Samples from Normal Populations with Known Variances
If the population variances are known to be σ12 andσ22 , then the
two-sided confidence interval for the difference of the population
means µ1 − µ2 with confidence level 1 − α is
!
r
r
σ12
σ22
σ12
σ22
X̄ − Ȳ + zα/2
+ ,
+
X̄ − Ȳ − zα/2
m
n
m
n
Samples from Normal Populations with Known Variances
In case of known population variances, the procedures for
hypothesis testing for the difference of the population means
µ1 − µ2 is similar to the one sample test for the population mean:
Null hypothesis H0 : µ1 − µ2 = ∆0
Test statistic value
z=
Alternative Hypothesis
Ha : µ1 − µ2 > ∆0
Ha : µ1 − µ2 < ∆0
Ha : µ1 − µ2 6= ∆0
(X̄ − Ȳ ) − ∆0
q
σ12
σ22
m + n
Rejection Region for Level α Test
z ≥ zα (upper-tailed)
z ≤ −zα (lower-tailed)
z ≥ zα/2 or z ≤ −zα/2 (two-tailed)
Large Size Samples
When the sample size is large, both X̄ and Ȳ are approximately
normally distributed, and
Z=
(X̄ − Ȳ ) − (µ1 − µ2 )
q
S12
S22
m + n
is approximately a standard normal rv.
Large Size Samples
In case both m and n are large (m, n > 30), the procedure for
constructing confidence interval and testing hypotheses for the
difference of two population means are similar to the one sample
case.
The two-sided confidence interval for the difference of the
population means µ1 − µ2 with confidence level 1 − α is
s
s


2
2
2
2
S1
S
S1
S
 X̄ − Ȳ − z
+ 2,
X̄ − Ȳ + zα/2
+ 2
α/2
m
n
m
n
Large Size Samples
In case both m and n are large (m, n > 30), the procedures for
hypothesis testing for the difference of the population means
µ1 − µ2 is :
Null hypothesis H0 : µ1 − µ2 = ∆0
Test statistic value
z=
Alternative Hypothesis
Ha : µ1 − µ2 > ∆0
Ha : µ1 − µ2 < ∆0
Ha : µ1 − µ2 6= ∆0
(X̄ − Ȳ ) − ∆0
q
S12
S22
m + n
Rejection Region for Level α Test
z ≥ zα (upper-tailed)
z ≤ −zα (lower-tailed)
z ≥ zα/2 or z ≤ −zα/2 (two-tailed)
Two-Sample t Test and C.I.
Theorem
When the population distributions are both normal, the
standardized variable
T =
(X̄ − Ȳ ) − (µX − µY )
q 2
S2
S
Y
X
m + n
has approximately a t distribution with df ν estimated from the
data by
ν=
s2
X
m
(s 2 /m)2
X
m−1
+
+
s2
Y
n
2
(s 2 /n)2
Y
n−1
(round ν down to the nearest integer.)
Two-Sample t Test and C.I.
The two-sample t confidence interval for µX − µY with
confidence level 100(1 − α)% is given by


s
s
sX2
sX2
sY2
sY2
(x̄ − ȳ ) − t
+ , (x̄ − ȳ ) + tα/2,ν
+ 
α/2,ν
m
n
m
n
A one-sided confidence bound can be obtained by replacing tα/2,ν
with tα,ν .
Two-Sample t Test and C.I.
The two-sample t test for testing H0 : µX − µY = ∆0 is as
follows:
(x̄ − ȳ ) − ∆0
Test statistic value: t = q 2
s
s2
X
Y
m + n
Alternative Hypothesis
Ha : µ X − µ Y > ∆ 0
Ha : µ X − µ Y < ∆ 0
Ha : µX − µY 6= ∆0
Rejection Region for Approximate Level α Test
t ≥ tα,ν (upper-tailed)
t ≤ tα,ν (lower-tailed)
t ≥ tα/2,ν or t ≤ tα/2,ν (two-tailed)
Analysis of Paired Data
Assumptions: The data consists of n independently selected pairs
(X1 , Y1 ), (X2 , Y2 ), . . . , (Xn , Yn ), with E [Xi ] = µ1 and E [Yi ] = µ2 .
Analysis of Paired Data
Assumptions: The data consists of n independently selected pairs
(X1 , Y1 ), (X2 , Y2 ), . . . , (Xn , Yn ), with E [Xi ] = µ1 and E [Yi ] = µ2 .
Let D1 = X1 − Y1 , D2 = X2 − Y2 , . . . , Dn = Xn − Yn , so the Di s
are the differences within pairs.
Analysis of Paired Data
Assumptions: The data consists of n independently selected pairs
(X1 , Y1 ), (X2 , Y2 ), . . . , (Xn , Yn ), with E [Xi ] = µ1 and E [Yi ] = µ2 .
Let D1 = X1 − Y1 , D2 = X2 − Y2 , . . . , Dn = Xn − Yn , so the Di s
are the differences within pairs.
Then the Di s are assumed to be normally distributed with mean
value µD and variance σD2 (this is usually a consequence of the Xi s
and Yi s themselves being normally distributed.)
