Concrete compressive strength

17.08.2020 Concrete compressive strength
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Concrete compressive strength
We have a numerical dataset with information about di erent batches of concrete. Each batch
is Strength of Concrete in speci c moment in time (number of days after prodaction). It is
known which and how much components are in of Concrete. There are information about
1030 batches in the dataset.
Cement Slag Ash Water Plasticizer Breakstone Sand Time Strength
540.0 0.0 0 162 2.5 1040.0 676.0 28 79.99
540.0 0.0 0 162 2.5 1055.0 676.0 28 61.89
332.5 142.5 0 228 0.0 932.0 594.0 270 40.27
332.5 142.5 0 228 0.0 932.0 594.0 365 41.05
198.6 132.4 0 192 0.0 978.4 825.5 360 44.30
Main components for production: Cement, Water, Sand, Breakstone. These components are
always using for Concrete production.
Additional components: Not all batches contain Slag, Ash or Plasticizer. Plasticizer is used to
add less water. If we add Ash, we can add less Cement. Furnace Slag in uences on the concrete
structure.
Time: There are bathes with the same composition of components, but with di erent number
of days. The Concrete Strength has been checked after 1,3, 7, 14, 28, 56, 90, 91, 100, 120, 180,
270, 360, 365 days after prodaction.There are a few data observations where Time is 1, 120
days in the dataset. We have exclude this observation from consideration. We have little data
with Time = 90, 91, 100 and Time = 270, 360, 365. We use 90% of all batches.
When, its strength has reached 99% in 28 days, still concrete continues to gain strength after
that period, but that rate of gain in compressive strength is very less compared to that in 28
days. So, data with Time = 90, 91 we have set Time = 90 and observations with Time = 270, 360,
365 we have set Time = 365. So, there are 1030 observations in the dataset, we use 1025.
1m3
1m3

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Our task: determine how a change in the proportions of di erent elements a ects the
commpresive strength of concrete.
Base analysis
Base vizualization
We can see strong dependence between Cement and Strength. The positive value of the slope
in the Linear Model (Strength~Cement) shows that If we add more Cement, Concrete Strength
increases. The negative values of the slope in the Linear Models (Strength~Water,
Strength~Sand) show that It is inverse dependence between Strength and Water; Strength and
Sand. The value of the slope in the Breakstone~Strength less than other, it shows that
Breakstone practically doesn’t a ect on Strength.

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We can see inverse dependence between Ash and Strength (the negative value of the slope).
The value of the slope in the Slag~Strength less than other, it shows that It is low dependence
between Slag and Strength. The positive value of the slope in the Plasticizer~Strength model
indicates that an increase in quantity Plasticizer improves Strength.
Correlation matrix
It shows that there isn’t big intercorrelation (>0.7) between explanatory variables and we don’t
need to delete any variables. It is easy to see that the most in uential components are Time
and Cement.We can see strong intercorrelation between Water and Plasticizer. If we add
Plasticizer, we can use less Water.
Before building a model we have checked for normality the dependent variable (Strength). This
Strength doesn’t Normally Distributed, but for Linear Regression this is an optional condition.
Building models
Hypothesis 1: Concrete compressive strength depends on all ingredients and Time.
##

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##
## Call:
## lm(formula = Strength ~ ., data = data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -28.970 -6.455 0.403 6.875 34.255
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -25.686533 27.145205 -0.946 0.344239
## Cement 0.120191 0.008657 13.883 < 2e-16 ***
## Slag 0.103565 0.010326 10.029 < 2e-16 ***
## Ash 0.088255 0.012815 6.887 9.99e-12 ***
## Water -0.146563 0.040955 -3.579 0.000362 ***
## Plasticizer 0.291271 0.094866 3.070 0.002195 **
## Breakstone 0.019292 0.009598 2.010 0.044692 *
## Sand 0.021507 0.010921 1.969 0.049199 *
## Time 0.101555 0.005176 19.620 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 10.55 on 1016 degrees of freedom
## Multiple R-squared: 0.6043, Adjusted R-squared: 0.6012
## F-statistic: 193.9 on 8 and 1016 DF, p-value: < 2.2e-16
Cement, Slag, Ash, Water, Time, Breakstone and Sand strongly a ect Strength. The R-squared is
0.6012, it indicates that 60% of the variation in the Strength is explained by the input variables.
Model 1 check. Conditionals of Gauss–Markov theorem.
1)Homoscedasticity of the dispersion of residues.
p = 0 < 0.05, it means that residual variance is related to model predicted values.Failed
3) An auto-correlation check ( The Durbin–Watson statistic).
p = 0 < 0.05, it means that it is auto-correlation. Failed
4)The normal distribution of residues (The Skewness-Kurtosis’s Test).

