the difference between two independent population means (standard deviations known or unknown)

Compute confidence interval limits for the difference between two independent population proportions

Determine the required sample size to estimate a mean or proportion within a specified margin of error

Ch. 8-2ECON 509, by Dr. M. Zainal

Chapter 8 8-2


Estimation: Additional Topics


Chapter Topics

Population Means,

Independent Samples

Population Means,

Dependent Samples

Sample Size Determination

Group 1 vs. independent

Group 2

Same group before vs. after

treatment

Finite Populations

Examples:

Population Proportions

Proportion 1 vs. Proportion 2

Large Populations

Confidence Intervals

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

Chapter 8 8-3


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

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

Chapter 8 8-4


Confidence Interval forMean Difference

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


(continued)

2

α)tP(t α/21,n1n

n

stME d

α/21,n

Dependent samples

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

ECON 509, by Dr. M. Zainal Ch. 8-8

Paired Samples Example

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

1 136 125 112 205 195 103 157 150 7 4 138 140 - 25 175 165 106 166 160 6

42

d = di

n

4.82

1n

)d(dS

2i

d

= 7.0

Dependent samples

Chapter 8 8-5


For a 95% confidence level, the appropriate t value is tn-1,/2 = t5,.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 cannot be 95% confident, given this limited data, that the weight loss program helps people lose weight

Dependent samples

Difference Between Two Means:Independent Samples

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:


Population means, independent

samples

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

x – y

Chapter 8 8-6


Difference Between Two Means:Independent Samples



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)

σx2 and σy

2 Known



samples

Assumptions:

Samples are randomly andindependently drawn

both population distributionsare normal

Population variances areknown

*σx2 and σy

2 known

σx2 and σy

2 unknown

Chapter 8 8-7


σx2 and σy

2 Known



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

Confidence Interval, σx

2 and σy2 Known



samples

The confidence interval for μx – μy is:

*

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

Chapter 8 8-8


σx2 and σy

2 Unknown,Assumed Equal



samples

Assumptions:


Populations are normallydistributed

Population variances areunknown 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

σx2 and σy




samples

(continued)

Forming interval estimates:

The population variances are assumed equal, so usethe two sample standard deviations and pool them toestimate σ

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

*σx2 and σy

2

assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2

assumed unequal

Chapter 8 8-9


σx2 and σy




samples

The pooled variance is

(continued)

* 2nn

1)s(n1)s(ns

yx

2yy

2xx2

p

σx2 and σy

2

assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2

assumed unequal


2 and σy2 Unknown, Equal


The confidence interval forμ1 – μ2 is:

Where

*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

Chapter 8 8-10


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 CPUyNumber Tested 17 14Sample mean 3004 2538Sample std dev 74 56


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

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

Chapter 8 8-11


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.

σx2 and σy

2 Unknown,Assumed Unequal



samples

Assumptions:


Populations are normallydistributed

Population variances areunknown and assumedunequal

*

σx2 and σy

2

assumed equal

σx2 and σy

2 known

σx2 and σy

2 unknown

σx2 and σy

2

assumed unequal

Chapter 8 8-12


σx2 and σy

2 Unknown,Assumed Unequal



samples

(continued)

Forming interval estimates:

The population variances areassumed unequal, so a pooledvariance is not appropriate

use a t value with degreesof freedom, where

σx2 and σy

2 known

σx2 and σy

2 unknown

*

σx2 and σy

2

assumed equal

σx2 and σy

2

assumed unequal1)/(n

n

s1)/(n

ns

)n

s()

ns

(

y

2

y

2y

x

2

x

2x

2

y

2y

x

2x

v


2 and σy2 Unknown, Unequal


The confidence interval forμ1 – μ2 is:

*

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

Chapter 8 8-13


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

Two Population Proportions

The random variable

is approximately normally distributed



(continued)

y

yy

x

xx

yxyx

n

)p(1p

n)p(1p

)p(p)pp(Z

ˆˆˆˆ

ˆˆ

Chapter 8 8-14


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/

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


Chapter 8 8-15



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


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)

(0.1012)1.645.70)(.52

n

)p(1p

n

)p(1pZ)pp(

y

yy

x

xxα/2yx

ˆˆˆˆˆˆ

Chapter 8 8-16




For the Mean

DeterminingSample Size

For theProportion

Large Populations

Finite Populations

For the Mean

For theProportion

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


Chapter 8 8-17




n

σzx α/2

n

σzME α/2

Margin of Error (sampling error)

For the Mean

Large Populations



n

σzME α/2

(continued)

2

22α/2

ME

σzn Now solve

for n to get

For the Mean

Large Populations

Chapter 8 8-18



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 population standard deviation, σ


(continued)

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

Chapter 8 8-19


Sample Size Determination:Population Proportion


n

)p(1pzp α/2

ˆˆˆ

n

)p(1pzME α/2

ˆˆ

Margin of Error (sampling error)

For theProportion

Large Populations



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

For theProportion

Large Populations

Chapter 8 8-20



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

Required Sample Size Example:Population Proportion

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


Chapter 8 8-21


Required Sample Size Example


Solution:

For 95% confidence, use z0.025 = 1.96

ME = 0.03

Estimate P(1 – P) = 0.25

So use n = 1068

(continued)

1067.11(0.03)

6)(0.25)(1.9

ME

z 0.25n

2

2

2

2α/2

Chapter Summary

Compared two dependent samples (paired samples) Formed confidence intervals for the paired difference

Compared two independent samples Formed confidence intervals for the difference between two

means, population variance known, using z Formed confidence intervals for the differences between two

means, population variance unknown, using t Formed confidence intervals for the differences between two

population proportions

Formed confidence intervals for the population variance using the chi-square distribution

Determined required sample size to meet confidence and margin of error requirements


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