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Statistical Inference: Confidence Intervals & Hypothesis Tests for Two Populations - Prof., Study notes of Data Analysis & Statistical Methods
An overview of statistical inference methods for determining the difference between two population proportions or variances. It covers the concepts of confidence intervals and hypothesis tests, including the expected values, formulas, and critical values for independent sampling. The document also includes examples of calculating confidence intervals and performing hypothesis tests using given data.
Typology: Study notes
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Chapter
- Part
B
Inferences
Based
on
Two
Samples:
Confidence
Intervals
and
Tests
of
Hypotheses
Identifying
the
Target
Parameter
p
‐
p
Difference
between
proportionsproportions
,
percentages,
fractions
or
rates;
compare
proportions
Qualitative
Data
•
Two
groups
may
or
may
not
have
similar
proportions
regarding
particular
characteristics.
•
We
can
make
inference
about
p
1
and
p
2
by
examining
and
.
•
Methods
based
on
large
sample
sizes!
2
1
ˆ
ˆ
p
p
Comparing
Two
Population
Proportions: Independent Sampling
The
Sampling
Distribution
of
2
1
ˆ
ˆ
p
p
−
Proportions:
Independent
Sampling
The
expected
value
of
the
sampling
distribution
is:
) ( ) ˆ ˆ ( 2
1
2
1
p
p
p
p
E
−
=
−
If
the
two
samples
are
independent,
the
standard
deviation
of the difference of their proportions is:
)
(
)
(
2
1
2
1
p
p
p
p
of
the
difference
of
their
proportions
is:
2
2
2
1
1
1
2
2
2
1
1 1 ) ˆ ˆ (
ˆ ˆ ˆ ˆ 2 1
n
q
p
n
q
p
n
q
p
n
q
p
p
p
- ≅ + = − σ
If
the
samples
are
large,
the
sampling
distribution
is
approximately
normal,
i.e.
2
1
2
1
pp
y
,
) , ( ~ ˆ ˆ
) ˆ ˆ ( 1 1 2 1 2
1
p p p p N p p
−
−
−
σ
Comparing
Two
Population
Proportions: Independent SamplingProportions:
Independent
Sampling
Large
Sample
100(
‐α
)%
Confidence
Interval
for
2
1
ˆ
ˆ
p
p
−
2 2 1 1 2 / 2 1 ˆ ˆ 2 / 2 1
) ˆ ˆ ( ) ˆ ˆ (
q p q p z p p z p p
p
p
+ ± − = ± −
α
α
σ
2
2
1
1
2 1 2 / 2 1 2 / 2 1
ˆ ˆ ˆ ˆ ) ˆ ˆ (
)
(
)
(
2
1
q p q p z p p
n n p p p p
p
p
+
±
−
≅
−
α
α
2 1 2 / 2 1
)
(
n
n
z
p
p
+
±
≅
α
Comparing
Two
Population
Proportions: Independent Sampling
One
‐
Tailed Test
Two
‐
Tailed Test
Proportions:
Independent
Sampling
Large
sample
test
of
hypothesis
about
p
1
–
p
2
One
‐
Tailed
Test
H
0
:
(
p
1
‐
p
2
)
=
D
0
H
a
:
(
p
1
‐
p
2
)
>
D
0
(
<
D
0
)
Two
‐
Tailed
Test
H
0
:
(
p
1
‐
p
2
)
=
D
0
H
a
:
(
p
1
‐
p
2
)
≠
D
0
a
(
p
1
p
2
)
0
(
0
)
Rejection
region:
|
z
o
|>
z
α
a
(
p
1
p
2
)
0
Rejection
region:
|
z
o
|
> z
α
/
Test
Statistic:
0
2
1
)
ˆ
ˆ
(
o
D
p
p
z
−
−
=
⎞ ⎟⎟⎠
⎛ ⎜⎜⎝
≅
=
−
) ˆ ˆ ( 2 1
1
1
ˆ
ˆ
ˆ
with
2
1
q
p
x
x
p
n
n
p
p
n
σ
Conditions:
The
two
samples
must
be
independent
of
each
other.
)
ˆ
ˆ (
2
1
p
p
o
−
σ
⎟⎠
⎜⎝
2 1 ) ( 2 1
2
1
n
n
n
n
n
n
p
p
n
The
sample
sizes
must
be
large,
i.e.
