Setting up, reading and describing the data.

setwd("C:/Users/admin/Downloads")
Store<-read.csv("Store24.csv")
attach(Store)
summary(Store)
##      store          Sales             Profit          MTenure      
##  Min.   : 1.0   Min.   : 699306   Min.   :122180   Min.   :  0.00  
##  1st Qu.:19.5   1st Qu.: 984579   1st Qu.:211004   1st Qu.:  6.67  
##  Median :38.0   Median :1127332   Median :265014   Median : 24.12  
##  Mean   :38.0   Mean   :1205413   Mean   :276314   Mean   : 45.30  
##  3rd Qu.:56.5   3rd Qu.:1362388   3rd Qu.:331314   3rd Qu.: 50.92  
##  Max.   :75.0   Max.   :2113089   Max.   :518998   Max.   :277.99  
##     CTenure              Pop             Comp          Visibility  
##  Min.   :  0.8871   Min.   : 1046   Min.   : 1.651   Min.   :2.00  
##  1st Qu.:  4.3943   1st Qu.: 5616   1st Qu.: 3.151   1st Qu.:3.00  
##  Median :  7.2115   Median : 8896   Median : 3.629   Median :3.00  
##  Mean   : 13.9315   Mean   : 9826   Mean   : 3.788   Mean   :3.08  
##  3rd Qu.: 17.2156   3rd Qu.:14104   3rd Qu.: 4.230   3rd Qu.:4.00  
##  Max.   :114.1519   Max.   :26519   Max.   :11.128   Max.   :5.00  
##     PedCount         Res          Hours24       CrewSkill    
##  Min.   :1.00   Min.   :0.00   Min.   :0.00   Min.   :2.060  
##  1st Qu.:2.00   1st Qu.:1.00   1st Qu.:1.00   1st Qu.:3.225  
##  Median :3.00   Median :1.00   Median :1.00   Median :3.500  
##  Mean   :2.96   Mean   :0.96   Mean   :0.84   Mean   :3.457  
##  3rd Qu.:4.00   3rd Qu.:1.00   3rd Qu.:1.00   3rd Qu.:3.655  
##  Max.   :5.00   Max.   :1.00   Max.   :1.00   Max.   :4.640  
##     MgrSkill        ServQual     
##  Min.   :2.957   Min.   : 57.90  
##  1st Qu.:3.344   1st Qu.: 78.95  
##  Median :3.589   Median : 89.47  
##  Mean   :3.638   Mean   : 87.15  
##  3rd Qu.:3.925   3rd Qu.: 99.90  
##  Max.   :4.622   Max.   :100.00

1)Use R to measure the mean and standard deviation of Profit.

mean(Profit)
## [1] 276313.6
sd(Profit)
## [1] 89404.08

2)Use R to measure the mean and standard deviation of MTenure.

mean(MTenure)
## [1] 45.29644
sd(MTenure)
## [1] 57.67155

3)Use R to measure the mean and standard deviation of CTenure.

mean(CTenure)
## [1] 13.9315
sd(CTenure)
## [1] 17.69752

Sorting and Subsetting data in R

attach(mtcars)
View(mtcars)
newdata <- mtcars[order(mpg),]# sort by mpg(ascending)
View(newdata)
newdata[1:5,]
##                      mpg cyl disp  hp drat    wt  qsec vs am gear carb
## Cadillac Fleetwood  10.4   8  472 205 2.93 5.250 17.98  0  0    3    4
## Lincoln Continental 10.4   8  460 215 3.00 5.424 17.82  0  0    3    4
## Camaro Z28          13.3   8  350 245 3.73 3.840 15.41  0  0    3    4
## Duster 360          14.3   8  360 245 3.21 3.570 15.84  0  0    3    4
## Chrysler Imperial   14.7   8  440 230 3.23 5.345 17.42  0  0    3    4
newdata <- mtcars[order(-mpg),] # sort by mpg (descending)
View(newdata)
detach(mtcars)

4) Use R to print the {StoreID, Sales, Profit, MTenure, CTenure} of the top 10 most profitable stores.