Paired t Test
The paired t test for testing H0 : µD = ∆0 (where D = X − Y is
the difference between the first and second observations within a
pair, and µD = µ1 − µ2 ) is as follows:
Test statistic value: t =
d̄ − ∆0
√
sD / n
(where d̄ and sD are the sample mean and standard deviation,
respectively, of the di s)
Alternative Hypothesis
Ha : µ D > ∆ 0
H a : µD < ∆ 0
Ha : µD 6= ∆0
Rejection Region for Approximate Level α Test
t ≥ tα,n−1 (upper-tailed)
t ≤ tα,n−1 (lower-tailed)
t ≥ tα/2,n−1 or t ≤ tα/2,n−1 (two-tailed)
Confidence Interval for µD
The Paired t CI for µD with confidence level 100(1 − α)% is
sD
sD
√
√
d̄ − tα/2,n−1 ·
, d̄ + tα/2,n−1 ·
n
n
A one-sided confidence bound results from retaining the relevant
sign and replacing tα/2,n−1 by tα,n−1 .
Analysis of Paired Data
Example: (Problem 40)
Lactation promotes a temporary loss of bone mass to provide adequate
amounts of calcium for milk production. The paper “Bone Mass Is
Recovered from Lactation to Postweaning in Adolescent Mothers with
Low Calcium Intakes” (Amer. J. Clinical Nutr., 2004: 1322-1326) gave
the following data on total body bone mineral content (TBBMC) (g) for
a sample both during lactation (L) and in the postweaning period (P).
L
P
1
1928
2126
2
2549
2885
3
2825
2895
4
1924
1942
5
1628
1750
6
2175
2184
7
2114
2164
8
2621
2626
9
1843
2006
10
2541
2627
Analysis of Paired Data
Example: (Problem 40)
Lactation promotes a temporary loss of bone mass to provide adequate
amounts of calcium for milk production. The paper “Bone Mass Is
Recovered from Lactation to Postweaning in Adolescent Mothers with
Low Calcium Intakes” (Amer. J. Clinical Nutr., 2004: 1322-1326) gave
the following data on total body bone mineral content (TBBMC) (g) for
a sample both during lactation (L) and in the postweaning period (P).
L
P
1
1928
2126
2
2549
2885
3
2825
2895
4
1924
1942
5
1628
1750
6
2175
2184
7
2114
2164
8
2621
2626
9
1843
2006
Does the data suggest that true average total body bone mineral content
during postweaning exceeds that during lactation by more than 25g?
10
2541
2627
Analysis of Paired Data
The differences for the sample are:
-198
-336
-70
-18
-122
-9
-50
-5
-163
with mean d̄ = −105.7 and standard deviation sD = 103.8
-86
Analysis of Paired Data
The differences for the sample are:
-198
-336
-70
-18
-122
-9
-50
-5
-163
with mean d̄ = −105.7 and standard deviation sD = 103.8
The Quantile-Quantile for this sample differences is
-86
Difference Between Population Proportions
Proposition
Let X ∼ Bin(m, p1 ) and Y ∼ Bin(n, p2 ) with X and Y
independent. Then
E [p̂1 − p̂2 ] = p1 − p2
so p̂1 − p̂2 is an unbiased estimator of p1 − p2 . Here p̂1 =
p̂2 = Yn .
Furthermore,
p1 q1 p2 q2
V [p̂1 − p̂2 ] =
+
m
n
where qi = 1 − pi .
X
m
and
Difference Between Population Proportions
Large-Sample Test Procedure
Null hypothesis: H0 : p1 − p2 = 0
Test statistic value (large samples):
p̂1 − p̂2
z=q
p̂q̂ m1 + n1
where p̂ =
X +Y
m+n
=
m
m+n p̂1
Alternative Hypothesis
Ha : p 1 − p 2 > 0
Ha : p 1 − p 2 < 0
Ha : p1 − p2 6= 0
+
n
m+n p̂2 .
Rejection Region for Approximate Level α Test
z ≥ zα (upper-tailed)
z ≤ zα (lower-tailed)
z ≥ zα/2 or z ≤ zα/2 (two-tailed)
Difference Between Population Proportions
Large-Sample Confidence Interval for p1 − p2
The 100(1 − α)% confidence interval for p1 − p2 is given by
!
r
r
p̂1 q̂1 p̂2 q̂2
p̂1 q̂1 p̂2 q̂2
(p̂1 − p̂2 ) − zα/2
+
, (p̂1 − p̂2 ) + zα/2
+
m
n
m
n
Difference Between Population Proportions
Example: (Problem 50)
Do teachers find their work rewarding and satisfying? The article
“Work-Related Attitudes” (Psychological Reports, 1991: 443-450)
reports the results of a survey of 395 elementary school teachers
and 266 high school teachers. Of the elementary school teachers,
224 said they were very satisfied with their jobs, whereas, 126 of
the high school teachers were very satisfied with their work. Is
there any difference between the proportion of all elementary
school teachers who are satisfied and all high school teachers who
are satisfied?