# Chapter 9 ```EF 507
QUANTITATIVE METHODS FOR ECONOMICS
AND FINANCE
FALL 2008
Chapter 9
1/49
Chapter Topics
Population
Means,
Dependent
Samples
Population
Means,
Independent
Samples
Population
Proportions
Population
Variance
Proportion 1 vs.
Proportion 2
Variance of a
normal distribution
Examples:
Same group
before vs. after
treatment
Group 1 vs.
independent
Group 2
2/49
1. Dependent Samples
Tests Means of 2 Related Populations
Dependent
samples



Paired or matched samples
Repeated measures (before/after)
Use difference between paired values:
di = xi - yi
 Eliminates Variation Among Subjects
 Assumptions:
 Both Populations Are Normally Distributed
3/49
Mean Difference
The ith paired difference is di , where
Dependent
samples
di = xi - yi
The point estimate for
the population mean
paired difference is d :
The sample
standard
deviation is:
n
d
d
i 1
i
n
n
Sd 
2
(d

d
)
 i
i1
n 1
n is the number of matched pairs in the sample
4/49
Confidence Interval for
Mean Difference
Dependent
samples
The confidence interval for difference
between population means, μd , is
d  t n1,α/2
Sd
Sd
 μd  d  t n1,α/2
n
n
Where
n = the sample size
(number of matched pairs in the paired sample)
5/49
Confidence Interval for
Mean Difference
(continued)
Dependent
samples
 The margin of error is
ME  t n1,α/2
sd
n
 tn-1,/2 is the value from the Student’s t
distribution with (n – 1) degrees of freedom
for which
α
P(t n1  t n1,α/2 ) 
2
6/49
Paired Samples Example
collect the following data:
Person
1
2
3
4
5
6
Weight:
Before (x)
After (y)
136
205
157
138
175
166
125
195
150
140
165
160
Difference, di
11
10
7
-2
10
6
42
di

d = n
= 7.0
Sd 
2
(d

d
)
 i
n 1
 4.82
7/49
Paired Samples Example
(continued)
 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
d  t n1,α/2
7  (2.571)
Sd
S
 μd  d  t n1,α/2 d
n
n
4.82
4.82
 μd  7  (2.571)
6
6
 1.94  μd  12.06
Since this interval contains zero, we can not be 95% confident, given
this limited data, that the weight loss program helps people lose weight
8/49
2. Difference Between Two Means
Population means,
independent
samples
Goal: Form a confidence interval
for the difference between two
population means, μx – μy
 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:
x–y
9/49
Difference Between Two Means
(continued)
Population means,
independent
samples
σx2 and σy2 known
Confidence interval uses z/2
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
Confidence interval uses a value
from the Student’s t distribution
10/49
2. σx2 and σy2 Known
Population means,
independent
samples
σx2 and σy2 known
σx2 and σy2 unknown
Assumptions:
*
 Samples are randomly and
independently drawn
 both population distributions
are normal
 Population variances are
known
11/49
σx2 and σy2 Known
(continued)
When σx and σy are known and
both populations are normal, the
variance of X – Y is
Population means,
independent
samples
2
σx2 and σy2 known
σx2 and σy2 unknown
*
σ 2X Y
2
σy
σx


nx
ny
…and the random variable
Z
(x  y)  (μX  μY )
2
σ 2x σ y

nX nY
has a standard normal distribution
12/49
Confidence Interval,
σx2 and σy2 Known
Population means,
independent
samples
σx2 and σy2 known
σx2 and σy2 unknown
(x  y)  z α/2
*
The confidence interval for
μx – μy is:
σ 2X σ 2Y
σ 2X σ 2Y

 μX  μY  (x  y)  z α/2

nx ny
nx ny
13/49
3.a) σx2 and σy2 Unknown, Assumed Equal
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
σx2 and σy2 known
 Populations are normally
distributed
σx2 and σy2 unknown
σx2 and σy2
assumed equal
*
 Population variances are
unknown but assumed equal
σx2 and σy2
assumed unequal
14/49
σx2 and σy2 Unknown,
Assumed Equal
(continued)
Forming interval
estimates:
Population means,
independent
samples
 The population variances
are assumed equal, so use
the two sample standard
deviations and pool them to
estimate σ
σx2 and σy2 known
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
 use a t value with
(nx + ny – 2) degrees of
freedom
15/49
σx2 and σy2 Unknown,
Assumed Equal
(continued)
Population means,
independent
samples
The pooled variance is
σx2 and σy2 known
σx2 and σy2 unknown
σx2 and σy2
assumed equal
*
sp2 
(n x  1)s 2x  (n y  1)s 2y
nx  ny  2
σx2 and σy2
assumed unequal
16/49
Confidence Interval,
σx2 and σy2 Unknown, Equal
σx2 and σy2 unknown
σx2 and σy2
assumed equal
*
The confidence interval for
μ1 – μ2 is:
sp2
sp2
σx2 and σy2
assumed unequal
(x  y)  t nx ny 2,α/2
Where
sp2 
sp2
nx

