Chap08 estimation additional topics

Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall
Statistics for
Business and Economics
7th Edition
Chapter 8
Estimation: Additional Topics
Ch. 8-1

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
Copyright © 2010 Pearson Education, Inc. Publishing as Prentice Hall Ch. 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
Ch. 8-3
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
Ch. 8-4
8.1

Mean Difference
The ith paired difference is di , where
di = xi - yi
The point estimate for
the population mean
paired difference is d :
n
d
d
n
1i
i

n is the number of matched pairs in the sample
1n
)d(d
S
n
1i
2
i
d




The sample
standard
deviation is:
Dependent
samples
Ch. 8-5

Confidence Interval for
Mean 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
S
tdμ
n
S
td d
α/21,nd
d
α/21,n  
Dependent
samples
Ch. 8-6

Mean 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
s
tME d
α/21,n
Dependent
samples
Ch. 8-7

 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(d
S
2
i
d





= 7.0
Ch. 8-8
Dependent
samples

 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
S
tdμ
n
S
td
d
d
d
α/21,nd
d
α/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
Ch. 8-9
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
Ch. 8-10
8.2

Population means,
independent
samples
Confidence interval uses z/2
Confidence interval uses a value
from the Student’s t distribution
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
(continued)
Ch. 8-11
Difference Between Two Means:
Independent Samples

σx
2 and σy
2 Known
Population means,
independent
samples
Assumptions:
 Samples are randomly and
independently drawn
 both population distributions
are normal
 Population variances are
known
*σx
2 and σy
2 known
σx
2 and σy
2 unknown
Ch. 8-12

σx
2 and σy
2 Known
Population means,
independent
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
2
y
x
2
x2
YX
n
σ
n
σ
σ 
(continued)
*
Y
2
y
X
2
x
YX
n
σ
n
σ
)μ(μ)yx(
Z



σx
2 and σy
2 known
σx
2 and σy
2 unknown
Ch. 8-13

Confidence Interval,
σx
2 and σy
2 Known
Population means,
independent
samples
The confidence interval for
μx – μy is:
*
y
2
Y
x
2
X
α/2YX
y
2
Y
x
2
X
α/2
n
σ
n
σ
z)yx(μμ
n
σ
n
σ
z)yx( 
σx
2 and σy
2 known
σx
2 and σy
2 unknown
Ch. 8-14

σx
2 and σy
2 Unknown,
Assumed Equal
Population means,
independent
samples
Assumptions:
independently drawn
 Populations are normally
distributed
unknown but assumed equal
*σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
Ch. 8-15

σx
2 and σy
2 Unknown,
Assumed Equal
Population means,
independent
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
*σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
Ch. 8-16

σx
2 and σy
2 Unknown,
Assumed Equal
Population means,
independent
samples
The pooled variance is
(continued)
* 2nn
1)s(n1)s(n
s
yx
2
yy
2
xx2
p



σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
Ch. 8-17

σx
2 and σy
2 Unknown, Equal
μ1 – μ2 is:
Where
*σx
2 and σy
2
assumed equal
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
y
2
p
x
2
p
α/22,nnYX
y
2
p
x
2
p
α/22,nn
n
s
n
s
t)yx(μμ
n
s
n
s
t)yx( yxyx
 
2nn
1)s(n1)s(n
s
yx
2
yy
2
xx2
p



Ch. 8-18

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 14
Sample mean 3004 2538
Sample std dev 74 56
Assume both populations are
normal with equal variances,
and use 95% confidence
Ch. 8-19

Calculating the Pooled Variance
        4427.03
1)141)-(17
5611474117
1)n(n
S1nS1n
S
22
y
2
yy
2
xx2
p 






(()1x
The pooled variance is:
The t value for a 95% confidence interval is:
2.045tt 0.025,29α/2,2nn yx

Ch. 8-20

Calculating the Confidence Limits
 The 95% confidence interval is
y
2
p
x
2
p
α/22,nnYX
y
2
p
x
2
p
α/22,nn
n
s
n
s
t)yx(μμ
n
s
n
s
t)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.
Ch. 8-21

σx
2 and σy
2 Unknown,
Assumed Unequal
Population means,
independent
samples
Assumptions:
independently drawn
 Populations are normally
distributed
unknown and assumed
unequal
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 known
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
Ch. 8-22

