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CHAPTER 9
Homework: 1,6,15,24,47,49,53,57
Sec 9.1: Sampling Distribution for the
difference of two sample means:
1. Sampling distribution for the difference of two
independent sample means:
(a). The central limit theory ensures that the
sampling distribution for the difference of two
independent sample means is approximately
normal for sufficiently large samples.
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(b). The difference of the two sample means is an
unbiased estimator for the difference of the two
population means.
(c). The standard deviation for the difference of two
independent sample means is
s d = s 12  s 22
n1
n2
where s1 and s2 are the population standard
deviations and n1 and n2 are the respective sample
sizes.
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2. Problem of two paired samples
Sometimes the two samples are paired. For
example, suppose x1, x2, .., xn are weights of n
people before they are enrolled onto a dietary
course, and y1,y2,..,yn are the corresponding
weights of these n people after they complete the
dietary course. Here the two samples are not
independent, because xi and yi are the weights of
the same person -- a paired measurements. We call
these two samples paired samples. In order to
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assess the usefulness of the dietary course, we
would be interested in making inference on the
mean weight reduction after completing the course,
d. For this, we can base our inference on y1-x1, y2x2,..,yn-xn, which can be regarded as a random
sample taken from the population of weight
reduction after completing the course. Therefore
the problem is reduced to a one-sample problem
which has already been dealt with in chapters 7 and
8. We will not discuss two-paired-sample problem
further in this chapter.
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Sec 9.2: Large sample inference for the difference
of two independent population means
(1) The confidence interval
Under the assumption that two random samples
taken independently from two populations have
large enough sample sizes, the 1- confidence
interval for difference of the two population means
is given by
x1  x2  z / 2 sd
where
sd = s12  s22
n1
n2
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(2) Hypotheses testing:
(a). Alternative Hypothesis:
(i). Two-Tailed Test: Ha: (m1  m2) d0
(ii). Right-Tailed Test: Ha: (m1  m2) >d0
(iii). Left-Tailed Test: Ha: (m1  m2) <d0
(b). Null Hypothesis:
(i). Two-Tailed Test: H0: (m1  m2) =d0
(ii). Right-Tailed Test: H0: (m1  m2) d0
(iii). Left-Tailed Test: H0: (m1  m2) d0
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(c). Test Statistic
x1  x2  d 0
Z=
sd
(d). Rejection Region
A size  test has the rejection region
(i).Two-Tailed Test: Z > Z/2or Z < -Z/2 ,
(ii). Right-Tailed Test: Z > Z ,
(iii). Left-Tailed Test: Z < -Z.
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(e). p-values of this test:
(i). Two-tailed test: p-value=2*P(z > |Z|)
(ii). Right-tailed test:p-value=P(z > Z)
(iii). Left-tailed test:p-value=P(z < Z)
A size  test rejects the null hypothesis if and only
if the p-value is less than , and this test always
reaches the same rejection/acceptance decision as
the test in (d).
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(EX 9.1) (Basic -- Large Sample Inference)
Two independent random samples were
selected from two independent populations. The
sample sizes, sample means, and sample variances
are as follows:
I
II
Sample Sizes
35
49
Sample Mean
12.7
7.4
Sample Variance
1.38
4.14
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(a). Can we use the large sample inference
procedure? Explain.
(b). Find a 95% confidence interval for the mean
difference.
(c). Test that the difference of these two means is
larger than 5 at =0.05. Report the p-value and
draw your conclusion.
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(EX 9.2) (Basic)
An experiment was conducted to compare two
diets A and B designed for weight reduction. Two
groups of 30 overweight dieters each was
randomly selected. One group was placed for diet
A and another on diet B and their weight loss was
recorded over a thirty-day period. The result is
presented in table 1.
Table I Weight Loss
Diet A
Diet B
sample mean
21.3
13.4
sample s.d.
2.6
1.9
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(a). Can we used the large sample inference
procedure? Explain.
(b). Find a 90% C.I. for the difference of mean
weight loss for these two diets.
(c). Does the difference larger than 9 at =0.01.
Report the p-value and draw your conclusion.
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Sec 9.3: Small sample inference for the
difference of two independent population
means
Assumptions:
two independent populations are normally
distributed;
the two populations have a common variance.
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The pooled estimator, Sp2 , of the common
variance is given by
2
2
(
n

1
)

S

(
n

1
)

