Question

3. An article in the Journal of Strain Analysis (Vol 18, No. 2, 1983) compares several procedures for predicting 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, Karlsruhe Method and Lehigh Method, are as follows in Table 2. Table 2: Girder Karlsruhe Method Lehigh Method S1/1 1.1860 1.0610 S2/1 1.1510 0.9920 S3/1 1.3220 1.0630 S4/1 1.3390 1.0620 S5/1 1.2000 1.0650 S2/1 1.4020 1.1780 S2/2 1.3650 1.0370 S2/3 1.5370 1.0860 S2/4 1.5590 1.0520 (a) Is there any evidence to support a claim that there is a difference in mean performance between the two methods? $alpha = 0.05$. (b) What is the p-value for the test in part (a)? (c) Construct a 95% confidence interval for the difference in mean shear strength between these two methods.

          3. An article in the Journal of Strain Analysis (Vol 18, No. 2, 1983) compares several procedures for predicting 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, Karlsruhe Method and Lehigh Method, are as follows in Table 2.
Table 2:
Girder Karlsruhe Method Lehigh Method
S1/1 1.1860 1.0610
S2/1 1.1510 0.9920
S3/1 1.3220 1.0630
S4/1 1.3390 1.0620
S5/1 1.2000 1.0650
S2/1 1.4020 1.1780
S2/2 1.3650 1.0370
S2/3 1.5370 1.0860
S2/4 1.5590 1.0520
(a) Is there any evidence to support a claim that there is a difference in mean performance between the two methods? $alpha = 0.05$.
(b) What is the p-value for the test in part (a)?
(c) Construct a 95% confidence interval for the difference in mean shear strength between these two methods.

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3. An article in the Journal of Strain Analysis (Vol 18, No. 2, 1983) compares several procedures for predicting 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, Karlsruhe Method and Lehigh Method, are as follows in Table 2.
Table 2:
Girder Karlsruhe Method Lehigh Method
S1/1 1.1860 1.0610
S2/1 1.1510 0.9920
S3/1 1.3220 1.0630
S4/1 1.3390 1.0620
S5/1 1.2000 1.0650
S2/1 1.4020 1.1780
S2/2 1.3650 1.0370
S2/3 1.5370 1.0860
S2/4 1.5590 1.0520
(a) Is there any evidence to support a claim that there is a difference in mean performance between the two methods? alpha = 0.05.
(b) What is the p-value for the test in part (a)?
(c) Construct a 95% confidence interval for the difference in mean shear strength between these two methods.

Added by Lu-S K.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert David Nguyen

10/17/2022

Step 1

- Null hypothesis (\(H_0\)): There is no difference in the mean performance between the Karlsruhe Method and the Lehigh Method. (\(\mu_K = \mu_L\)) - Alternative hypothesis (\(H_1\)): There is a difference in the mean performance between the Karlsruhe Method and Show more…

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Transcript

00:01 Hi, i'm david and i'm here to have your answer your question.

00:03 In the question here, we are going to discuss about the hybrid testing.

00:07 Here we're given the table of the two methods, so let me begin up here.

00:12 Let me make it smaller.

00:15 And from the table here, this will be the method 1 and the method 2, and i will define the d, that will be the method 1, we minus the method 2, and it will do the calculation here.

00:31 We will get equal to the 1 -60 minus 0 6 .10 equal to the 1 -2 -50 and this one we get 11510 minus this one will be equal to the zupon so this one 1 .151 minus 099 to 0 again equal to zabon 59 to 0 again equal to zabon 159 and and then the next one has the 1 .322 minus 1 .063 equal to the 0 .29.

01:17 1 .339 minus 1 .062 equal to the 0 .27.

01:26 1 .2 minus 1 .065 equal to the 0 .135.

01:34 1 .402 minus 1 .178 equal to the 0 .224.

01:43 1 .365 minus 1 .037 equal to the 0 .365 minus 1 .037 equal to the 0 .38.

01:52 1 .537 minus 1 .086 equal to the 0 .451 .559 minus 1 .0 .52.

02:05 Equal to the zubon 5 -07.

02:09 Now the question a here, we need to perform the test where we have the known hypothesis, then the mean different mu -d, equal to the 0 versus the alternative hypothesis, then the new -d will not equal to the 0.

02:24 Notice that, this will be the 2 -tel test.

02:28 Here, the first thing we need to do is we need to find the n, and then the d -bar, and the sd.

02:35 So we have to pass the data into the calculator.

02:39 We will have the 0 .125, 0159, 0159, 027, 135, 01225, 01224, 01528, 01545451, 0151515151, 01507.

02:57 Then we have an equal to the 9, 1, 2, 3, 4, 5, 6, 7, 8, 9.

03:03 And we have the d bar equal to the 02739 and the sd equal to the 0 .115...

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