15
ˆ
and
15
ˆ
15
ˆ
and
15
ˆ
2
2
2
2
1
1
1
1
≥
≥
≥
≥
q
n
p
n
q
n
p
n
A
i
ti t d
id d t
l
t
th
i
l
f
it
t
th
li
ti
f
Example
A
scientist
d
ecided
t
o
evaluate
th
e
survival
of
mites
t
o
th
e
application
of
a
new
insecticide.
Because
it
is
known
that
males
and
females
can
have
different
reactions
to
this
chemical
individuals
of
each
sex
were
l
t d Sh
l
t d
h i di id
l
it
i t
ll
i
f
li d
selected.
Sh
e
l
ocated
each
i
ndividual
mite
i
nto
a
small
piece
of
rug,
applied
the
recommended
dose
of
insecticide
and
after
hour
the
status
of
the
insect
(alive/dead)
was
recorded.
The
summary
table
is
shown
below:
Alive
Dead
TOTAL
Male
20
30
50
Female
5
45
50
TOTAL
25
75
100
a)
Obtain
a
90%
confidence
interval
for
the
population
difference
in
proportion
of
dead
mites
between
females
and
males.
b)
Conduct
a
test
to
evaluate
if
there
is
significant
difference
in
the
proportion
of
dead
mites
between
females
and
males.
Use
α
=
0.10.
Example
Alive
Dead
TOTAL
Male
20
30
50
Female
5
45
50
TOTAL
25
75
100
a)
Obtain
a
90%
confidence
interval
for
the
population
difference
in
=
=
f
p
.
0
ˆ
α
proportion
of
dead
mites
between
females
and
males.
=
m
ˆ p
=
=
m
m
f
f
q
p
q
p
ˆ ˆ ˆ ˆ ˆ σ
=
=
−
m
f
p
p
n
n
m
f^
ˆ
ˆ
σ
=
±
−
−
m
f^
p
p
m
f
z
p
p
ˆ
ˆ
2 /
ˆ
)
ˆ
ˆ
(
σ
α
m
f
Example
Alive
Dead
TOTAL
Male
20
30
50
Female
5
45
50
TOTAL
25
75
100
a)
Obtain
a
90%
confidence
interval
for
the
population
difference
in
.
0
.
0
/
ˆ
=
=
=
f
p
α
proportion
of
dead
mites
between
females
and
males.
.
0
/
ˆ
=
=
m
p
0812
0
40 .
0
60 .
0
10 .
0
90
. 0 ˆ ˆ ˆ ˆ ˆ
= × + × = + =
m
m
f
f
q
p
q
p
σ
0812 .
0
50
50
ˆ
ˆ
= + = + = −
m
f
p
p
n
n
m
f
σ
.
0
.
)
.
0
. 0 ( ˆ ) ˆ ˆ (
ˆ
ˆ
2 /
× ± − = ± −
−
m
f^
p
p
m
f
z
p
p
σ
α
]
.
0
;
.
0
[
.
0
.
0
=
±
=
m
f^
p
p
f
Example
Alive
Dead
TOTAL
Male
20
30
50
Female
5
45
50
TOTAL
25
75
100
b)
Conduct
a
test
to
evaluate
if
there
is
significant
difference
in
the
ti
f d
d
it
b t
f
l
d
l
U
0 10
60 .
0
50
/
30
ˆ
10 .
0
90 .
0
50
/
45
ˆ
=
=
=
=
=
f
p p
α
H
0
:
p
f
=
p
m
H
a
:
p
roportion
of
d
ead
mites
b
etween
f
emales
and
males.
U
se
α
=
0
.10.
60 .
0
50
/
30
m
p
a
:
=
=
m
f
m
f
n
n
n
x
x
ˆ p
)
ˆ
ˆ
(
=
⎞⎟ ⎟⎠
⎛⎜ ⎜⎝
=
−
m f n n p p n n q p m f
1
1
ˆ
ˆ
ˆ
)
ˆ
ˆ (
σ
=
−
=
−
)
ˆ
ˆ ( ˆ
)
ˆ
ˆ
(
m
f^
p
p
m
f
o
p
p
z
σ
Example
Alive
Dead
TOTAL
Male
20
30
50
Female
5
45
50
TOTAL
25
75
100
b)
Conduct
a
test
to
evaluate
if
there
is
significant
difference
in
the
ti
f d
d
it
b t
f
l
d
l
U
0 10
60 .