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Residues are distributed normally. Sucess

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Conclusion: Not all conditions of the Gauss–Markov theorem are satis ed. Our hypothesis is
false, so Concrete compressive strength doesn’t depend on all ingredients and Time and we
can’t use our model.
We have made some changes: built Linear Model (Strength ~log(Time)), using regression
coe cients, counted predicted values of Strength and after subtraction predicted Strength
from primary Strength, we have got values Strength without Time.

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Hypothesis 2: Concrete compressive strength depends on all ingredients, besides Time.
##
## Call:
## lm(formula = strengthWithoutTime ~ Slag + Ash + Plasticizer +
## Breakstone + Sand + Cement + Water, data = data)
##
## Residuals:
## -23.7942 -4.4161 -0.5019 3.9322 29.9414
##
## Coefficients:
## (Intercept) -84.013818 18.711087 -4.490 7.93e-06 ***
## Slag 0.113116 0.007079 15.978 < 2e-16 ***
## Ash 0.092166 0.008777 10.500 < 2e-16 ***
## Plasticizer 0.147213 0.065461 2.249 0.0247 *
## Breakstone 0.029070 0.006616 4.394 1.23e-05 ***
## Sand 0.033932 0.007501 4.524 6.79e-06 ***
## Cement 0.133024 0.005965 22.301 < 2e-16 ***
## Water -0.122236 0.028234 -4.329 1.64e-05 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
Cement, Slag, Ash, Water, Time, Breakstone and Sand strongly a ect Strength.The R-squared is
Model 2 check. Conditionals of Gauss–Markov theorem

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p = 0 < 0.05, it means that residual variance is related to model predicted values. Failed
Residues don’t have the normal distribution. Failed

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Conclusion: Our hypothesis is false. Model Strength~Time is wrong, because Conditionals of
Gauss–Markov theorem are failed.
Let’s divide the dataset into two parts: rst part includes observations with main and additional
components; second part includes observations only with main components. We have build
two models for each sets.

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Hypothesis 3: Concrete compressive strength depends on only Main components.
##
## Call:
## lm(formula = strengthWithoutTime ~ Cement + Water + Breakstone +
## Sand, data = data_base)
##
## Residuals:
## -13.0673 -2.8142 -0.0592 2.0592 19.8428
##
## Coefficients:
## (Intercept) 76.471423 28.553201 2.678 0.00802 **
## Cement 0.099861 0.009644 10.355 < 2e-16 ***
## Water -0.348389 0.047123 -7.393 3.9e-12 ***
## Breakstone -0.023489 0.008764 -2.680 0.00798 **
## Sand -0.027102 0.012582 -2.154 0.03244 *
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
Cement, Slag, Ash, Water, Breakstone and Sand strongly a ect Strength, especially Cement and
Water.The R-squared is 0.8289, it indicates that 82% of the variation in the Strength is explained
by the input variables.
p = 0.0004 < 0.05, it means that residual variance is related to model predicted values. Failed
p = 0.016 < 0.05, it means that it is auto-correlation. Failed

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Conclusion: Our hypothesis is false. We can’t use model 3. Model Strength~Time is wrong,
because Conditionals of Gauss–Markov theorem are failed.
Hypothesis 4: Concrete compressive strength depends on all components (There are
additional components in all observations).
##
## Call:
## lm(formula = strengthWithoutTime ~ Cement + Slag + Ash + Water +
## Plasticizer + Breakstone + Sand, data = data_all)
##
## Residuals:
## -23.210 -4.804 -0.552 4.169 28.024
##
## Coefficients:
## (Intercept) -1.216e+02 2.181e+01 -5.578 3.32e-08 ***
## Cement 1.433e-01 6.953e-03 20.618 < 2e-16 ***
## Slag 1.130e-01 8.595e-03 13.149 < 2e-16 ***
## Ash 9.070e-02 1.046e-02 8.675 < 2e-16 ***
## Water -7.444e-02 3.202e-02 -2.324 0.0203 *
## Plasticizer 9.944e-02 7.315e-02 1.359 0.1744
## Breakstone 4.513e-02 7.919e-03 5.699 1.68e-08 ***
## Sand 4.719e-02 8.473e-03 5.570 3.47e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##

