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3. An article shows data to compare several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied eight times to a girder, are shown in the following table. We wish to determine whether there is any difference between the two methods. Assume the same standard deviations. (Note that we will perform unpaired comparison. We do not perform a paired test.) Karlsruhe Method | Lehigh Method 1.224487 | 1.140994 1.170828 | 1.207858 1.229321 | 1.190368 1.200906 | 1.148473 1.28087 | 1.167855 1.201107 | 1.179984 1.287458 | 1.196009 1.186672 | 1.154533 1.227283 | 1.166998 1.211963 | 1.212248 3.1 (2 points) Draw the normal probability plots for two samples using Microsoft Excel Data Analysis Toolpak. What are their R-squared? 3.2 (2 points) For a sample standard deviation, use Excel function STDEV.S(), not STDEV.P(). Fill in the blanks below: Karlsruhe Method | Lehigh Method ?? = | ?? = s? = | s? = n? = | n? = 3.3. (2 points) Do the data suggest that the two methods provide the same mean shear strength? Use ? = 0.05. Find the P-value rounded to 6 decimal places (using Microsoft Excel.) 3.4. (2 points) Find a 95% confidence interval on the mean difference between the two methods and use it to answer the question in 3.3.

          3. An article shows data to compare several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied eight times to a girder, are shown in the following table. We wish to determine whether there is any difference between the two methods. Assume the same standard deviations. (Note that we will perform unpaired comparison. We do not perform a paired test.)

Karlsruhe Method | Lehigh Method
1.224487 | 1.140994
1.170828 | 1.207858
1.229321 | 1.190368
1.200906 | 1.148473
1.28087 | 1.167855
1.201107 | 1.179984
1.287458 | 1.196009
1.186672 | 1.154533
1.227283 | 1.166998
1.211963 | 1.212248

3.1 (2 points) Draw the normal probability plots for two samples using Microsoft Excel Data Analysis Toolpak. What are their R-squared?

3.2 (2 points) For a sample standard deviation, use Excel function STDEV.S(), not STDEV.P(). Fill in the blanks below:

Karlsruhe Method | Lehigh Method
?? = | ?? =
s? = | s? =
n? = | n? =

3.3. (2 points) Do the data suggest that the two methods provide the same mean shear strength? Use ? = 0.05. Find the P-value rounded to 6 decimal places (using Microsoft Excel.)

3.4. (2 points) Find a 95% confidence interval on the mean difference between the two methods and use it to answer the question in 3.3.

3. An article shows data to compare several methods for predicting the shear strength for steel plate girders. Data for two of these methods, the Karlsruhe and Lehigh procedures, when applied eight times to a girder, are shown in the following table. We wish to determine whether there is any difference between the two methods. Assume the same standard deviations. (Note that we will perform unpaired comparison. We do not perform a paired test.)

Karlsruhe Method | Lehigh Method
1.224487 | 1.140994
1.170828 | 1.207858
1.229321 | 1.190368
1.200906 | 1.148473
1.28087 | 1.167855
1.201107 | 1.179984
1.287458 | 1.196009
1.186672 | 1.154533
1.227283 | 1.166998
1.211963 | 1.212248

3.1 (2 points) Draw the normal probability plots for two samples using Microsoft Excel Data Analysis Toolpak. What are their R-squared?

3.2 (2 points) For a sample standard deviation, use Excel function STDEV.S(), not STDEV.P(). Fill in the blanks below:

Karlsruhe Method | Lehigh Method
?? = | ?? =
s? = | s? =
n? = | n? =

3.3. (2 points) Do the data suggest that the two methods provide the same mean shear strength? Use ? = 0.05. Find the P-value rounded to 6 decimal places (using Microsoft Excel.)

3.4. (2 points) Find a 95% confidence interval on the mean difference between the two methods and use it to answer the question in 3.3.

Added by Raul P.

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