0
50
/
30
ˆ
10 .
0
90 .
0
50
/
45
ˆ
=
=
=
=
=
f
p p
α
H
0
:
p
f
=
p
m
H
a
:
p
f
≠
p
m
p
roportion
of
d
ead
mites
b
etween
f
emales
and
males.
U
se
α
=
0
.10.
60 .
0
50
/
30
m
p
a
:
p
f
≠
p
m
75 . 0
100
45
30
ˆ
=
=
=
m
f
m
f
n
n
n
x
x
p
)
ˆ
ˆ
(
0866 . 0
50
1
50
1
25 . 0
75
. 0 1 1 ˆ ˆ ˆ
)
ˆ
ˆ (
=
⎞⎟ ⎠
⎛⎜ ⎝
×
=
⎞⎟ ⎟⎠
⎛⎜ ⎜⎝
=
−
m f n n p p n n q p m f σ
464 . 3
0866 . 0
60 .
0
90 . 0
ˆ
)
ˆ
ˆ
(
)
ˆ
ˆ (
=
−
=
−
=
−
m
f^
p
p
m
f
o
p
p
z
σ
3.464 = |
z
o
|
> z
α
/
=
Hence, we reject
H
0
.
Determining
the
Sample
Size
•
Calculations
are
different
for
two
samples
as
we
have
two
sources
of
variability
!
•
With
a
g
iven
level
of
confidence,
and
a
specified
sampling
error
(
E
)
,
g
p
p
g
(
)
it
is
possible
to
calculate
the
required
sample
size.
‐
Typically,
n
1
=
n
2
=
n
•
Sample
size
needed
to
estimate
(
μ
1
‐
μ
2
)
‐
Based
on:
)
,
(
~
2 1 2 1 2 1 x
x
N
x
x
−
−
−
σ
μ
μ
‐
Given
α
and
the
sampling
error
(
SE
)
required:
(^22)
2 1
2
2 /
)
(
)
(
z
σ
σ
α
(^22)
2 1
2
2
2 /
)
(
)
(
2
1
t
n
n
σ
σ
α
−
−
‐
Estimates of
σ
1
2
and
σ
2
2
will be needed.
2
2 1 2 / 2 1
)
(
)
(
SE
z
n
n
σ
σ
α
=
=
2
2
1
2 /
2
1
)
(
)
(
SE
t
n
n
σ
σ
α
=
=
Estimates
of
σ
1
and
σ
2
will
be
needed.
‐
For
small
sample
sizes,
we
use
t
α
/2,
n1+n
‐
2
and
an
iterative
process.
A
i
d
d
i
h
ff
d i
l
f
Example
An
experiment
was
d
one
to
d
etermine
the
effect
on
d
airy
cattle
of
a
diet
supplement.
Two
groups
were
studied:
standard
diet
and
liquid
whey
diet
.
Suppose
that
we
test
the
hypothesis
of
difference
in
mean
hay
consumption.
Determine
the
approximate
number
of
cattle
that
should
be
included
in
each
g
roup
if
we
want
a
margin
error
of
at
most
±0.
for
a
95%
g
p
g
confidence
interval.
Previous
experimentation
has
shown
σ
to
be
approximately
0.8. 50
0
.
0
±
=
=
α^ SE
.
0
.
0
2
1
≅
=
=
±
σ
σ
SE σ
Example
0
0
±
α
SE
.
0
.
0
.
0
2
1
≅
=
=
±
=
=
σ
σ
σ
α
SE
)
(
)
(
(^22)
2 1
2
2 /
+
z
σ
σ
α
.
)
.
0
.
0
(
.
)
(
)
(
2
2
2
2
2
1
2
2 1 2 / 2 1
≅
=
+
=
=
+
=
=
=
n
n
SE
z
n
n
n
σ
σ
α
0
.
(^2)
2
1
If
we
do
this
experiment
we
might
want
to
use
a
t
distribution
with
df = n
- n
2 = 38
hence
t
= 2 024
and
df
=
n
1
+
n
2
– 2
=
,
hence
t
0.025,
38
=
.
and 21
97 .
20
0 50
)
80 .
0
80 .
0 (
024 .