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Cement, Slag, Ash, Water, Breakstone and Sand strongly a ect Strength.The R-squared is

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Conclusion: Our hypothesis is false. We can’t use model 4. Model Strength~Time is wrong,
because Conditionals of Gauss–Markov theorem are failed.
All our models are wrong, so we have to apply a di erent analysis approach. We found out and
deleted dublicate data (with the same compositon, Time and Strength) and wrong batches. We
have got 988 batches. After that, we have grouped batches with di erent composotion, but the
same Time. As a result, we have got groups with: Time = 3, 7, 14, 28. Batches with Time = 180,

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270, 360, 365 were united in one group with Time = 300;with Time = 90, 91, 100 were united in
one group with Time = 95, because we have found approximately the same values of Strength
for batches with Time = 180, 270, 360, 365 and for Time = 90, 91, 100.
We have created 6 datasets with 125, 117, 62, 417, 86, 122, 59 observations, where Time is
respectively 3, 7, 14, 28, 56, 95 (90, 91, 100), 300 (180, 270, 360, 365).
Time NumberOfBatches
3 125
7 117
14 62
28 417
56 86
95 122
300 59
At rst, we have checked the normal distribution of Strength for di erent groups. It was
successful for group with Time = 56, 95, 300, it means that in these datasets Stregth has the
normal distribution. But, we will build model for all datasets, because it’s uncritical for Linear
Model that predicted variable doesn’t have the normal distribution.
Time = 3 Hypothesis: Concrete compressive strength depends on all components.
##
## Call:
## lm(formula = Strength ~ Cement + Sand + Breakstone + Ash + Slag,
## data = d3)
##

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## Residuals:
## -9.3248 -1.9449 -0.0407 2.2557 8.6397
##
## Coefficients:
## (Intercept) -1.398e+02 1.341e+01 -10.422 < 2e-16 ***
## Cement 1.268e-01 5.281e-03 24.015 < 2e-16 ***
## Sand 6.790e-02 7.167e-03 9.474 3.30e-16 ***
## Breakstone 5.815e-02 7.301e-03 7.964 1.11e-12 ***
## Ash 9.115e-02 7.446e-03 12.241 < 2e-16 ***
## Slag 9.537e-02 7.130e-03 13.377 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## F-statistic: 145 on 5 and 119 DF, p-value: < 2.2e-16
Cement, Slag, Ash, Breakstone and Sand strongly a ect Strength.The R-squared is 0.8531, it
indicates that 85% of the variation in the Strength is explained by the input variables.
Conditionals of Gauss–Markov theorem 1)Homoscedasticity of the dispersion of residues.
p = 0.02,it means that it isn’t auto-correlation. Sucess
Residues have the normal distribution. Sucess

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Conclusion: Our hypothesis is false. Because one contitional of Gauss–Markov theorem is
failed.
##
## Call:

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## Call:
## lm(formula = Strength ~ Cement + Water + Plasticizer + Slag -
## 1, data = d7)
##
## Residuals:
## -15.0102 -2.8381 -0.4443 2.5889 13.4414
##
## Coefficients:
## Cement 0.105573 0.005572 18.947 < 2e-16 ***
## Water -0.076393 0.010230 -7.468 1.84e-11 ***
## Plasticizer 0.590486 0.091078 6.483 2.44e-09 ***
## Slag 0.052515 0.005285 9.936 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
Cement, Slag, Water, Plasticizer strongly a ect Strength.The R-squared is 0.9749, it indicates
that 97.5% of the variation in the Strength is explained by the input variables.

