Sets 28 and 29
Instead of drawing samples from one population, we may take random
samples from two populations so that we can carry out some sort of
comparison.
We may wish to compare
μ
1
and
μ
2
, the population means for populations
1
and
2
. We do so by examining the difference,
μ
1

μ
2
.
Example:
Suppose we wish to compare the
μ
1
, the mean lead content
(in
ppm
) per
mL
in Victoria tap water with
μ
2
, the mean lead content
(in
ppm
) per
mL
in Vancouver tap water.
•
If the means are equal, then
μ
1

μ
2
= 0
.
•
If the means are different, then
μ
1

μ
2
6
= 0
.
•
If the lead content is higher in Victoria, then
μ
1

μ
2
>
0
.
•
If the lead content is higher in Vancouver, then
μ
1

μ
2
<
0
.
•
If the lead content is higher in Victoria by at least
4
ppm
than in
Vancouver, then
μ
1

μ
2
>
4
•
If the lead content is higher in Vancouver by at least
2
ppm
than in
Victoria, then
μ
1

μ
2
<

2
1
The point estimate for
μ
1

μ
2
we will use is
x
1

x
2
.
The following pivotal quantities apply to a variety of cases where two
samples are drawn
independently
from
two
populations.
As before, any confidence interval we construct will have the form
estimate
±
(c.v.)(e.s.e.)
All of the pivotal quantities below have the form
estimate

parameter
e
.
s
.
e
.
.
Large Sample Size Procedures:
Z
=
(
x
1

x
2
)

(
μ
1

μ
2
)
s
s
2
1
n
1
+
s
2
2
n
2
Assumptions:
•
Independent random samples from two populations.
•
Both
sample sizes are large (
n
1
≥
40
, n
2
≥
40)
and the population
standard deviations are unknown.
•
Populations may have any distribution.
2
Example:
We wish to compare the cube compressive strength (in
N/mm
2
)
of two types of concrete. The summary statistics are as follows:
sample size
sample mean
sample sd
Type A
70
31.9
1.4
Type B
50
35.6
2.1
Test the research hypothesis that the two types of concrete have different
mean cube compressive strengths.
3
Small Sample Size Procedures:
In the case where
n
1
or
n
2
(or both)
are small, there are two choices of test statistic. This choice depends on
whether or not we assume that
σ
1
=
σ
2
.
While there are hypothesis tests that we could carry out to investigate
whether or not
σ
1
=
σ
2
, we will use the following rule for our class:
•
Using
s
1
, s
2
, calculate the
larger
standard deviation divided by the
smaller
•
If this value is
less than 1.4
, we assume
σ
1
=
σ
2
. If it is
greater
than 1.4
, we assume
σ
1
6
=
σ
2
.
Note that we only need to decide whether or not
σ
1
=
σ
2
when the sample
size is
not
large.
4
Pooled Procedures:
t
n
1
+
n
2

2
=
(
x
1

x
2
)

(
μ
1

μ
2
)
s
(
n
1

1)
s
2
1
+ (
n
2

1)
s
2
2
n
1
+
n
2

2
1
n
1
+
1
n
2
Assumptions:
•
Independent random samples from two populations. At least one of
the sample sizes is small, and the population standard deviations are
unknown.
•
We know that (or assume that)
σ
1
=
σ
2
•
Both populations have normal (or approximately normal) distribu
tion.
Comments:
The value
(
n
1

1)
s
2
1
+ (
n
2

1)
s
2
2
n
1
+
n
2

2
is sometimes denoted
by
s
2
p
, and is called the
pooled variance estimate
.
Recall that for pooled procedures, we assumed that
σ
1
=
σ
2
. The value
of
s
2
p
is the estimate for both
σ
2
1
and
σ
2
2
.
5
Unpooled Procedures:
t
γ
=
(
x
1

x
2
)

(
μ
1

μ
2
)
s
s
2
1
n
1
+
s
2
2
n
2
where
γ