Newbold_chap09.ppt

Chap 9-1
Statistics 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
1
i
i



n is the number of matched pairs in the sample
1
n
)
d
(d
S
n
1
i
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
t
d
μ
n
S
t
d d
α/2
1,
n
d
d
α/2
1,
n 
 



Dependent
samples

 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
Mean Difference
(continued)
2
α
)
t
P(t α/2
1,
n
1
n 
 

n
s
t
ME d
α/2
1,
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 =
 di
n
4.82
1
n
)
d
(d
S
2
i
d





= 7.0

 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
12.06
μ
1.94
6
4.82
(2.571)
7
μ
6
4.82
(2.571)
7
n
S
t
d
μ
n
S
t
d
d
d
d
α/2
1,
n
d
d
α/2
1,
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 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)

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
x
2
Y
X
n
σ
n
σ
σ 


(continued)
*
Y
2
y
X
2
x
Y
X
n
σ
n
σ
)
μ
(μ
)
y
x
(
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/2
Y
X
y
2
Y
x
2
X
α/2
n
σ
n
σ
z
)
y
x
(
μ
μ
n
σ
n
σ
z
)
y
x
( 








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)
* 2
n
n
1)s
(n
1)s
(n
s
y
x
2
y
y
2
x
x
2
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/2
2,
n
n
Y
X
y
2
p
x
2
p
α/2
2,
n
n
n
s
n
s
t
)
y
x
(
μ
μ
n
s
n
s
t
)
y
x
( y
x
y
x








 



2
n
n
1)s
(n
1)s
(n
s
y
x
2
y
y
2
x
x
2
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)
14
1)
-
(17
56
1
14
74
1
17
1)
n
(n
S
1
n
S
1
n
S
2
2
y
2
y
y
2
x
x
2
p 













(
(
)
1
x
The pooled variance is:
The t value for a 95% confidence interval is:
2.045
t
t 0.025
,
29
α/2
,
2
n
n y
x





Calculating the Confidence Limits
 The 95% confidence interval is
y
2
p
x
2
p
α/2
2,
n
n
Y
X
y
2
p
x
2
p
α/2
2,
n
n
n
s
n
s
t
)
y
x
(
μ
μ
n
s
n
s
t
)
y
x
( y
x
y
x








 



14
4427.03
17
4427.03
(2.054)
2538)
(3004
μ
μ
14
4427.03
17
4427.03
(2.054)
2538)
(3004 Y
X 








515.31
μ
μ
416.69 Y
X 


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  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
,
Y
X
y
2
y
x
2
x
α/2
,
n
s
n
s
t
)
y
x
(
μ
μ
n
s
n
s
t
)
y
x
( 







 

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
Where

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)
y
x p
p ˆ
ˆ 

Population
proportions
(continued)
 The random variable
is approximately normally distributed
y
y
y
x
x
x
y
x
y
x
n
)
p
(1
p
n
)
p
(1
p
)
p
(p
)
p
p
(
Z
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ








Population
proportions
The confidence limits for
Px – Py are:
y
y
y
x
x
x
y
x
n
)
p
(1
p
n
)
p
(1
p
Z
)
p
p
(
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ 2
/




 

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
(1
p
n
)
p
(1
p
y
y
y
x
x
x





 ˆ
ˆ
ˆ
ˆ
0.52
50
26
px 

ˆ
0.70
40
28
py 

ˆ
(continued)
For 90% confidence, Z/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
(1
p
n
)
p
(1
p
Z
)
p
p
(
y
y
y
x
x
x
α/2
y
x








ˆ
ˆ
ˆ
ˆ
ˆ
ˆ

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
1
n
σ
1)s
(n 



follows a chi-square distribution
with (n – 1) degrees of freedom
(continued)
The chi-square value denotes the number for which
2
,
1
n 
 
α
)
P( 2
α
,
1
n
2
1
n 
 
 χ
χ

Population Variance
Population
Variance
The (1 - )% confidence interval
for the population variance is
2
/2
-
1
,
1
n
2
2
2
/2
,
1
n
2
1)s
(n
σ
1)s
(n
α
α χ
χ 





(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
  = 0.05, so use the the chi-square values with area
0.025 in each tail:
probability
α/2 = .025
2
16
2
16
= 28.85
6.91
28.85
2
0.975
,
16
2
/2
-
1
,
1
n
2
0.025
,
16
2
/2
,
1
n






χ
χ
χ
χ
α
α
2
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
,
1
n
2
2
2
/2
,
1
n
2
1)s
(n
σ
1)s
(n
α
α χ
χ 





6.91
1)(74)
(17
σ
28.85
1)(74)
(17 2
2
2




12683
σ
3037 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 - )
 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
z
x α/2

n
σ
z
ME α/2

Margin of Error
(sampling error)

For the
Mean
Determining
Sample Size
n
σ
z
ME α/2

(continued)
2
2
2
α/2
ME
σ
z
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 - ), which
determines the z/2 value
 The acceptable margin of error (sampling error), ME
 The 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)
219.19
5
(45)
(1.645)
ME
σ
z
n 2
2
2
2
2
2
α/2



So the required sample size is n = 220

n
)
p
(1
p
z
p α/2
ˆ
ˆ
ˆ 

n
)
p
(1
p
z
ME α/2
ˆ
ˆ 

Determining
Sample Size
For the
Proportion
Margin of Error
(sampling error)

Determining
Sample Size
For the
Proportion
2
2
α/2
ME
z
0.25
n 
Substitute
0.25 for
and solve for
n to get
(continued)
n
)
p
(1
p
z
ME α/2
ˆ
ˆ 

cannot
be larger than
0.25, when =
0.5
)
p
(1
p ˆ
ˆ 
p̂
)
p
(1
p ˆ
ˆ 

 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

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
z
0.25
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

Newbold_chap09.ppt

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