3. n article in the Journal of Strain Analysis (vol.18, no. 2. 1983) compares se
ID: 387342 • Letter: 3
Question
3. n article in the Journal of Strain Analysis (vol.18, no. 2. 1983) compares several procedures for preacting the shear strength for steel plate girders. Data for nine girders in the form of the ratio of predicted to observed load for two of these procedures, the Karlsruhe and Lehigh methods, are as follows: Girder Karlsruhe Method Lehigh Method Difference Difference'2 SI/ 1.186 S2/1 15 S3/1 1.322 S4I 1.339 SS/1 1.200 S2/1 1402 S2/2 1.365 S2/3 1.537 S2/4 1.559 0.015625 0.025281 0.067081 0.076729 0.018225 0.050176 0.107584 0.203401 0.257049 1.061 0.992 1.062 1.065 1.178 1.037 1.086 1.052 0.125 0.159 0.259 0277 0.135 0224 0.328 0.451 0.507 to support a claim that there is a difference in mean performance 2 between the two methods? Use alpha -0.05. (you can use paired t-test)
Explanation / Answer
Let us look at the predicted value
Karlsruhe Method
Lehigh Method
Difference
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
1.200
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
Validating it with paired t test
Calculating mean and standard deviation of difference:
Mean or Average = (0.125+0.159+0.259+0.277+0.135+0.224+0.328+0.451+0.507)/9 = 0.273
Standard deviation of difference: 0.135
SE(d)= Sd/n^1/2 = 0.135 /(9)^1/2
= 0.045
So, we have:
T= 0.273 / 0.045 =6.06
As alpha = 0.05
Therefore 95% confidence interval
D+- t* Sd/(n)^1/2
Where t is the 2.5% point of the t-distribution with 8 degree of freedom is 2.5% of 1.86 = 0.04
0.135 +-(0.045 X 0.04)
At 95% confidence interval
= (0.136 , 0.133)
Therefore there is difference in mean performance of both method
Karlsruhe Method
Lehigh Method
Difference
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
1.200
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
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