Determine whether there is any difference (on the average) for the two methods.
Girder
Karlsruhe Method
Lehigh Method
Difference d
j
S1/1
1.186
1.061
0.125
S2/1
1.151
0.992
0.159
S3/1
1.322
1.063
0.259
S4/1
1.339
1.062
0.277
S5/1
1.2
1.065
0.135
S2/1
1.402
1.178
0.224
S2/2
1.365
1.037
0.328
S2/3
1.537
1.086
0.451
S2/4
1.559
1.052
0.507
Sec 104 Paired
t
Test
Example 1011
Shear Strength of Steel Girder
30
The sevenstep procedure is:
1. Parameter of interest:
The parameter of interest is the difference in mean
shear strength for the two methods.
2. Null hypothesis:
H
0
: μ
D
= 0
3. Alternative hypothesis:
H
1
: μ
D
≠
0
4. Test statistic:
The test statistic is
5. Reject
H
0
if:
Reject
H
0
if the
P
value is <0.05.
6. Computations:
The sample average and standard deviation of the
differences
d
j
are
and
s
d
= 0.1350, and so the test statistic is
7. Conclusions:
Because
t
0.0005.8
= 5.041 and the value of the test statistic
t
0
= 6.15 exceeds this value, the
P
value is less than 2(0.0005) = 0.001.
Therefore, we conclude that the strength prediction methods yield different
results.
Interpretation:
The data indicate that the Karlsruhe method produces, on the
average, higher strength predictions than does the Lehigh method.
n
s
d
t
d
/
0
=
2769
.
0
=
d
15
.
6
9
/
1350
.
0
2769
.
0
/
0
=
=
=
n
s
d
t
d
Sec 104 Paired
t
Test
A Confidence Interval for
μ
D
from Paired Samples
31
If
and
s
D
are the sample mean and standard deviation of
the difference of
n
random pairs of normally distributed
measurements, a 100(1 
α
)% confidence interval on the
difference in means 31 μ
D
= μ
1
 μ
2
is
where
t
α
/2,
n
1
is the upper
α
/2% point of the
t
distribution
with
n
 1 degrees of freedom.
d
n
s
t
d
n
s
t
d
D
n
D
D
n
/
/
1
/2,
1
,
/2

α

α
+
≤
μ
≤

(1025)
Sec 104 Paired
t
Test
Example 1012
Parallel Park Cars
32
The journal
Human Factors
(1962, pp. 375–380) reported a study in which
n
= 14
subjects were asked to parallel park two cars having very different wheel bases
and turning radii. The time in seconds for each subject was recorded and is
given in Table 104. From the column of observed differences, we calculate
d
= 1.21 and
s
D
= 12.68. Find the 90% confidence interval for μ
D
= μ
1
 μ
2.
(
)
(
)
21
.
7
79
.
4
14
/
68
.
12
771
.
1
21
.
1
14
/
68
.
12
771
.
1
21
.
1
/
/
≤
μ
≤

+
≤
μ
≤

+
≤
μ
≤

0.05,13
0.05,13
D
D
D
D
D
n
s
t
d
n
s
t
d
Notice that the confidence interval on μ
D
includes zero. This implies that, at the 90%
level of confidence, the data do not support
the claim that the two cars have different
mean parking times μ
1
and μ
2
. That is, the
value μ
D
= μ
1
 μ
2
= 0 is not inconsistent
with the observed data.
Subject
1(
x
1
j
)
2(
x
2
j
)
(
d
j
)
1
37.0
17.8
19.2
2
25.8
20.2
5.6
3
16.2
16.8
0.6
4
24.2
41.4
17.2
5
22.0
21.4
0.6
6
33.4
38.4
5.0
7
23.8
16.8
7.0
8
58.2
32.2
26.0
9
33.6
27.8
5.8
10
24.4
23.2
1.2
11
23.4
29.6
6.2
12
21.2
20.6
0.6
13
36.2
32.2
4.0
14
29.8
53.8
24.0
Table 104
Sec 104 Paired
t
Test
105.1 The
F
Distribution
105 Inferences on the Variances of Two Normal Populations
We wish to test the hypotheses:
33
2
2
2
1
1
2
2
2
1
0
:
:
σ
≠
σ
σ
=
σ
H
H
Let
W and Y
be independent chisquare random variables with
u
and
v
degrees of
freedom respectively. Then the ratio
(1028)
has the probability density function
(1029)
and is said to follow the distribution with
u
degrees of freedom in the numerator
and
v
degrees of freedom in the denominator. It is usually abbreviated as
F
u,v
.
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 Fall '15
 Normal Distribution, Standard Deviation, Variance, Statistical hypothesis testing