Inference on Two Population
Proportions
1
Overview
Comparing
Means
Two sample
problem
Comparing
Paired
samples
σ1 and σ2
known
Independent
sample
σ1 and σ2
unknown
Proportions
Comparing
Variances
s12
m1
Sample 1
x11, x12, ,,,, x1n1
s 22
m2
Sample 2
x21, x22, ,,,, x2n2
2

Suppose that the two independent random samples of sizes n1 and
n2 are taken from two populations.

Let X1 and X2 represent the number of observations that belong to
the class of interest in samples 1 and 2, respectively.

The estimators of the population proportions
have approximate normal distributions.

The quantity
has approximately a standard normal distribution, N(0,1)

The approximation as reasonable as long as x1, n1-x1, x2 and n2-x2 are
all larger than 5
3
Confidence Intervals

100(1-a)% confidence interval for P1- P2

100(1-a)% upper confidence bound for P1- P2
-1 

100(1-a)% lower confidence bound for P1- P2
 P1 - P2  1
4
Example Legal agreements have been reached whereby if 10% or more of
the building tiles are cracked, then the construction company that
originally installed the tiles must help pay for the building repair costs.
Buildings A revealed a total of 406 cracked tiles out of 6000 tiles.
Another group of buildings, buildings B, of which tiles were cemented
into place with a different resin mixture than that used on building A.
The construction engineers are interested in investigating whether the
two types of resin mixture have different expansion and contraction
properties which affect the chances of the tiles becoming cracked. A
sample of 2000 tiles on buildings B is examined and 83 are found to be
cracked. Let pA and pB be the probabilities that a tile on buildings A
and B becomes cracked, respectively. Construct a 99% confidence
interval for the difference in the probabilities.
5
=0.0120
=0.0404
A two-sided 99% CI for pA-pB is (0.0120, 0.0404).
Note: This CI contains only positive values. => “pA > pB” (at the 99%
confidence level 99%). That is, the resin mix employed on building B
seems to be better than the resin mixture employed on buildings A.
In fact, we are 99% confident that the resin mixture on building A has
a probability of causing a tile to crack between 1.20% and 4.04%
larger than the resin mixture on buildings B.
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Hypothesis Testing

Test statistics for testing H0: P1= P2
: (pooled estimate)
Note: Since the null hypothesis specifies that P1= P2 , it is appropriate to
employ a pooled estimate of the common success probability.


Alternative hypothesis H1: P1≠ P2

Rejection Region:
Z0 > za/2 or Z0 < - za/2

P-value: 2 P( Z0 > | z0 | )
Alternative hypothesis H1: P1 > P2

Alternative hypothesis H1: P1 <P2

Rejection Region: : Z0 > za

Rejection Region: : Z0 < -za

P-value: P( Z0 > z0 )

P-value: P( Z0 < z0 )
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Example When polling the agreement with the statement “ The city mayor
is doing a good job.” the local newspaper is also interested in how a
person’s support for this statement may depend upon his or her age.
Therefore the pollsters also gather information on the ages of the
respondents in their random sample. The polling results consist of
n=952 people aged 18 to 39 of whom x=627 agree with the statement,
and m=1043 people aged at least 40 of whom y=421 agree with the
statement. Does the strength of support for the statement differ
between the two age groups? Use the significance level 0.01.
Let pA be the proportion of the younger group who agree with the
statement and pB be that of the older group. Then the estimates are
We are testing
versus
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The pooled estimate of a common proportion is
The test statistic value is
The rejection region is
Since Our z0 falls in the R.R, we reject Ho at a=0.01. The poll has
demonstrated a difference in agreement with the statement
between the two age groups.
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Example (Montgomery Page 235) Two different types of injection-molding
machines are used to form plastic parts. A part is considered defective if it
has excessive shrinkage or is discolored. Two random samples, each of
size 300, are selected, and 15 defective parts are found in the sample
from machine 1 whereas 8 defective parts are found in the sample from
machine 2.
A.
Is it reasonable to conclude that both machines produce the same
fraction of defective parts, using a=0.05?
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B.
Calculate the P-value for this test.
C.
Construct a 95% confidence interval on the difference in the two
fractions defective.
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Inference on the Ratio of
Variances of Two Normal
Populations
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The Chi Square Distribution