ny
 μX  μY  (x  y)  t nx ny 2,α/2
nx

sp2
ny
(n x  1)s 2x  (n y  1)s 2y
nx  ny  2
17/49
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
Number Tested
17
Sample mean
3,004
Sample std dev
74
CPUy
14
2,538
56
Assume both populations are
normal with equal variances,
and use 95% confidence
18/49
Calculating the Pooled Variance
The pooled variance is:
2
2




n

1
S

n

1
S

17  1742  14  1562
x
x
y
y
2
S 

p
(n x  1)  (ny  1)
(17 - 1)  (14  1)
 4427.03
The t value for a 95% confidence interval is:
tnx ny 2 , α/2  t 29 , 0.025  2.045
19/49
Calculating the Confidence Limits
 The 95% confidence interval is
(x  y)  t nx ny 2,α/2
(3004  2538)  (2.054)
sp2
nx

sp2
ny
 μX  μY  (x  y)  t nx ny 2,α/2
sp2
nx

sp2
ny
4427.03 4427.03
4427.03 4427.03

 μX  μY  (3004  2538)  (2.054)

17
14
17
14
416.69  μX  μY  515.31
We are 95% confident that the mean difference in
CPU speed is between 416.69 and 515.31 Mhz.
20/49
3.b) σx2 and σy2 Unknown, Assumed Unequal
Assumptions:
Population means,
independent
samples
 Samples are randomly and
independently drawn
σx2 and σy2 known
 Populations are normally
distributed
σx2 and σy2 unknown
 Population variances are
unknown and assumed
unequal
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
21/49
σx2 and σy2 Unknown,
Assumed Unequal
(continued)
Forming interval estimates:
Population means,
independent
samples
 The population variances are
assumed unequal, so a pooled
variance is not appropriate
σx2 and σy2 known
 use a t value with  degrees
of freedom, where
σx2 and σy2 unknown
2
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
*
 s2x
s2y 
( )  ( )
n y 
 n x
v
2
2
2 2


s
 sx 
  /(n x  1)   y  /(n y  1)
n 
 nx 
 y
22/49
Confidence Interval,
σx2 and σy2 Unknown, Unequal
σx2 and σy2 unknown
σx2 and σy2
assumed equal
σx2 and σy2
assumed unequal
(x  y)  t ,α/2
*
The confidence interval for
μ1 – μ2 is:
2
2
s2x s y
s2x s y

 μX  μY  (x  y)  t ,α/2

nx ny
nx ny
Where
v
 s2x
s2y 
( )  ( )
n y 
 n x
2
2
 s2 
 s2x 
  /(n x  1)   y  /(n y  1)
n 
 nx 
 y
2
23/49
Two Population Proportions
Population
proportions
Goal: Form a confidence interval for
the difference between two
population proportions, Px – Py
Assumptions:
Both sample sizes are large (generally at
least 40 observations in each sample)
The point estimate for
the difference is
pˆ x  pˆ y
24/49
Two Population Proportions
(continued)
Population
proportions
 The random variable
Z
(pˆ x  pˆ y )  (p x  p y )
pˆ x (1 pˆ x ) pˆ y (1 pˆ y )

nx
ny
is approximately normally distributed
25/49
Confidence Interval for
Two Population Proportions
Population
proportions
The confidence limits for
Px – Py are:
(pˆ x  pˆ y )  Z / 2
pˆ x (1 pˆ x ) pˆ y (1 pˆ y )

nx
ny
26/49
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
27/49
Example:
Two Population Proportions
(continued)
Men:
ˆp x  26  0.52
50
Women:
ˆp y  28  0.70
40
pˆ x (1 pˆ x ) pˆ y (1 pˆ y )
0.52(0.48) 0.70(0.30)



