11 Shear Strength of Steel Girder An article in the Journal of Strain Analysis [1983, Vol. 18(2)]
Example 10-11 Shear Strength of Steel Girder An article in the Journal of Strain Analysis [1983, Vol. 18(2)]
reports a comparison of several methods for predicting the shear strength for steel plate girders.
Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied to nine specifi c girders, are shown in Table 10-3. We wish to determine whether there is any difference (on the average) for the two methods.
The seven-step procedure is applied as follows:
1. Parameter of interest: The parameter of interest is the difference in mean shear strength for the two methods—say,
μ D = 1 − 2 = 0.
2. Null hypothesis: H 0 : μ= D
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 <0 05 .
6. Computations: The sample average and standard deviation of the differences d j are t 0 = . and s 6 08 d = 0 1350 . , and
so the test statistic is
7. Conclusion: Because t . 0 0005 8 = 5 041 and the value of the test statistic t 0 = . exceeds this value, the P-value is less than 6 15
(. 2 0 0005 ) = . 0 001 . Therefore, we conclude that the strength prediction methods yield different results. Practical Interpretation: Specifi cally, the data indicate that the Karlsruhe method produces, on the average, greater
strength predictions than does the Lehigh method. This is a strong conclusion.
Software can perform the paired t-test. Typical output for Example 10-10 follows:
Paired T for Karlsruhe–Lehigh
SE Mean
95 CI for mean difference: (0.173098, 0.380680) T-test of mean difference = 0 (vs. not = 0): T-value = 6.15, P-value = 0.000
Chapter 10Statistical Inference for Two Samples
TABLE t 10-3 Strength Predictions for Nine Steel Plate Girders (Predicted LoadObserved Load)
Girder
Karlsruhe Method
Lehigh Method
Difference d j
The results essentially agree with the manual calculations. In addition to the hypothesis test results. Most computer software report a two-sided CI on the difference in means. This Cl was
found by constructing a single-sample CI on μ D . We provide the details later. Paired Versus Unpaired Comparisons
In performing a comparative experiment, the investigator can sometimes choose between the paired experiment and the two-sample (or unpaired) experiment. If n measurements are to be made on each population, the two-sample t-statistic is
n which would be compared to t 2 n − , and of course, the paired t-statistic is
S D n which is compared to t n −1 . Notice that because
D = ∑ = = − = X 1 − 2
the numerators of both statistics are identical. However, the denominator of the two-sample
t-test is based on the assumption that X 1 and X 2 are independent. In many paired experiments,
a strong positive correlation ρ exists for X 1 and X 2 . Then it can be shown that
assuming that both populations X 1 and X 2 . Furthermore, S 2 2 have identical variances σ D n esti-
mates the variance of D. Whenever a positive correlation exists within the pairs, the denomi- nator for the paired t-test will be smaller than the denominator of the two-sample t-test. This can cause the two-sample t-test to considerably understate the signifi cance of the data if it is incorrectly applied to paired samples.
Although pairing will often lead to a smaller value of the variance of X 1 − , it does 2
have a disadvantage—namely, the paired t-test leads to a loss of n −1 degrees of freedom in
Section 10-4A Nonparametric Test for the Difference in Two Means
comparison to the two-sample t-test. Generally, we know that increasing the degrees of free- dom of a test increases the power against any fi xed alternative values of the parameter.
So how do we decide to conduct the experiment? Should we pair the observations or not? Although this question has no general answer, we can give some guidelines based on the pre- ceding discussion.
1. If the experimental units are relatively homogeneous (small σ) and the correlation within pairs is small, the gain in precision attributable to pairing will be offset by the loss of degrees of freedom, so an independent-sample experiment should be used.
2. If the experimental units are relatively heterogeneous (large σ) and there is large positive correlation within pairs, the paired experiment should be used. Typically, this case occurs when the experimental units are the same for both treatments; as in Example 10-11, the same girders were used to test the two methods.
Implementing the rules still requires judgment because σ and ρ are never known precisely. Furthermore, if the number of degrees of freedom is large (say, 40 or 50), the loss of n − 1 of them for pairing may not be serious. However, if the number of degrees of freedom is small (say, 10 or 20), losing half of them is potentially serious if not compensated for by increased precision from pairing.
Confi dence Interval for l D To construct the confi dence interval for μ D = 1 − , note that 2
D −μ D T =
S n
follows a t distribution with n − 1 degrees of freedom. Then, because P t ( − α2, n − 1 ≤≤ T t ) =
1 − α, we can substitute for T in the preceding expression and perform the necessary steps to
isolate μ D = 1 − for the inequalities. This leads to the following 100 1 2 ( − α confi dence )
interval on μ 1 −. 2
Confi dence Interval
for l D from Paired
If d and s D are the sample mean and standard deviation of the difference of n random
Samples
pairs of normally distributed measurements, a 100 1 ( − α) confi dence interval on the
difference in means l D = 1 − 2 is
d − t α 2 ,n − 1 s D n ≤μ≤+ d t α 2 ,n − 1 s D n (10-25)
where t 2 a , n − 1 is the upper α 2 point of the t distribution with n − 1 degrees of
freedom.
2 This confi dence interval is also valid for the case in which 2 σ
1 D 2 estimates ≠ because s
σ D = VX ( 1 − X 2 ) . Also, for large samples (say, n ≥ 30 pairs), the explicit assumption of nor-
mality is unnecessary because of the central limit theorem.