attach(Store)
## The following objects are masked from Store (pos = 3):
## 
##     Comp, CrewSkill, CTenure, Hours24, MgrSkill, MTenure,
##     PedCount, Pop, Profit, Res, Sales, ServQual, store, Visibility
newdata1 <-Store[order(-Profit),][1:10 ,]
data.frame(
StoreID = newdata1$store,
Sales = newdata1$Sales,
Profit = newdata1$Profit,
MTenure = newdata1$MTenure,
CTenure = newdata1$CTenure 
)
##    StoreID   Sales Profit   MTenure    CTenure
## 1       74 1782957 518998 171.09720  29.519510
## 2        7 1809256 476355  62.53080   7.326488
## 3        9 2113089 474725 108.99350   6.061602
## 4        6 1703140 469050 149.93590  11.351130
## 5       44 1807740 439781 182.23640 114.151900
## 6        2 1619874 424007  86.22219   6.636550
## 7       45 1602362 410149  47.64565   9.166325
## 8       18 1704826 394039 239.96980  33.774130
## 9       11 1583446 389886  44.81977   2.036961
## 10      47 1665657 387853  12.84790   6.636550

5)Use R to print the {StoreID, Sales, Profit, MTenure, CTenure} of the bottom 10 least profitable stores.

newdata1 <-Store[order(Profit),][1:10 ,]
data.frame(
StoreID = newdata1$store,
Sales = newdata1$Sales,
Profit = newdata1$Profit,
MTenure = newdata1$MTenure,
CTenure = newdata1$CTenure 
)
##    StoreID   Sales Profit     MTenure   CTenure
## 1       57  699306 122180  24.3485700  2.956879
## 2       66  879581 146058 115.2039000  3.876797
## 3       41  744211 147327  14.9180200 11.926080
## 4       55  925744 147672   6.6703910 18.365500
## 5       32  828918 149033  36.0792600  6.636550
## 6       13  857843 152513   0.6571813  1.577002
## 7       54  811190 159792   6.6703910  3.876797
## 8       52 1073008 169201  24.1185600  3.416838
## 9       61  716589 177046  21.8184200 13.305950
## 10      37 1202917 187765  23.1985000  1.347023

6)Use R to draw a scatter plot of Profit vs. MTenure

library(car)
scatterplot(x = Store$MTenure ,  y = Store$Profit)

7)Use R to draw a scatter plot of Profit vs. CTenure

library(car)
scatterplot(x = Store$CTenure ,  y = Store$Profit)

8)Use R to construct a Correlation Matrix for all the variables in the dataset.

matrix <- cor(Store)
round(matrix,2)
##            store Sales Profit MTenure CTenure   Pop  Comp Visibility
## store       1.00 -0.23  -0.20   -0.06    0.02 -0.29  0.03      -0.03
## Sales      -0.23  1.00   0.92    0.45    0.25  0.40 -0.24       0.13
## Profit     -0.20  0.92   1.00    0.44    0.26  0.43 -0.33       0.14
## MTenure    -0.06  0.45   0.44    1.00    0.24 -0.06  0.18       0.16
## CTenure     0.02  0.25   0.26    0.24    1.00  0.00 -0.07       0.07
## Pop        -0.29  0.40   0.43   -0.06    0.00  1.00 -0.27      -0.05
## Comp        0.03 -0.24  -0.33    0.18   -0.07 -0.27  1.00       0.03
## Visibility -0.03  0.13   0.14    0.16    0.07 -0.05  0.03       1.00
## PedCount   -0.22  0.42   0.45    0.06   -0.08  0.61 -0.15      -0.14
## Res        -0.03 -0.17  -0.16   -0.06   -0.34 -0.24  0.22       0.02
## Hours24     0.03  0.06  -0.03   -0.17    0.07 -0.22  0.13       0.05
## CrewSkill   0.05  0.16   0.16    0.10    0.26  0.28 -0.04      -0.20
## MgrSkill   -0.07  0.31   0.32    0.23    0.12  0.08  0.22       0.07
## ServQual   -0.32  0.39   0.36    0.18    0.08  0.12  0.02       0.21
##            PedCount   Res Hours24 CrewSkill MgrSkill ServQual
## store         -0.22 -0.03    0.03      0.05    -0.07    -0.32
## Sales          0.42 -0.17    0.06      0.16     0.31     0.39
## Profit         0.45 -0.16   -0.03      0.16     0.32     0.36
## MTenure        0.06 -0.06   -0.17      0.10     0.23     0.18
## CTenure       -0.08 -0.34    0.07      0.26     0.12     0.08
## Pop            0.61 -0.24   -0.22      0.28     0.08     0.12
## Comp          -0.15  0.22    0.13     -0.04     0.22     0.02
## Visibility    -0.14  0.02    0.05     -0.20     0.07     0.21
## PedCount       1.00 -0.28   -0.28      0.21     0.09    -0.01
## Res           -0.28  1.00   -0.09     -0.15    -0.03     0.09
## Hours24       -0.28 -0.09    1.00      0.11    -0.04     0.06
## CrewSkill      0.21 -0.15    0.11      1.00    -0.02    -0.03
## MgrSkill       0.09 -0.03   -0.04     -0.02     1.00     0.36
## ServQual      -0.01  0.09    0.06     -0.03     0.36     1.00