 0.1012
nx
ny
50
40
For 90% confidence, Z/2 = 1.645
28/49
Example:
Two Population Proportions
(continued)
The confidence limits are:
(pˆ x  pˆ y )  Zα/2
pˆ x (1  pˆ x ) pˆ y (1  pˆ y )

nx
ny
 (0.52  0.70)  1.645(0.1012)
so the confidence interval is
-0.3465 &lt; Px – Py &lt; -0.0135
Since this interval does not contain zero we are 90% confident that the
two proportions are not equal
29/49
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
30/49
Confidence Intervals for the
Population Variance
(continued)
Population
Variance
The random variable

2
n1
(n  1)s

2
σ
2
follows a chi-square distribution
with (n – 1) degrees of freedom
2

The chi-square value n1,  denotes the number for which
P( χn21  χn21, α )  α
31/49
Confidence Intervals for the
Population Variance
(continued)
Population
Variance
The (1 - )% confidence interval
for the population variance is
(n  1)s
(n  1)s
2
σ  2
2
χn1, α/2
χn1, 1 - α/2
2
2
32/49
Example
You are testing the speed of a computer processor. You
collect the following data (in Mhz):
CPUx
Sample size
17
Sample mean
3,004
Sample std dev
74
Assume the population is normal.
Determine the 95% confidence interval for σx2
33/49
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:
2
χ n21, α/2  χ16
, 0.025  28.85
χ
probability
α/2 =0 .025
2
n 1, 1 - α/2
χ
2
16 , 0.975
 6.91
probability
α/2 = 0.025
216 = 6.91
216 = 28.85
216
34/49
Calculating the Confidence Limits
 The 95% confidence interval is
2
(n  1)s 2
(n

1)s
2

σ
 2
2
χn1, α/2
χn1, 1 - α/2
2
(17  1)(74) 2
(17

1)(74)
 σ2 
28.85
6.91
3,037  σ2  12,683
Converting to standard deviation, we are 95%
confident that the population standard deviation of
CPU speed is between 55.1 and 112.6 Mhz
35/49
Sample PHStat Output
36/49
Sample PHStat Output
(continued)
Input
Output
37/49
Sample Size Determination
Determining
Sample Size
For the
Mean
For the
Proportion
38/49
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
39/49
Sample Size Determination
Determining
Sample Size
For the
Mean
x  z α/2
σ
n
Margin of Error
(sampling error)
ME  z α/2
σ
n
40/49
Sample Size Determination
(continued)
Determining
Sample Size
For the
Mean
ME  z α/2
σ
n
Now solve
for n to get
z σ
n
2
ME
2
α/2
2
41/49
Sample Size Determination
(continued)
 To determine the required sample size for
the mean, you must know:
z σ
n
ME 2
2
α/2
2
 The desired level of confidence (1 - ), which
determines the z/2 value
 The acceptable margin of error (sampling
error), ME
 The standard deviation, σ
42/49
Required Sample Size Example
If  = 45, what sample size is needed to
estimate the mean within &plusmn; 5 with 90%
confidence?
z σ
(1.645) (45)
n

 219.19
2
2
ME
5
2
α/2
2
2
2
So the required sample size is n = 220
(Always round up)
43/49
Sample Size Determination
Determining
Sample Size
For the
Proportion
pˆ  z α/2
pˆ (1 pˆ )
n
ME  z α/2
pˆ (1  pˆ )
n
Margin of Error
(sampling error)
44/49
Sample Size Determination
(continued)
Determining
Sample Size
ME  z α/2
For the
Proportion
pˆ (1  pˆ )
n
pˆ (1 pˆ ) cannot
be larger than
0.25, when p̂ =
0.5
Substitute
0.25 for pˆ (1 pˆ )
and solve for
n to get
0.25 z
n
2
ME
2
α/2
45/49
Sample Size Determination
(continued)
 The sample and population proportions, p̂ 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
46/49
Required Sample Size Example
How large a sample would be necessary
to estimate the true proportion defective in
a large population within &plusmn;3%, with 95%
confidence?
47/49
Required Sample Size Example
(continued)
Solution:
For 95% confidence, use z0.025 = 1.96
ME = 0.03
Estimate P(1 – P) = 0.25
2
α/2
2
0.25 z
n
ME
2
(0.25)(1.96)

 1,067.11
2
(0.03)
So use n = 1,068
48/49
PHStat Sample Size Options
49/49
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