Chap09

Chap 9-1Statistics for Business and Economics, 6e © 2007 Pearson Education, Inc.
Chapter 9
Estimation: Additional Topics
Statistics for
Business and Economics
6th
Edition

Statistics for Business and
Economics, 6e © 2007 Pearson
Education, Inc. Chap 9-2
Chapter Goals
After completing this chapter, you should be able to:
 Form confidence intervals for the mean difference 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
 Create confidence intervals for a population variance
 Find chi-square values from the chi-square distribution table
 Determine the required sample size to estimate a mean or
proportion within a specified margin of error

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

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

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

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

 The margin of error is
 tn-1,α/2 is the value from the Student’s t
distribution with (n – 1) degrees of freedom
for which
Mean Difference
(continued)
2
α
)tP(t α/21,n1n => −−
n
s
tME d
α/21,n−=
Dependent
samples

 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 =
Σ di
n
4.82
1n
)d(d
S
2
i
d
=
−
−
=
∑
= 7.0

 For a 95% confidence level, the appropriate t value is tn-
1,α/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

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:

Difference Between Two Means
Population means,
independent
samples
Confidence interval uses zα/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)

Population means,
independent
samples
σx
2
and σy
2
Known
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

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
σx
2
and σy
2
Known

Population means,
independent
samples
The confidence interval for
μx – μy is:
Confidence Interval,
σx
2
and σy
2
Known
*
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

Population means,
independent
samples
σx
2
and σy
2
Unknown,
Assumed Equal
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

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
σx2 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
σx
2
and σy
2
Unknown,
Assumed Equal

μ1 – μ2 is:
Where
*
σx
2
and σy
2
Unknown, Equal
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
−+
−+−
=

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

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
==−+

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.

Population means,
independent
samples
σx
2
and σy
2
Unknown,
Assumed Unequal
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

Population means,
independent
samples
σx
2
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 ν 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

μ1 – μ2 is:
*
σx
2
and σy
2
Unknown, Unequal
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( ++−<−<+−− νν
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

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 −

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
−
+
−
−−−
=

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/
−
+
−
±− α

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, Zα/2 = 1.645

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

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

Population Variance
Population
Variance
The random variable
2
2
2
1n
σ
1)s(n−
=−χ
follows a chi-square distribution
with (n – 1) degrees of freedom
(continued)
The chi-square value denotes the number for which
2
,1n αχ −
α)P( 2
α,1n
2
1n => −− χχ

Population Variance
Population
Variance
The (1 - α)% confidence interval
for the population variance is
2
/2-1,1n
2
2
2
/2,1n
2
1)s(n
σ
1)s(n
αα χχ −−
−
<<
−
(continued)

Example
You are testing the speed of a computer processor.
You collect the following data (in Mhz):
CPUx
Sample size 17
Sample mean 3004
Sample std dev 74
Assume the population is normal.
Determine the 95% confidence interval for σx
2

Finding the Chi-square Values
 n = 17 so the chi-square distribution has (n – 1) = 16
degrees of freedom
 α = 0.05, so use the the chi-square values with area
0.025 in each tail:
probability
α/2 = .025
χ2
16
χ2
16
= 28.85
6.91
28.85
2
0.975,16
2
/2-1,1n
2
0.025,16
2
/2,1n
==
==
−
−
χχ
χχ
α
α
χ2
16 = 6.91
probability
α/2 = .025

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
2
2
2
/2,1n
2
1)s(n
σ
1)s(n
αα χχ −−
−
<<
−
6.91
1)(74)(17
σ
28.85
1)(74)(17 2
2
2
−
<<
−
12683σ3037 2
<<

Sample PHStat Output

Sample PHStat Output
Input
Output
(continued)

Sample Size Determination
For the
Mean
Determining
Sample Size
For the
Proportion

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 - α)
 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

For the
Mean
Determining
Sample Size
n
σ
zx α/2±
n
σ
zME α/2=
Margin of Error
(sampling error)

For the
Mean
Determining
Sample Size
n
σ
zME α/2=
(continued)
2
22
α/2
ME
σz
n =Now solve
for n to get

 To determine the required sample size for the
mean, you must know:
 The desired level of confidence (1 - α), which
determines the zα/2 value
 The acceptable margin of error (sampling error), ME
 The standard deviation, σ
(continued)

Required Sample Size Example
If σ = 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

n
)p(1p
zp α/2
ˆˆ
ˆ −
±
n
)p(1p
zME α/2
ˆˆ −
=
Determining
Sample Size
For the
Proportion
Margin of Error
(sampling error)

Determining
Sample Size
For the
Proportion
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 −

 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 - α), which determines the
critical zα/2 value
 The acceptable sampling error (margin of error), ME
 Estimate P(1 – P) = 0.25
(continued)
pˆ

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
===

PHStat Sample Size Options

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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