9)Use R to measure the correlation between Profit and MTenure

pm <- cor(Profit , MTenure)
round(pm , 2)
## [1] 0.44

10)Use R to measure the correlation between Profit and CTenure.

pc <- cor(Profit , CTenure)
round(pc , 2)
## [1] 0.26

11)Use R to construct the following Corrgram based on all variables in the dataset

library(corrgram)
corrgram(Store, upper.panel=panel.pie)

12)Run a Pearson’s Correlation test on the correlation between Profit and MTenure. What is the p-value?

cor.test(Profit , MTenure)
## 
##  Pearson's product-moment correlation
## 
## data:  Profit and MTenure
## t = 4.1731, df = 73, p-value = 8.193e-05
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.2353497 0.6055175
## sample estimates:
##       cor 
## 0.4388692

13)Run a Pearson’s Correlation test on the correlation between Profit and CTenure. What is the p-value?

cor.test(Profit , CTenure)
## 
##  Pearson's product-moment correlation
## 
## data:  Profit and CTenure
## t = 2.2786, df = 73, p-value = 0.02562
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.03262507 0.45786339
## sample estimates:
##       cor 
## 0.2576789

14)Run a regression of Profit on {MTenure, CTenure Comp, Pop, PedCount, Res, Hours24, Visibility}

fit <- lm(Profit ~ MTenure + CTenure + Comp + Pop + PedCount + Res + Hours24 + Visibility  ,data=Store)
summary(fit)
## 
## Call:
## lm(formula = Profit ~ MTenure + CTenure + Comp + Pop + PedCount + 
##     Res + Hours24 + Visibility, data = Store)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -105789  -35946   -7069   33780  112390 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   7610.041  66821.994   0.114 0.909674    
## MTenure        760.993    127.086   5.988 9.72e-08 ***
## CTenure        944.978    421.687   2.241 0.028400 *  
## Comp        -25286.887   5491.937  -4.604 1.94e-05 ***
## Pop              3.667      1.466   2.501 0.014890 *  
## PedCount     34087.359   9073.196   3.757 0.000366 ***
## Res          91584.675  39231.283   2.334 0.022623 *  
## Hours24      63233.307  19641.114   3.219 0.001994 ** 
## Visibility   12625.447   9087.620   1.389 0.169411    
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 56970 on 66 degrees of freedom
## Multiple R-squared:  0.6379, Adjusted R-squared:  0.594 
## F-statistic: 14.53 on 8 and 66 DF,  p-value: 5.382e-12

15)List the explanatory variable(s) whose beta-coefficients are statistically significant (p < 0.05).

Explanatory variable(s) whose beta-coefficients are statistically significant are - MTenure , CTenure, Pop ,Comp, PedCount , Res , Hours24

16)List the explanatory variable(s) whose beta-coefficients are not statistically significant.

Explanatory variable(s) whose beta-coefficients are not statistically significant are - Visibility

17)What is expected change in the Profit at a store, if the Manager’s tenure i.e. number of months of experience with Store24, increases by one month?

round(summary(fit)$coefficients["MTenure",1], digits=0)
## [1] 761

18)What is expected change in the Profit at a store, if the Crew’s tenure i.e. number of months of experience with Store24, increases by one month?

round(summary(fit)$coefficients["CTenure",1], digits=0)
## [1] 945

19)Executive Summary

1)The most profitable store is with ID:74 and the least profitable store is ID:57

2)The correlation between Profit and MTenure are 0.44 while of that between Profit and CTenure is 0.26.

3)Pearson coefficient suggests that the value of p<0.05 which means the hypothesis is true.

4)The regression coefficient suggests that the value of p is significiant which says it is a good fit model.

5)R square value is:0.6379.It means that 63.79% of variations in the dependent variable can be explained by the independent variable.

6)Adjusted R square value is 0.594.It means 59.4% variation in the dependent variable can be explained by the independent variable also the value decreases as we add no of independent variables to it.

7)Explanatory variable(s) whose beta-coefficients are statistically significant are - MTenure , CTenure, Pop , PedCount , Res , Hours24 while that whose beta-coefficients are not statistically significant is the Visibility variable.