σx
2 and σy
2 Unknown,
Assumed Unequal
Population means,
independent
samples
(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
σx
2 and σy
2 known
σx
2 and σy
2 unknown
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2
assumed unequal
1)/(n
n
s
1)/(n
n
s
)
n
s
()
n
s
(
y
2
y
2
y
x
2
x
2
x
2
y
2
y
x
2
x
























v
Ch. 8-23

σx
2 and σy
2 Unknown, Unequal
μ1 – μ2 is:
*
σx
2 and σy
2
assumed equal
σx
2 and σy
2 unknown
σx
2 and σy
2
assumed unequal
y
2
y
x
2
x
α/2,YX
y
2
y
x
2
x
α/2,
n
s
n
s
t)yx(μμ
n
s
n
s
t)yx(  
1)/(n
n
s
1)/(n
n
s
)
n
s
()
n
s
(
y
2
y
2
y
x
2
x
2
x
2
y
2
y
x
2
x
























vWhere
Ch. 8-24

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 
Ch. 8-25
8.3

 The random variable
is approximately normally distributed
Population
proportions
(continued)
y
yy
x
xx
yxyx
n
)p(1p
n
)p(1p
)p(p)pp(
Z
ˆˆˆˆ
ˆˆ





Ch. 8-26

Population
proportions
The confidence limits for
Px – Py are:
y
yy
x
xx
yx
n
)p(1p
n
)p(1p
Z)pp(
ˆˆˆˆ
ˆˆ 2/



 
Ch. 8-27

Example:
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

Example:
Men:
Women:
0.1012
40
0.70(0.30)
50
0.52(0.48)
n
)p(1p
n
)p(1p
y
yy
x
xx



 ˆˆˆˆ
0.52
50
26
px ˆ
0.70
40
28
py ˆ
(continued)
For 90% confidence, Z/2 = 1.645
Ch. 8-29

Example:
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(1p
Z)pp(
y
yy
x
xx
α/2yx





ˆˆˆˆ
ˆˆ
Ch. 8-30

Sample Size Determination
For the
Mean
Determining
Sample Size
For the
Proportion
Ch. 8-31
Large
Populations
Finite
Populations
For the
Mean
For the
Proportion
8.4

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

n
σ
zx α/2
n
σ
zME α/2
Margin of Error
(sampling error)
Ch. 8-33
For the
Mean
Large
Populations

n
σ
zME α/2
(continued)
2
22
α/2
ME
σz
n Now solve
for n to get
Ch. 8-34
For the
Mean
Large
Populations

 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)
Ch. 8-35

Required Sample Size Example
If  = 45, what sample size is needed to
estimate the mean within ± 5 with 90%
confidence?
(Always round up)
219.19
5
(45)(1.645)
ME
σz
n 2
22
2
22
α/2

So the required sample size is n = 220
Ch. 8-36

Sample Size Determination:
Population Proportion
n
)p(1p
zp α/2
ˆˆ
ˆ 

n
)p(1p
zME α/2
ˆˆ 

Margin of Error
(sampling error)
Ch. 8-37
For the
Proportion
Large
Populations

2
2
α/2
ME
z0.25
n 
Substitute
0.25 for
and solve for
n to get
(continued)
n
)p(1p
zME α/2
ˆˆ 

cannot
be larger than
0.25, when =
0.5
)p(1p ˆˆ 
pˆ
)p(1p ˆˆ 
Ch. 8-38
For the
Proportion
Large
Populations

 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)
pˆ
Ch. 8-39

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

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
z0.25
n 2
2
2
2
α/2

Ch. 8-41

Finite Populations
Finite
Populations
For the
Mean









1N
nN
n
σ
)XVar(
2
A finite population
correction factor is added:
1. Calculate the required
sample size n0 using the
prior formula:
2. Then adjust for the finite
population:
2
22
α/2
0
ME
σz
n 
1)-(Nn
Nn
n
0
0


8.5

Finite Populations
Finite
Populations
For the
Proportion









1N
nN
n
P)P(1-
)pVar( ˆ
A finite population
correction factor is added:
1. Solve for n:
2. The largest possible value
for this expression
(if P = 0.25) is:
3. A 95% confidence interval
will extend ±1.96 from
the sample proportion
P)P(11)σ(N
P)NP(1
n 2
p



ˆ
0.251)σ(N
P)0.25(1
n 2
p



ˆ
p
σˆ

Example: Sample Size to
Estimate Population Proportion
(continued)
p
σˆ
How large a sample would be necessary to
estimate within ±5% the true proportion of
college graduates in a population of 850
people with 95% confidence?

Solution:
 For 95% confidence, use z0.025 = 1.96
 ME = 0.05
So use n = 265
(continued)
Ch. 8-45
0.02551σ0.05σ1.96 pp
 ˆˆ
264.8
0.25551)(849)(0.02
)(0.25)(850
0.251)σ(N
0.25N
n 22
p
max 




ˆ

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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