Civil engineers often use the straight-line equatin y=A0+A1x, to model the relat

Question

Civil engineers often use the straight-line equatin y=A0+A1x, to model the relationship between the mean shear strenth of masonry joints and precompression stress,x. To test this theory, a series of stress tests were performed on solid bricks arranged in triplets and joined with mortar. The precompression stress was varied for each triplet and the ultimate shear load just before failure was recorded. The stress results for n=7 triplet test is shows in the table along with an SAS printout regression analysis.

1.75

                                          Analysis of Varience

Source          DF          Sum of Squares          Mean Square          FValue          Prob>F

Model           1                2.39555                      2.39555                47.732           0.0010

Error             5                0.25094                     0.05019

C Total          6                2.64649

Root MSE             0.22403         R-Sqaure    0.9052

Dep Mean             2.32857         Adj R-sq      0.8862

CV                       9.62073

                                       Parameter Estimates

Variable          DF          Parameter Estimate          Standard Error          T for HO: Parameter=0 Prob> I T I

Intercept          1                  1.191930                     0.18503093                           6.442              0.0013

X                     1                  0.987157                     0.14288331                          6.909               0.0010

a) Give a practical interpretation of the estimate of y-intercept of the least squares line.

b) Give a practical interpretation of the slope of the least squares line

c) Give a practical interpretation of r^2 = .905

d) Find r

Triplet Test 1 2 3 4 5 6 7 Shear Strength (tons), y 1.00 2.18 2.24 2.41 2.59 2.82 3.06 Precomp. Stress (tons), x 0 0.60 1.20 1.33 1.43 1.75

1.75

Explanation / Answer

a)y-intercept=1.192,tells us that even when the precompression stress is 0,we find a shear strength of 1.192.

It can also be interpreted as the starting value,the value you may have started with. Or the initial value(if t=0)

b)slope=0.987,it tells us that for every 1ton increase in precompression stress,the shear strength increases by 0.987tons.

c)R^2=0.905 tells us that about 90.5% of the response variation is explained by the model.

d)r=sqrt(0.905)=0.951

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