Business Case yang ingin dijawab adalah Bagaimana kita dapat memprediksi nilai Sulphide Sulfur

1 Library

library(dplyr)
library(GGally)
library(MLmetrics)
library (lmtest)
library(car)

2 LOAD DATA

Load data intercept (hanya data ore) yang memiliki irisan antara Total Sulphur, Sulphide Megaskopis dan Nilai Sulphide Sulphur

ss<-read.csv("Data Input/AllZone_database_Resume_Rstudio_081019_R1.csv")
ss_z1<-ss[ss$Zone=="Z1",c(1,5,8,9,16,18,19,21,23)]
ss_z2<-ss[ss$Zone=="Z2",c(1,5,8,9,16,18,19,21,23)]
ss_z3<-ss[ss$Zone=="Z3",c(1,5,8,9,16,18,19,21,23)]
ss_z4<-ss[ss$Zone=="Z4",c(1,5,8,9,16,18,19,21,23)]
ss_z68<-ss[ss$Zone=="Z68",c(1,5,8,9,16,18,19,21,23)]
head(ss,10)

3 Data Cleansing

3.1 Cek Struktur Data

glimpse(ss)

## Rows: 93
## Columns: 25
## $ Zone         <chr> "Z1", "Z1", "Z1", "Z1", "Z2", "Z2", "Z2", "Z2", "Z2", "Z2~
## $ holeid       <chr> "PDH294", "PDH309", "PDH309", "PDH309", "PDH299", "PDH299~
## $ from         <dbl> 53.90, 46.80, 48.10, 63.35, 25.90, 27.10, 26.50, 61.75, 1~
## $ to           <dbl> 54.90, 47.60, 49.10, 66.35, 27.10, 28.40, 27.50, 62.45, 1~
## $ Cu_pc        <dbl> 0.97200, 0.00740, 0.00850, 0.01150, 0.02840, 0.05590, 0.0~
## $ Cu_ppm       <dbl> 9720.0, 74.0, 85.0, 115.0, 284.0, 559.0, 90.0, 1070.0, 64~
## $ INTERCEPT_AU <chr> "125", "125", "125", "120", "22", "22", "22", "23", "23",~
## $ Agppm        <dbl> 27.0, 10.0, 8.0, 0.5, 22.0, 98.0, 3.0, 80.0, 35.0, 62.0, ~
## $ Auppm        <dbl> 5.020, 0.590, 0.700, 0.740, 0.540, 1.480, 0.745, 0.920, 0~
## $ ALT          <chr> "SIL", "PRO", "PRO", "PRO", "SIL", "SIL", "IAR", "SIL", "~
## $ ALT_CODE     <int> 1, 5, 5, 5, 1, 1, 4, 1, 4, 1, 1, 1, 1, 1, 1, 4, 4, 4, 1, ~
## $ INTERVAL     <dbl> 6.50, 1.80, 1.00, 13.25, 2.50, 2.50, 1.00, 0.70, 10.80, 1~
## $ OXDOM        <chr> "High Sulphide", "Low Sulphide", "Low Sulphide", "Low Sul~
## $ S2S_Charc    <chr> "High Sulphide", "High Sulphide", "High Sulphide", "Low S~
## $ Sample_ID    <chr> "DC-022401-Z1", "DC-023007-Z1", "DC-023009-Z1", "DC-02301~
## $ AuCn_Meta    <dbl> 2.170, 0.170, 0.005, 0.005, 0.360, 1.020, 0.005, 0.430, 0~
## $ INTERVAL_2   <dbl> 1.0, 0.8, 1.0, 3.0, 1.2, 1.3, 1.0, 0.7, 2.0, 1.2, 1.0, 1.~
## $ S2S          <dbl> 3.91, 2.31, 2.59, 1.07, 3.15, 3.48, 2.27, 2.83, 1.36, 3.7~
## $ S_pc_Meta    <dbl> 4.14, 2.33, 2.61, 1.15, 3.27, 3.59, 2.33, 2.90, 1.38, 3.8~
## $ INTERVAL_3   <dbl> 1.0, 0.8, 1.0, 3.0, 1.2, 1.3, 1.0, 0.7, 2.0, 1.2, 1.0, 1.~
## $ SULPHIDE     <dbl> 5.8, 0.3, 0.4, 0.3, 4.0, 4.0, 1.0, 1.5, 1.3, 2.2, 1.9, 1.~
## $ Interval_4   <dbl> 1.0, 0.8, 1.0, 3.0, 1.2, 1.3, 1.0, 0.7, 2.0, 1.2, 1.0, 1.~
## $ S_pc_Expl    <dbl> 3.80, 2.25, 2.53, 1.10, 3.00, 3.61, 2.10, 2.86, 1.41, 3.9~
## $ Sppm         <int> 38000, 22500, 25300, 11000, 30000, 36100, 21000, 28600, 1~
## $ Rec_Indic    <dbl> 0.43, 0.29, 0.01, 0.01, 0.67, 0.69, 0.01, 0.47, 0.90, 0.5~

3.2 Modifikasi tipe data dan pemilihan variable

ss_clean <- ss %>% 
  mutate_at(vars(Zone,S2S_Charc),as.factor) %>% 
  select(c(Zone,S2S_Charc,S2S,S_pc_Meta,S_pc_Expl,Cu_pc,Rec_Indic))

Mengubah tipe data Zone,S2S_Charc menjadi factor
Filter variable yang akan dianalisa Zone,S2S_Charc,S2S,S_pc_Meta,S_pc_Expl,Cu_pc,Rec_Indic

