The data below represent the results of a test for the strength of an asphalt co

Question

The data below represent the results of a test for the strength of an asphalt concrete mix. The test consisted of applying a compressive force on the top of different sample specimens. Two responses occurred: the stress and strain at which a sample specimen failed The factors relate to mixture proportions, rates of speed at which the force temperature. Higher values of the response variables indicate stronger materials. The variables are: XI: percent binder (the amount of asphalt in the mixture). X2: loading rate (the speed at which the force was applied). X3: ambient temperature. YI: the stress at which the sample specimen failed, and Y2: the strain at which the specimen failed. Perform separate regressions to relate stress and strain to the factors of the experiment. Check the residuals for possible specification errors. Calculate and plot the residuals (errors) against the fitted values. Normal-probability plot. Interpret all results.

Explanation / Answer

first we have to read the data into R.

so we mada a dataframe df in which all the data is read into R like below

df=read.csv('C:/Users/hp/Desktop/chegg/cheg1.csv')
> df
x1 x2 x3 y1 y2
1 5.3 0.02 77 42 3.20
2 5.3 0.02 32 481 0.73
3 5.3 0.02 0 543 0.16
4 6.0 2.00 77 609 1.44
5 7.8 0.20 77 444 3.68
6 8.0 2.00 104 194 3.11
7 8.0 2.00 77 593 3.07
8 8.0 2.00 32 977 0.19
9 8.0 2.00 0 872 0.00
10 8.0 0.02 104 35 5.86
11 8.0 0.02 77 96 5.97
12 8.0 0.02 32 663 0.29
13 8.0 0.02 0 702 0.04
14 10.0 2.00 77 518 2.72
15 12.0 0.02 77 40 7.35
16 12.0 0.02 32 627 1.17
17 12.0 0.02 0 683 0.14
18 12.0 0.02 104 22 15.00
19 14.0 0.02 77 35 11.80

then we perform two regeression for dependent variable y1 and y2 like this

reg1=lm(df[,4]~df[,1]+df[,2]+df[,3])
> reg1

Call:
lm(formula = df[, 4] ~ df[, 1] + df[, 2] + df[, 3])

Coefficients:
(Intercept) df[, 1] df[, 2] df[, 3]
700.618 -1.526 175.984 -6.697

> reg2=lm(df[,5]~df[,1]+df[,2]+df[,3])
> reg2

Call:
lm(formula = df[, 5] ~ df[, 1] + df[, 2] + df[, 3])

Coefficients:
(Intercept) df[, 1] df[, 2] df[, 3]
-5.61130 0.66754 -1.23535 0.07319

to find residual error we give the command and will get

residual1=resid(reg1)
1 2 3 4 5 6
-138.3719746 -0.7432004 -153.0516277 81.2478005 235.7651139 -149.8781060
7 8 9 10 11 12
68.2991585 150.9279327 -168.3804945 39.5700943 -80.2526412 185.3761330
13 14 15 16 17 18
10.0677057 -3.6494834 -130.1499251 155.4788491 -2.8295781 32.6728104
19
-132.0985670

residual2=resid(reg2)
1 2 3 4 5 6 7
-0.3377460 0.4858880 2.2580277 -0.1190242 -1.3042396 -1.7602904 0.1758900
8 9 10 11 12 13 14
0.5895240 2.7416637 -1.4562924 0.6298880 -1.7564779 0.3356618 -1.5091959
15 16 17 18 19
-0.6602837 -3.5466497 -2.2345099 5.0135359 2.4546305

if we give the command

plot(residual1,df[,4]) and plot(residual2,df[,5])

we get the required graph residual versus dependent variable

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