Let X1, …, Xn be a random sample from a normal distribution with
unknown mean m and unknown variance s2. The quantity
has a chi square distribution with n-1 degrees of freedom, abbreviated as
c 2 n-1

The probability density function of a chi-square random variable
is
where k is the number of degrees of freedom and
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Figure 4-19. Probability
density functions of
several chi-square
distributions
Figure 4-20. Percentage point of
the chi-square distribution
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The F Distribution



Let W and Y be independent chi-square random variables with u
and n degrees of freedom, respectively. Then the ratio
has the F distribution with u degrees of freedom in the numerator
and n degrees of freedom in the denominator.
It is usually abbreviated as Fu,n
The probability density function of an F distribution is given by
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Figure 5-4. Probability density
functions of two F distributions
Figure 5-5. Upper and lower
percentage points of the F
distribution
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F분포표:
F0.05,n1,n2
17
F분포표:
F0.10,n1,n2
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Example (Page 230) For an F distribution, find the following:






f
f
f
f
f
f
0.25, 5,10
0.75, 5,10
0.10, 24,9
0.90, 24,9
0.95, 5,10
0.05, 5,10
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
Let X11, X12, …, X1n1 be a random sample from N(m1 , s12 ) and
let X21, X22, …, X2n2 be a random sample from N(m2 , s22 )
Assume that both normal populations are independent.
Let S12 and S22 be the sample variances.

Then the ratio


has an F-distribution with n1-1 numerator degrees of freedom and
n2-1 denominator degrees of freedom.
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Confidence Interval

100(1-a)% confidence interval for s12/s22

100(1-a)% upper confidence bound for s12/s22

100(1-a)% lower confidence bound for s12/s22
21
Example (Montgomery Page 229) A company manufactures impellers for use
in jet-turbine engines. One of the operations involves grinding a particular
surface finish on a titanium alloy component. Two different grinding
processes can be used and both processes can produce parts at identical
mean surface roughness. The manufacturing engineer would like to select
the process having the least variability in surface roughness. A random
sample of n1=11 parts from the first process results in a sample standard
deviation s1=5.1 microinches. A random sample of n2=16 parts from the
second process results in a sample standard deviation s2=4.7 microinches.
Find a 90% CI on the ratio of the two variances s12/s22.
,where
A 90% CI on s12/s22 is
22
Hypothesis Testing




Test statistics for testing H0: s12 = s22
Alternative hypothesis H1: s12 ≠ s22

Rejection Region:
F0 > fa/2, n1-1,n2-1 OR F0 < f 1-a/2, n1-1,n2-1

P-value: 2 min {P( F0 > f0 ), P( F0 < f0 ) }
Alternative hypothesis H1: s12 >s22

Rejection Region: : F0 > fa,n1-1,n2-1

P-value: P( F0 > f0 )
Alternative hypothesis H1: s12 <s22

Rejection Region: : F0 < f1-a, n1-1,n2-1

P-value: P( F0 < f0 )
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Figure The F distribution for the test of
H0: s12 = s22 with critical values for (a) H1: s12 ≠ s22 (b) H1: s12 >s22 and (c) H1: s12 <s22
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Example (Montgomery 227) Oxide layers on semiconductor wafers are
etched in a mixture of gases to achieve the proper thickness. The
variability in the thickness of these oxide layers is a critical characteristic
of the wafer, and low variability is desirable for subsequent processing
steps. Two different mixtures of gases are being studied to determine
whether one is superior in reducing the variability of oxide thickness.
Sixteen wafers are etched in each gas. The sample standard deviations of
oxide thickness are s1=1.96 angstroms and s2=2.13 angstroms,
respectively.
Is there any evidence that either gas is preferable? Use a=0.05.
Rejection Region:
Conclusion: Since f0 does not fall into the R.R., we cannot reject H0
at a=0.05. Therefore, there is no strong evidence to indicate that
either gas is preferable.
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