BUS1200-6
1
Part 6
Comparison of Two Populations
1.
Inferences about the Difference between
Two Normal Population Means:
Independent Samples
2.
Inferences about the Difference between
Two Normal Population Means: Paired
Samples
Section 6.1
Inferences about the
Difference between Two Normal
Population Means: Independent Samples
Assumption
The two populations follow
N(
µ
1
,
σ
1
2
) and N(
µ
2
,
σ
2
2
) respectively.
Two random samples, one from each of the
two populations, are drawn independently
.
Let
n
1
and
n
2
be the two sample sizes,
1
X
and
2
X
be the two sample means, and
S
1
2
and
S
2
2
be the two sample variances. It can be proved
BUS1200-6
2
that
2
1
X
X
−
follows
⎟
⎠
⎞
⎜
⎝
⎛
+
−
2
2
2
1
2
1
2
1
,
N
n
n
σ
σ
µ
µ
.
Therefore
2
1
X
X
−
is an unbiased point
estimator of
µ
1
−
µ
2
.
Case 1:
σ
1
2
and
σ
2
2
are known
Interval estimation
The 100(1
−
α
)% confidence interval for
µ
1
−
µ
2
is
⎠
⎞
⎜
⎜
⎝
⎛
+
+
−
+
−
−
2
2
2
1
2
1
2
/
2
1
2
2
2
1
2
1
2
/
2
1
,
n
n
z
x
x
n
n
z
x
x
σ
σ
σ
σ
α
α
.
Example 1
Construction workers earned an
average of $551 per week and manufacturing
workers earned an average of $487 per week.
The
above
mean
earnings
are
actually
obtained from samples of 500 and 700
workers taken from the two groups of
workers respectively. Further assume that the
standard deviations of weekly earnings of the
two groups of workers are $66 and $60
respectively. Construct a 95% confidence
BUS1200-6
3
interval for the difference between the mean
weekly earnings of the two groups of workers.
[Solution]
Refer to weekly earnings of all
construction workers as population 1 and
those
of
all
manufacturing
workers
as
population 2. The respective samples are then
samples 1 and 2. From the given information,
n
1
=
500,
1
x
=
551,
σ
1
=
66,
n
2
=
700,
2
x
=
487,
σ
2
=
60,
α
=
0.05.
2
1
x
x
−
=
551
−
487
=
64,
700
60
500
66
960
.
1
2
2
2
2
2
1
2
1
2
/
+
=
+
n
n
z
σ
σ
α
=
7.30,
64
−
7.30
=
56.70,
64
+
7.30
=
71.30.
Thus, the 95% confidence interval for the
difference in the mean weekly earnings of all
construction workers and all manufacturing
workers is ($56.70, $71.30).
Hypothesis test
The test statistic is
(
)
2
2
2
1
2
1
2
1
n
n
X
X
σ
σ
+
−
.
Let
α
be the level of significance.
BUS1200-6
4
H
0
H
1
Reject
H
0
if and only if
µ
1
=
µ
2
or
µ
1
≥
µ
2
µ
1
<
µ
2
(
)
2
2
2
1
2
1
2
1
n
n
x
x
σ
σ
+
−
≤
−
z
α
µ
1
=
µ
2
or
µ
1
≤
µ
2
µ
1
>
µ
2
(
)
2
2
2
1
2
1
2
1
n
n
x
x
σ
σ
+
−
≥
z
α
µ
1
=
µ
2
µ
1
≠
µ
2
2
2
2
1
2
1
2
1
n
n
x
x
σ
σ
+
−
≥
z
α
/
2
Example 2
Reconsider the previous example.
Test at the 1% significance level if the mean
weekly earnings of the two groups of workers
are significantly different.
[Solution]
H
0
:
µ
1
=
µ
2
,
H
1
:
µ
1
≠
µ
2
,
α
=
0.01.
The value of the test statistic is
700
60
500
66
487
551
2
2
2
2
2
1
2
1
2
1
+
−
=
+
−
n
n
x
x
σ
σ
=
17.19.
Since |17.19|
≥
2.576
=
z
0.005
, we reject
H
0
.
Thus we conclude that there is a significant

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