5.15 a) Using the Minitab output, Pvalues = 0.001.
Since P = 0.001 < 0.05, therefore, the null hypothesis
H
0
:
μ
1
−
μ
2
=
0
can be rejected at the
0.05 level of significance.
b) It is a twosided test because the alternative hypothesis is
H
1
:
μ
1
−
μ
2
≠
0.
c)
Parameter of interest
: We are interested in determining if
μ
1
−
μ
2
=
2
.
using μ
1
∧
μ
2
.
Null Hypothesis
:
H
0
:
μ
1
−
μ
2
=
2
Alternative Hypothesis
:
H
1
:
μ
1
−
μ
2
≠
2
Test Statistic
:
t
0
=
´
X
1
−
´
X
2
−
2
s
p
√
1
n
1
+
1
n
2
standard deviation S
p
=
√
(
n
1
−
1
)
S
1
2
+
(
n
2
−
1
)
S
2
2
n
1
+
n
2
−
2
Assume
α
=
0.05
Reject
H
0
if
:
t
0
>
t
α
2
,n
1
+
n
2
−
2
∨
t
0
<−
t
α
2
, n
1
+
n
2
−
2
Computations
: Using Minitab output, the estimated values for standard deviation is
S
p
=
2.1277. Since
´
X
1
=
50.19,
´
X
2
=
52.52,
∧
n
1
=
n
2
=
20.
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:
t
0
=
50.19
−
52.52
−
2
2.1277
√
1
20
+
1
20
= 6.435
Conclusion
: Using excel function = T.INV.2T (a,
n
1
+
n
2
−
2
¿
.n
1
+
n
2
−
2
= 38.
.
t
0.025,38
= 2.024. Since
t
0
=−
6.435
<−
2.024
=−
t
0.025,38
,
we reject the null hypothesis
H
0
:
μ
1
−
μ
2
=
2
in favor of the Alternative hypothesis
H
1
:
μ
1
−
μ
2
≠
2
at the 0.05 level of
significance. Also using the Pvalue approach to perform the test, the decision rule which is
H
0
if P
<
0.05
.
the null hypothesis will be rejected at the 0.05 level of significance.
d) We obtained that the null hypothesis
H
0
would be rejected at the 0.05 level of significance
if the hypothesis had been
H
0
:
μ
1
−
μ
2
=
2
versus
H
1
:
μ
1
−
μ
2
≠
2.
if the hypotheses had been
H
0
:
μ
1
−
μ
2
=
2
versus H
1
:
μ
1
−
μ
2
<
2,
the Tvalue for the onesided test would remain unchanged
but the Pvalue would be that half of the twosided Pvalue. Therefore, the null hypothesis
H
0
would alsobe rejectedat the
0.05
levelof significance when
testing
H
0
:
μ
1
−
μ
2
=
2
versus
H
1
:
μ
1
−
μ
2
<
2.
e) UCL =
´
X
1
−
´
X
2
+
t
a, n
1
+
n
2
−
2
s
p
√
1
n
1
+
1
n
2
= 50.19 52.52 + 1.68 (2.1277)
√
1
20
+
1
20
= 1.1956
f) P = 2P (
T
38
> 6.435)
= T.DIST.2T (6.435, 38)
= 0.0000
5.17a
) Parameter of Interest:
μ
single
−
μ
double
=
0
Null Hypothesis
:
H
0
:
:
μ
single
−
μ
double
=
0
Alternative Hypothesis
:
H
1
::
μ
single
−
μ
double
≠
0
Test Statistic
:
t
0
=
´
´
X
1
−
´
´
X
2
−
2
sp
√
1
n
1
+
1
n
2
Assuming
α = 0.05
Reject
Computation:
66.385
−
45.278
8,26182
√
1
15
+
1
15
=
6.997
P
=
T
−
DIST
.2
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 Fall '17
 Dr Bheem

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