S
1
2
2 .
S 2p = 1
n1  n2  2
where n1 and n2 are the sample sizes.
The standard deviation of the difference of the two
sample mean is given by
1
2 1
S ( x1  x2 ) = S p (  ) .
n1 n2
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1. Confidence Interval
The (1)*100% confidence interval for the
population mean difference is given by
( x1  x 2 )  t / 2 ,n1  n2 2 , s( x1  x2 ) .
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2 Hypotheses Testing
(a). Alternative Hypothesis:
(i). Two-Tailed Test: Ha: (m1  m2) d0
(ii). Right-Tailed Test: Ha: (m1  m2) >d0
(iii). Left-Tailed Test: Ha: (m1  m2) <d0
(b). Null Hypothesis:
(i). Two-Tailed Test: H0: (m1  m2) =d0
(ii). Right-Tailed Test: H0: (m1  m2) d0
(iii). Left-Tailed Test: H0: (m1  m2) d0
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(c) Test statistics is given by
( x1  x 2 )  d 0
tc =
.
s( x1  x2 )
(d) Rejection region
A size  has the the rejection region
(i).Two-Tailed Test: t c> t/2,n1+n2-2 or t c< -t/2,n1+n2-2 ,
(ii). Right-Tailed Test: t c > t,n1+n2-2 ,
(iii). Left-Tailed Test: t c < -t,n1+n2-2.
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(e) p-values of this test
(i). Two-tailed test: p-value=2*P(tn1+n2-2 > |tc|)
(ii). Right-tailed test:p-value=P(tn1+n2-2 > tc)
(iii). Left-tailed test:p-value=P(tn1+n2-2 < tc)
 test rejects the null hypothesis if and
only if the p-value is smaller than .
A size
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(EX 9.3)
In a study of iron deficiency among infants,
samples of infants following different feeding
regimens were compared. One group contained
breast-fed infants, while the children in another
group were fed a standard baby formula without
any iron supplements. Here are summary results
on blood hemoglobin levels at 12 months of age.
Group
Breast-fed
Formula
n
mean
s.d.
23
19
13.3
12.4
1.7
1.8
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(a). Set up a test to test that the mean hemoglobin
level is higher among breast-fed babies?
(i). State the null and alternative hypothesis.
(ii). Carry out an appropriate test.
(iii). What kind assumptions do you need to
carry out your test?
(iv). Give the P-value and state your
conlusions at  = 0.05.
(b). Give a 95% C.I. for the mean difference in
hemoglobin level between these two populations
of infants.
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(EX 9.4)
The following table presented the
summary statistics from an experiment for
decreasing the blood pressure.
Group
Treatment
n
1
2
Calcium
Placebo
10
11
21
mean
decrease
s.d.
5.0
-.273
8.743
5.901
(a). Set up a test to test that the mean blood pressure
decreased is larger in Calcuim group?
(i). State the null and alternative hypotheses.
(ii). Carry out an appropriate test.
(iii). What kind assumptions do you need?
(iv). Give the P-value and state your
conlusions at  = 0.05.
(b). Give a 95% C.I. for the mean difference in
blood presure decreased between the two
populations.
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Sec 9.4: Large sample inference for the
difference of two independent population
proportions
We need to assume that the two samples are selected
from two independent binomial populations and the
both sample sizes are sufficiently large.
The difference of the two sample proportions is
approximately normally distributed with mean (p1-p2)
and variance
p1 (1  p1 ) p 2 (1  p 2 )
2
Sp  p =

.
1
2
n1
n2
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1. Confidence interval
A (1)*100% confidence interval for the difference
of two population proportions is given by
( p 1  p 2 )  z /2  S ( p 1 p 2 ).
where
S (p 1 p 2 ) =
p 1 (1  p 1 ) p 2 (1  p 2 )

.
n1
n2
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2 Hypotheses Testing
(a). Alternative Hypothesis:
(i). Two-Tailed Test: Ha: (m1  m2) 0
(ii). Right-Tailed Test: Ha: (m1  m2) >0
(iii). Left-Tailed Test: Ha: (m1  m2) <0
(b). Null Hypothesis:
(i). Two-Tailed Test: H0: (m1  m2) =0
(ii). Right-Tailed Test: H0: (m1  m2) 0
(iii). Left-Tailed Test: H0: (m1  m2) 0
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(c) Test Statistics:
p1  p 2
Z=
s p1  p 2
(d) Rejection Region of this test:
A size  test has the rejection region
(i).Two-Tailed Test: Z > Z/2or Z < -Z/2 ,
(ii). Right-Tailed Test: Z > Z ,
(iii). Left-Tailed Test: Z < -Z.
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(e) p-values of this test:
(i). Two-tailed test: p-value=2*P(z > |Z|)
(ii). Right-tailed test:p-value=P(z > Z)
(iii). Left-tailed test:p-value=P(z < Z)
A size  test rejects the null hypothesis if and only
if the p-value is smaller than  .
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(EX 9.5) (Basic)
Samples of n1 = n2 = 500 observations were
selected from two independent binomial
populations, and x1 = 120 and x2 = 147 were
observed.
(a). Find the 95% C.I. for the difference of the two
proportions.
(b). Are these two proportions different at 0.05
level?
(c). What assumptions do we need to use a large
sample inference procedure?
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(EX 9.6). (Applications)
A sampling of political candidates -- 200
randomly chosen from the west and 200 from the
east – was classified according to whether the
candidate received backing by a national labor
union and whether the candidate won. A summary
of the data is shown below.
West
East
Winners (union)
120
142
(a). Find a 95% C.I. interval for the difference of the
proportions of union-backed winners in the west
verse the east.
(b). Is this difference significant at =0.05?
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