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An article in the Journal of Strain Analysis (1983, Vol. 18, No. 2) compares 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 specific girders, are shown in the accompanying Table. Construct a 95% confidence interval on the difference in mean shear strength for the two methods. Round your answer to four decimal places (e.g. 98.7654). 0.1698 <= ?_D <= 0.3793 Girder Karlsruhe Method Lehigh Method Difference d_j S1/1 1.194 1.052 0.142 S2/1 1.156 0.988 0.168 S3/1 1.324 1.070 0.254 S4/1 1.338 1.067 0.271 S5/1 1.194 1.056 0.138 S2/1 1.407 1.182 0.225 S2/2 1.358 1.034 0.324 S2/3 1.536 1.081 0.455 S2/4 1.552 1.044 0.508

          An article in the Journal of Strain Analysis (1983, Vol. 18, No. 2) compares 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 specific girders, are shown in the accompanying Table. Construct a 95% confidence interval on the difference in mean shear strength for the two methods. Round your answer to four decimal places (e.g. 98.7654).

0.1698 <= ?_D <= 0.3793

Girder Karlsruhe Method Lehigh Method Difference d_j
S1/1 1.194 1.052 0.142
S2/1 1.156 0.988 0.168
S3/1 1.324 1.070 0.254
S4/1 1.338 1.067 0.271
S5/1 1.194 1.056 0.138
S2/1 1.407 1.182 0.225
S2/2 1.358 1.034 0.324
S2/3 1.536 1.081 0.455
S2/4 1.552 1.044 0.508

Show more…

An article in the Journal of Strain Analysis (1983, Vol. 18, No. 2) compares 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 specific girders, are shown in the accompanying Table. Construct a 95% confidence interval on the difference in mean shear strength for the two methods. Round your answer to four decimal places (e.g. 98.7654).

0.1698 <= ?D <= 0.3793

Girder Karlsruhe Method Lehigh Method Difference dj
S1/1 1.194 1.052 0.142
S2/1 1.156 0.988 0.168
S3/1 1.324 1.070 0.254
S4/1 1.338 1.067 0.271
S5/1 1.194 1.056 0.138
S2/1 1.407 1.182 0.225
S2/2 1.358 1.034 0.324
S2/3 1.536 1.081 0.455
S2/4 1.552 1.044 0.508

Added by Kathryn H.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Paul A.

02/28/2022

Step 1

Given that the sum of the differences in the last column is 2.485 and there are 9 data points, the mean difference is: \[ \bar{d} = \frac{2.485}{9} = 0.276 \] Show more…

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An article in the Journal of Strain Analysis (1983, Vol. 18, No. 2) compares 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 specific girders, are shown in the accompanying Table. Construct a 95% confidence interval on the difference in mean shear strength for the two methods. Round your answer to four decimal places (e.g. 98.7654). 0.1698 <= μ_D <= 0.3793 Girder Karlsruhe Method Lehigh Method Difference d_j S1/1 1.194 1.052 0.142 S2/1 1.156 0.988 0.168 S3/1 1.324 1.070 0.254 S4/1 1.338 1.067 0.271 S5/1 1.194 1.056 0.138 S2/1 1.407 1.182 0.225 S2/2 1.358 1.034 0.324 S2/3 1.536 1.081 0.455 S2/4 1.552 1.044 0.508

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