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Conclusion: Our hypothesis is false. Because Homoscedasticity of the dispersion of residues is
failed.
##
## Call:
## lm(formula = Strength ~ Cement + Water + Sand + Plasticizer +
## Ash + Slag, data = d14)
##
## Residuals:
## -11.9751 -2.4389 -0.3798 2.8660 7.9254
##
## Coefficients:
## (Intercept) -42.39184 20.43416 -2.075 0.04272 *
## Cement 0.16392 0.01685 9.726 1.50e-13 ***
## Water -0.09603 0.04460 -2.153 0.03570 *
## Sand 0.03811 0.01422 2.680 0.00968 **
## Plasticizer 0.60109 0.20272 2.965 0.00447 **
## Ash 0.10555 0.02217 4.760 1.45e-05 ***
## Slag 0.14019 0.02108 6.650 1.41e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##

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Cement, Water, Slag, Ash, Plasticizer, Sand strongly a ect Strength.The R-squared is 0.7824, it
indicates that 78.2% of the variation in the Strength is explained by the input variables.
p = 0.051289, it means that residual variance isn’t related to model predicted values.Sucess

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Conclusion: Our hypothesis is true. It means that Least squares is better for this data and we
can use this model.
##
## Call:

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## Call:
## lm(formula = Strength ~ Cement + Water + Sand + Breakstone +
## Ash + Slag, data = d28)
##
## Residuals:
## -20.646 -4.046 -1.357 3.436 27.835
##
## Coefficients:
## (Intercept) -79.647940 27.560586 -2.890 0.00406 **
## Cement 0.165211 0.009482 17.423 < 2e-16 ***
## Water -0.090868 0.035966 -2.526 0.01190 *
## Sand 0.049274 0.011697 4.213 3.11e-05 ***
## Breakstone 0.034402 0.009934 3.463 0.00059 ***
## Ash 0.104380 0.014159 7.372 9.36e-13 ***
## Slag 0.139517 0.011267 12.383 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
Cement, Slag, Water, Plasticizer, Sand, Breakstone, Ash strongly a ect Strength.The R-squared
is 0.7574, it indicates that 75% of the variation in the Strength is explained by the input
variables.
p = 0,it means that it is auto-correlation. Failed

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Conclusion: Our hypothesis is false. All condidtionals of Gauss–Markov theorem are failed. We
have to split into several parts this dataset and build the model again.
##
## Call:
## lm(formula = Strength ~ Cement + Water + Sand + Ash + Slag -
## 1, data = d56)
##
## Residuals:
## -18.7469 -4.0671 -0.8282 4.7413 16.7102
##
## Coefficients:
## Cement 0.103363 0.008088 12.779 < 2e-16 ***
## Water -0.140737 0.029632 -4.749 8.69e-06 ***
## Sand 0.032485 0.005855 5.548 3.53e-07 ***
## Ash 0.117464 0.017629 6.663 2.99e-09 ***
## Slag 0.161311 0.012577 12.826 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
Cement, Water, Slag, Ash, Sand strongly a ect Strength.The R-squared is 0.9848, it indicates
that 98% of the variation in the Strength is explained by the input variables.

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can use this model.
##
## Call:

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## Call:
## lm(formula = Strength ~ Cement + Water + Ash + Slag, data = d95)
##
## Residuals:
## -18.8475 -3.0299 -0.7854 3.3528 16.1291
##
## Coefficients:
## (Intercept) 40.938765 7.004618 5.845 4.68e-08 ***
## Cement 0.110679 0.007575 14.611 < 2e-16 ***
## Water -0.202701 0.028434 -7.129 9.00e-11 ***
## Ash 0.113753 0.015122 7.522 1.20e-11 ***
## Slag 0.109286 0.008733 12.514 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
Cement, Water, Slag, Ash strongly a ect Strength.The R-squared is 0.7755, it indicates that 77%
of the variation in the Strength is explained by the input variables.

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can use this model.
##
## Call:
## lm(formula = Strength ~ Cement + Sand + Breakstone + Slag, data = d300)
##
## Residuals:
## -14.974 -4.686 -1.429 6.205 14.222
##
## Coefficients:
## (Intercept) 89.06778 31.56549 2.822 0.00667 **
## Cement -0.15165 0.05226 -2.902 0.00536 **
## Sand -0.17691 0.04058 -4.360 5.89e-05 ***
## Breakstone 0.14330 0.02781 5.152 3.73e-06 ***
## Slag -0.15096 0.05292 -2.852 0.00614 **
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## F-statistic: 18.74 on 4 and 54 DF, p-value: 1.034e-09
Cement, Water, Slag, Ash strongly a ect Strength.The R-squared is 0.5503 , it indicates that 55%
of the variation in the Strength is explained by the input variables.

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p = 0,it means that it is auto-correlation. Failed

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Conclusion: Our hypothesis is false. Because An auto-correlation check is failed.

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If we use model for Time = 14 for all data, we get such result.

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