2
2
2
2
2
2
1
≅ = + = = =
n
n
n
t
0.025,
40
=
0
.
2
How
about
the
power of
the
test?
This
needs
to
be
checked
too!
(later)
Determining
the
Sample
Size
•
Sample
size
needed
to
estimate
(
p
1
‐
p
2
)
‐
Based
on:
) , ( ~ ˆ ˆ
) ˆ ˆ ( 1 1 2 1 2
1
p p p p N p p
−
−
−
σ
‐
Given
α
and
the
sampling
error
(
E
)
required:
)
,
(
) ( 1 1 2 1 2
1
p p p p p p
2
2
2 2 1 1 2 2 / 2 1
)
(
)
(
SE
q p q p z n n
+
=
=
α
‐
Estimates
of
p
1
and
p
2
will
be
needed!
Use
conservative
values
of:
p
1
=
p
2
=
0.
For small sample sizes we do not use t but we need the general
‐
For
small
sample
sizes
,
we
do
not
use
t
,
but
we
need
the
general
condition:
1
ˆ
d
1
ˆ
15
ˆ
and
15
ˆ
1
1
1
1
≥
≥
q
n
p
n
1
5
ˆ
and
15
ˆ
2
2
2
2
≥
≥
q
n
p
n
Identifying
the
Target
Parameter
σ
/
σ
Ratio
of
variances;
difference
in
variability
or
spread;
compare
variation
Quantitative
Data
•
Since
variances
do
not
follow
a
normal
distribution
(due
to
the squaring involved) we use a different technique forthe
squaring
involved)
we
use
a
different
technique
for
inferences
about
variances.
Comparing
Two
Population
Variances:
Independent Sampling
•
Goal:
compare
variances
between
populations
2
Independent
Sampling
Parameter:
(equal
to
when
variances
are
equal)
Estimator:
(observed ratio of sample variances)
(^2122)
σ σ
2 1 s
Estimator:
(observed
ratio
of
sample
variances)
•
Distribution
of
(a
multiple)
of
this
estimator
(under
the
assumption
h
h
l d
)
(^22) 1 s
that
we
h
ave
normal
d
ata)
is:
df
and
df
with
~
(^22)
2 1
2 1
2 1
−
=
−
=
=
n n F s s s
σ
•
Special case: If
then
df
df
~
2 1
2
2
−
=
−
=
=
n n F s σ σ
df
and
df
with
2
2
1
1
(^22)
2 1
(^22)
(^22)
=
=
=
n
n
F
s
σ
σ
σ
Special
case:
If
then
df
df
~
2
2
1
1
(^22)
2
1
=
=
=
n n F s σ σ
Comparing
Two
Population
Variances:
Independent Sampling
•
If
the
ratio
of
sample
variances
gets
too
far
from
(in
either
direction) the likelihood that both samples share the same
Independent
Sampling
direction)
the
likelihood
that
both
samples
share
the
same
population
variance
drops.
Properties
of
the
F
‐
distribution
•
Take
on
positive
density
over
the
range (
∞
)
range
(
,
∞
)
•
Cannot
take
on
negative
values.
•
Non
‐
symmetric
(skewed
right).
•
Indexed by two degrees ofIndexed
by
two
degrees
of
freedom
(df
1
and
df
2
).
Comparing
Two
Population
Variances:
Independent SamplingIndependent
Sampling
(
‐α
)100%
Confidence
Interval
for
(^22)
2 1
/
σ
σ
Ob
i
i
f
l
i
F
2
/
2
(
/
)
2
•
Ob
tain
ratio
of
sample
variances
F
o
=
s
1
2
/
s
2
2
=
(
s
1
/
s
2
)
2
.
(for
convenience
the
larger
sample
variance
is
put
in
the
numerator)
•
For
a
given
α
obtain:
F
F
(l
/
i
th
t il)
F
U
=
F
a
/2,
n
‐
1,
n
‐
1
(leaves
α
/
i
n
th
e
upper
t
ail)
F
L
=
F
1
‐
a
/2,
n
‐
1,
n
‐
1
=
1/
F
a
/2,
n
‐
1,
n
‐
1
(leaves
α/
in
the
lower
tail)
•
Compute (
‐
α
)%
CI:
Compute
(
α
)%
CI:
⎤ ⎥ ⎦
⎡ ⎢ ⎣
U
L
F
s s
F
s s
2 2 1
2 2 1
,
•
Conclude
population
variances
are unequal if interval does not
⎦
⎣
s
s
2
2
are
unequal
if
interval
does
not
contain
Example
(Part
I)
C
ffi i l
d
b
h
l
h
f i
i
l
Company
officials
are
concerned
about
the
l
ength
of
time
a
particular
drug
retains
its
potency.
