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Choose the One Alternative That Best Completes the Statement or Answers

Question 126

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Choose the one alternative that best completes the statement or answers the question.
-Civil engineers often use the straight-line equation, E(y) =β0+β1x\mathrm { E } ( \mathrm { y } ) = \beta _ { 0 } + \beta _ { 1 } \mathrm { x } , to model the relationship between the mean shear strength E(y) \mathrm { E } ( \mathrm { y } ) of masonry joints and precompression stress, xx . 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 (called the shear strength) was recorded. The stress results for n=7\mathrm { n } = 7 triplet tests is shown in the accompanying table followed by a SAS printout of the regression analysis.
 Triplet Test 1234567 Shear Strength (tons) , y1.002.182.242.412.592.823.06 Precomp. Stress (tons) , x00.601.201.331.431.751.75\begin{array}{l|ccccccc}\text { Triplet Test } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\\hline \text { Shear Strength (tons) , } \mathrm{y} & 1.00 & 2.18 & 2.24 & 2.41 & 2.59 & 2.82 & 3.06 \\\hline \text { Precomp. Stress (tons) , } \mathrm{x} & 0 & 0.60 & 1.20 & 1.33 & 1.43 & 1.75 & 1.75\end{array}

 Analysis of Variance \text { Analysis of Variance }

 Sum of  Mean  Source  DF  Squares  Square  F Value  Prob > F  Model 12.395552.3955547.7320.0010 Error 50.250940.05019 C Total 62.64649\begin{array}{lccccc} & {\text { Sum of }} & \text { Mean } \\\text { Source } & \text { DF } & \text { Squares } & \text { Square } & \text { F Value } & \text { Prob > F } \\\text { Model } & 1 & 2.39555 & 2.39555 & 47.732 & 0.0010 \\\text { Error } & 5 & 0.25094 & 0.05019 & & \\\text { C Total } & 6 & 2.64649 & & &\end{array}

 Root MSE 0.22403 R-square 0.9052 Dep Mean 2.32857 Adj R-sq 0.8862 C.V. 9.62073\begin{array}{llll}\text { Root MSE } & 0.22403 & \text { R-square } & 0.9052 \\\text { Dep Mean } & 2.32857 & \text { Adj R-sq } & 0.8862 \\\text { C.V. } & 9.62073 & &\end{array}

 Parameter Estimates  Parameter  Standard  T for HO:  Variable  DF  Estimate  Error  Parameter =0 Prob >T INTERCEP 11.1919300.185030936.4420.0013X10.9871570.142883316.9090.0010\begin{array}{l}\text { Parameter Estimates }\\\begin{array}{llllll} & & \text { Parameter } & \text { Standard } & \text { T for HO: } & \\\text { Variable } & \text { DF } & \text { Estimate } & \text { Error } & \text { Parameter }=0 & \text { Prob }>|\mathrm{T}| \\\text { INTERCEP } & 1 & 1.191930 & 0.18503093 & 6.442 & 0.0013 \\\mathrm{X} & 1 & 0.987157 & 0.14288331 & 6.909 & 0.0010\end{array}\end{array}

Give a practical interpretation of R2R ^ { 2 } , the coefficient of determination for the least squares model.


A) About 91%91 \% of the total variation in the sample of yy -values can be explained by (or attributed to) the linear relationship between shear strength and precompression stress.
B) In repeated sampling, approximately 91%91 \% of all similarly constructed regression lines will accurately predict shear strength.
C) We expect to predict the shear strength of a triplet test to within about 91 ton of its true value.
D) We expect about 91%91 \% of the observed shear strength values to lie on the least squares line.

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