3.3 Cek Missing Value

colSums(is.na(ss_clean))

##      Zone S2S_Charc       S2S S_pc_Meta S_pc_Expl     Cu_pc Rec_Indic 
##         0         0         0         0         0         0         0

Tidak terdapat missing value

3.4 EDA

ggcorr(ss_clean,label = T)

Melihat korelasi antar variable, dan hasilnya : - S2S berkorelasi sangat kuat dengan S_pc_Meta dan S-pc_Expl

4 Membuat Model

model_ss <- lm(S2S~S_pc_Meta + S_pc_Expl,data = ss_clean)
summary(model_ss)

## 
## Call:
## lm(formula = S2S ~ S_pc_Meta + S_pc_Expl, data = ss_clean)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.38296 -0.01228  0.02907  0.06260  0.08255 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.08189    0.01677  -4.882 4.53e-06 ***
## S_pc_Meta    1.03763    0.03803  27.281  < 2e-16 ***
## S_pc_Expl   -0.04263    0.03928  -1.085    0.281    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1006 on 90 degrees of freedom
## Multiple R-squared:  0.996,  Adjusted R-squared:  0.9959 
## F-statistic: 1.11e+04 on 2 and 90 DF,  p-value: < 2.2e-16

Model ini menggunakan variable dengan korelasi kuat terhadap target, yaitu : - S_pc_Meta - S_pc_Expl

5 Prediksi

predict_ss <- predict(object = model_ss,newdata = ss_clean)
head(predict_ss)

##        1        2        3        4        5        6 
## 4.051908 2.239874 2.518474 1.064495 3.183274 3.489311

6 Evaluasi Model

6.1 Adj. R-Square

summary(model_ss)

## 
## Call:
## lm(formula = S2S ~ S_pc_Meta + S_pc_Expl, data = ss_clean)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.38296 -0.01228  0.02907  0.06260  0.08255 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.08189    0.01677  -4.882 4.53e-06 ***
## S_pc_Meta    1.03763    0.03803  27.281  < 2e-16 ***
## S_pc_Expl   -0.04263    0.03928  -1.085    0.281    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1006 on 90 degrees of freedom
## Multiple R-squared:  0.996,  Adjusted R-squared:  0.9959 
## F-statistic: 1.11e+04 on 2 and 90 DF,  p-value: < 2.2e-16

R-squared = 0.996

6.2 MAPE

MAPE(y_pred = predict_ss,y_true = ss_clean$S2S)

## [1] 4.540517

7 Asumsi Liniear

-Linearity

hist(model_ss$residuals)

Variabel predictor linear terhadap target

Normal Residual

shapiro.test(model_ss$residuals)

## 
##  Shapiro-Wilk normality test
## 
## data:  model_ss$residuals
## W = 0.69922, p-value = 1.653e-12

karena p-value (1.65 x 10e-12) < alpha (0.05), maka tolak H0 Residual tidak berdistribusi Normal

-Homosdescity

bptest(model_ss)

## 
##  studentized Breusch-Pagan test
## 
## data:  model_ss
## BP = 6.1597, df = 2, p-value = 0.04597

karena p-value (0.045) < alpha (0.05), maka tolak H0 Tidak memenuhi asumsi Homosdescity

No Multikolinearity

vif(model_ss)

## S_pc_Meta S_pc_Expl 
##  32.30393  32.30393

karena vif > 10, variabel predictor saling berkorelasi Tidak memenuhi asumsi No Multicolinierity

8 Interpretasi

summary(model_ss)

## 
## Call:
## lm(formula = S2S ~ S_pc_Meta + S_pc_Expl, data = ss_clean)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.38296 -0.01228  0.02907  0.06260  0.08255 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -0.08189    0.01677  -4.882 4.53e-06 ***
## S_pc_Meta    1.03763    0.03803  27.281  < 2e-16 ***
## S_pc_Expl   -0.04263    0.03928  -1.085    0.281    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1006 on 90 degrees of freedom
## Multiple R-squared:  0.996,  Adjusted R-squared:  0.9959 
## F-statistic: 1.11e+04 on 2 and 90 DF,  p-value: < 2.2e-16

ModelRegresi = -0.08189 + 1.03763 * S_pc_Meta - 0.04263 * S_pc_Expl a. Model regesi yang diperoleh:

s2s = -0.08189 + 1.03763 * S_pc_Meta - 0.04263 * S_pc_Expl

Intercept: Ketika S_pc_Meta dan S_pc_Expl = 0, s2s yang diperoleh adalah -0.08189
Tanda koefisen slope: S_pc_Meta dan s2s berhubungan positif, S_pc_Expl dan s2s berhubungan negatif
Slope: setiap kenaikan/penurunan S_pc_Meta sebesar 1 akan meyebabakan s2s turun/naik sebesar 1.03763, setiap kenaikan/penurunan S_pc_Expl sebesar 1 akan meyebabakan s2s turun/naik sebesar 0.04263

Pengaruh variabel prediktor terhadap target

H0: S_pc_Meta tidak mempengaruhi s2s
H1: S_pc_Meta mempengaruhi s2s

Tolak H0 jika p-value < alpha (5%). Berdasarkan output di atas diperoleh p-value (<0.0000000000000002), sehingga dapat disimpulkan bahwa S_pc_Meta signifikan memepengaruhi s2s

Goodness of fit

multiple R-Squared = 0.996, S_pc_Meta dapat menggambarkan/mendeskripsikan s2s sebesar 99.6%

Sulphide Sulphur

Bayu Respati Y.

06/05/2020