A
random
sample
(
sample
)
of
bottles
of
the
product
is
drawn
from
the
current
production
and
analyzed
for
(
)
potency.
A
second
sample
(
sample
)
of
bottles
is
obtained,
stored
for
year,
and
then
analyzed.
The
summary
readings
are:
.
0
.
2
2 2 2
=
≈
= =
s
n^ x
σ
.
0
.
1
1 1 1
=
≈
= =
s
n^ x
σ
.
0
2
2
s
σ
.
0
1
1
s
σ
Obtain
a
95%
confidence
interval
for
the
ratio
of
the
population
variances,
and
use
it
to
test
its
equality.
Example
.
0
.
2
2 2 2
=
≈
= =
s
n^ x
σ
.
0
.
1
1 1 1
=
≈
= =
s
n^ x
σ
2
2
1
1
=
=
(^222) 1
0
s s
F
F
U
=
F
a
/2, n1-1, n2-
=
F
L
=
F
1-
a
/2,
n1-1, n2-
=
⎤
⎡
2
2
=
⎤ ⎥ ⎦
⎡ ⎢ ⎣
U
L
F
s s
F
s s
(^222) 1
(^2122)
,
F
‐
distribution
Critical values of theCritical
values
of
the
F
‐
distribution
(
α
=
0.025)
Example
.
0
.
2
2 2 2
=
≈
= =
s
n^ x
σ
.
0
.
1
1 1 1
=
≈
= =
s
n^ x
σ
2
2
1
1
.
.
0
.
0
2 2
(^2122)
0
=
=
=
s s
F
F
U
=
F
a
/2, n1-1, n2-1
=
F
0.025, 49, 49
= 1.762
F
L
=
F
1-
a
/2,
n1-1, n2-1
= 1/
F
0.025, 49, 49
= 1/ 1.762 = 0.567
2
2
⎤
⎡
[
]
[
]
133 .
3 ,
009 .
1
762 .
1
778 .
1 ,
567 .
0
778 .
1
,
(^222) 1
(^2122)
=
×
×
=
⎤ ⎥ ⎦
⎡ ⎢ ⎣
U
L
F
s s
F
s s
H
:
σ
2
/
σ
2
= 1
H
:
σ
2
=
σ
2
H
0
:
σ
1
/
σ
2
= 1
H
a
:
σ
1
2
/
σ
2
2
≠
H
0
:
σ
1
=
σ
2
H
a
:
σ
1
2
≠
σ
2
2
Because 1.0 is not contained in our 95% confidence interval
then we reject
H
0
!
Example
data
drug;
data
drug;
input
Sample
$
Resp
@@;
datalines; 1
10.74
2
9.50
1
10.03
2
10.11
1
10.68
2
9.31
proc
ttest
data=drug
alpha=0.05;
class
Sample;
var
Resp;
run;
... 1
10.73
2
9.87
1
10.10
2
9.73
1
10.67
2
9.65
;
≅
1.778
F
‐
test
of
Hypotheses
for
Equal
One
‐
Tailed
Test
Two
‐
Tailed
Test
Population
Variances
[
]
(^22)
2 1
(^22)
2 1
(^22)
2 1
0
: :
σ
σ
σ
σ
σ
σ
>
= <
a
H H
(^22)
2 1
(^22)
2 1
0
: :
σ
σ
σ
σ
= ≠
a
H H
Test
statistic:
Test
statistic:
(^22)
2 1
2 2 2
0
:
σ
σ
<
=
a
H
if
s
F
(^2) max
s
F
(^22)
2 1
2 1 2
0
2
1
(^21)
0
:
σ
σ
>
=
a a
H
if
s s
F
f
s
max^2 min
0
s
F
=
Rejection
region:
Rejection
region:
2 s
α
F
F
>
0
2 /
0
α
F
F
>
(
p
‐
values
can
also
be
calculated).
(for
convenience
the
larger
sample
variance
is
put
in
the
numerator)