Question

Please include the R code for each individual question.

Save PDF to My Note The article "The Undrained Strength of Some Thawed Permafrost Soils" (Canadian Geotech. J., 1979: 420-427) contained the accompanying data on y shear strength of sandy soil (kPa), xl depth (m), and x2 water content (%) Obs Depth Content Strength 8.9 31.5 14.7 2 36.6 27.0 48.0 3 36.8 25.9 25.6 46.1 39.1 10.0 56.9 39.216.0 66.9 38.3 16.8 77.3 33.9 20.7 88.4 33.8 38.8 9 6.5 27.9 16.9 10 8.0 33.1 27.0 114.5 26.3 16.0 12 9.9 37.0 24.9 13 2.9 34.6 7.3 142.0 36.4 12.8 a) Perform regression to predict y from xi, x2, x3-x1"2, x4-x2n2, and x5 = x1 *x2; and write down the coefficients of the various b) Can you interpret the regression coeficients? Explain c) Compute RA2 and explain what it says about goodness-of-fit ("in English") d) Compute s e, and interpret ("in English") c) Froduce the residual plot (residuals vs. *predicted* y), and explain what it suggests, if any5ki2 avt a f) Now perform regression to predict y from x1 and x2 only ) Compute RA2 and explain what it says about goodness-of-fit h) Compare the above two RA2 values. Does the comparison suggest that at least one of the higher-order terms in the regression eqn provides useful information about strength? i) Compute s_e for the model in part f, and compare it to that in part d. What do you conclude?

Answer 1

R- commands with out put

> y=c(14.7,48.0,25.6,10.0,16.0,16.8,20.7,38.8,16.9,27.0,16.0,24.9,7.3,12.8)
> y
[1] 14.7 48.0 25.6 10.0 16.0 16.8 20.7 38.8 16.9 27.0 16.0 24.9 7.3 12.8
> x1=c(8.9,36.6,36.8,6.1,6.9,6.9,7.3,8.4,6.5,8.0,4.5,9.9,2.9,2.0)
> x1
[1] 8.9 36.6 36.8 6.1 6.9 6.9 7.3 8.4 6.5 8.0 4.5 9.9 2.9 2.0
> x2=c(31.5,27.0,25.9,39.1,39.2,38.3,33.9,33.8,27.9,33.1,26.3,37.0,34.6,36.4)
> x2
[1] 31.5 27.0 25.9 39.1 39.2 38.3 33.9 33.8 27.9 33.1 26.3 37.0 34.6 36.4
> x3=x1*x1
> x4=x2*x2
> x5=x1*x2
> fit=lm(y~x1+x2+x3+x4+x5)
> fit

Call:
lm(formula = y ~ x1 + x2 + x3 + x4 + x5)

Coefficients:
(Intercept) x1 x2 x3 x4 x5  
-140.22976 -16.47521 12.82710 0.09555 -0.24339 0.49864  

> summary(fit)

Call:
lm(formula = y ~ x1 + x2 + x3 + x4 + x5)

Residuals:
Min 1Q Median 3Q Max
-8.349 -2.188 0.279 1.649 13.613

Coefficients:
Estimate Std. Error t value Pr(>|t|)  
(Intercept) -140.22976 136.13743 -1.030 0.3331  
x1 -16.47521 9.07116 -1.816 0.1069  
x2 12.82710 8.25854 1.553 0.1590  
x3 0.09555 0.07206 1.326 0.2214  
x4 -0.24339 0.12744 -1.910 0.0925 .
x5 0.49864 0.23543 2.118 0.0670 .
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 7.023 on 8 degrees of freedom
Multiple R-squared: 0.7561, Adjusted R-squared: 0.6037
F-statistic: 4.961 on 5 and 8 DF, p-value: 0.02307

> Residuals=resid(fit)
> Residuals
1 2 3 4 5 6 7
-8.3487300 1.5754583 -1.4933155 -1.1941138 1.6740011 0.1382374 -2.4192665
8 9 10 11 12 13 14
13.6129620 1.3376151 2.9644709 0.4196648 -5.1782487 -7.9658627 4.8771276
> predicted=fitted(fit)
> predicted
1 2 3 4 5 6 7 8
23.048730 46.424542 27.093316 11.194114 14.325999 16.661763 23.119267 25.187038
9 10 11 12 13 14
15.562385 24.035529 15.580335 30.078249 15.265863 7.922872
> plot(Residuals,predicted)
> fit2=lm(y~x1+x2)
> fit2

Call:
lm(formula = y ~ x1 + x2)

Coefficients:
(Intercept) x1 x2  
14.8893 0.6607 -0.0284  

> summary(fit2)

Call:
lm(formula = y ~ x1 + x2)

Residuals:
Min 1Q Median 3Q Max
-12.867 -4.475 -1.526 3.878 19.321

Coefficients:
Estimate Std. Error t value Pr(>|t|)  
(Intercept) 14.8893 23.2447 0.641 0.5349  
x1 0.6607 0.2737 2.414 0.0344 *
x2 -0.0284 0.6423 -0.044 0.9655  
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 9.019 on 11 degrees of freedom
Multiple R-squared: 0.447, Adjusted R-squared: 0.3465
F-statistic: 4.446 on 2 and 11 DF, p-value: 0.03845

____________________________________________________________________________________________

a)
Coefficients:
(Intercept) x1 x2 x3 x4 x5  
-140.22976 -16.47521 12.82710 0.09555 -0.24339 0.49864  
b)
Interpritation-
Intercept=-140.22976 indicates that the value of y at the point where x1=x2=x3=x4=x5=0
slope(b1)=-16.47521 indicates that for unit change in x1, y decreases with -16.47521.
slope(b2)=12.82710 indicates that for unit change in x1, y increases with 12.82710.
slope(b3)=0.09555 indicates that for unit change in x1, y increases with 0.09555.
slope(b4)=-0.24339 indicates that for unit change in x1, y decreases with -0.24339.
slope(b5)=0.49864 indicates that for unit change in x1, y increases with 0.49864.
c)
Rsquare = 0.7561 indicates that adequacy of the fitted model is much good.
d)
Se = 7.023 indicates that predictive quality of a regression model with lower Se.
e)see plot
f)
g)
R-squared = 0.447 indicates that adequacy of the fitted model is good.
h)
Rsquare from first model is larger than the R square from the second model suggest that one of the higher order terms in the regression equation provides useful information about strengths i.e. strengrh increases.
i)
Standard error of the second model is 9.019 which is larger the first model (Se = 7.023 ) indicates predictive quality of first regression model is larger.

0 10 Residuals