Task 2a
Based on thereading of the case, following are the crucial issues that are being faced by the management of store24.
1.How to increase store level employees retention.
2.How to know the financial impact of increasing tenures of crew or manager.
Questions
Based on the above stated issues, following
How tenure is related to financial performance of stores?
Correlation between the impact of manager and store performance actually related?
How tenure of manager is related to financial performance of stores?
With the available data how to predict performance of different program like, increasing wages, new training, bonus schemes etc.?
How site-locations factors related to store level financial performance
Task2c
store.df= read.csv("Store24.csv")
library(psych)
describe(store.df)[,c(2:5,8,9)]
## n mean sd median min max
## store 75 38.00 21.79 38.00 1.00 75.00
## Sales 75 1205413.12 304531.31 1127332.00 699306.00 2113089.00
## Profit 75 276313.61 89404.08 265014.00 122180.00 518998.00
## MTenure 75 45.30 57.67 24.12 0.00 277.99
## CTenure 75 13.93 17.70 7.21 0.89 114.15
## Pop 75 9825.59 5911.67 8896.00 1046.00 26519.00
## Comp 75 3.79 1.31 3.63 1.65 11.13
## Visibility 75 3.08 0.75 3.00 2.00 5.00
## PedCount 75 2.96 0.99 3.00 1.00 5.00
## Res 75 0.96 0.20 1.00 0.00 1.00
## Hours24 75 0.84 0.37 1.00 0.00 1.00
## CrewSkill 75 3.46 0.41 3.50 2.06 4.64
## MgrSkill 75 3.64 0.41 3.59 2.96 4.62
## ServQual 75 87.15 12.61 89.47 57.90 100.00
Task 2d
Use R to measure the mean and standard deviation of Profit.
store.df= read.csv("Store24.csv")
mean(store.df$Profit)
## [1] 276313.6
Mean of Profit is 276313.6
sd(store.df$Profit)
## [1] 89404.08
Standard Deviation of profit is 89404.08
Use R to measure the mean and standard deviation of MTenure.
mean(store.df$MTenure)
## [1] 45.29644
Mean of MTenure is 45.29644
sd(store.df$MTenure)
## [1] 57.67155
Standard Deviation of MTenure is 57.67155
Use R to measure the mean and standard deviation of CTenure.
mean(store.df$CTenure)
## [1] 13.9315
Mean of CTenure is 13.9315
sd(store.df$CTenure)
## [1] 17.69752
Standard Deviation of CTenure is 17.69752
Task 2e
attach(mtcars)
View(mtcars)
newdata <- mtcars[order(mpg),] # sort by mpg (ascending)
View(newdata)
newdata[1:5,] # see the first 5 rows
## 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)
Task 2f
Top 5 most profitable stores
top_profit=store.df[order(-store.df$Profit),]
top_profit[1:10,1:5]
## store Sales Profit MTenure CTenure
## 74 74 1782957 518998 171.09720 29.519510
## 7 7 1809256 476355 62.53080 7.326488
## 9 9 2113089 474725 108.99350 6.061602
## 6 6 1703140 469050 149.93590 11.351130
## 44 44 1807740 439781 182.23640 114.151900
## 2 2 1619874 424007 86.22219 6.636550
## 45 45 1602362 410149 47.64565 9.166325
## 18 18 1704826 394039 239.96980 33.774130
## 11 11 1583446 389886 44.81977 2.036961
## 47 47 1665657 387853 12.84790 6.636550
bottom_profit= store.df[order(store.df$Profit), ]
bottom_profit[1:10,1:5]
## store Sales Profit MTenure CTenure
## 57 57 699306 122180 24.3485700 2.956879
## 66 66 879581 146058 115.2039000 3.876797
## 41 41 744211 147327 14.9180200 11.926080
## 55 55 925744 147672 6.6703910 18.365500
## 32 32 828918 149033 36.0792600 6.636550
## 13 13 857843 152513 0.6571813 1.577002
## 54 54 811190 159792 6.6703910 3.876797
## 52 52 1073008 169201 24.1185600 3.416838
## 61 61 716589 177046 21.8184200 13.305950
## 37 37 1202917 187765 23.1985000 1.347023
Task 2g
Draw a scatter plot of Profit vs. MTenure.
scatter.smooth(store.df$MTenure,store.df$Profit, pch=16, col="black", lpars = list(col="red", lwd=3, lty=3))
abline(h=0, v=0, col="green")
Task 2h
Use R to draw a scatter plot of Profit vs. CTenure.
scatter.smooth(store.df$CTenure,store.df$Profit, pch=16, col="black", lpars = list(col="red", lwd=3, lty=3))
abline(h=0, v=0, col="green")
Task 2i
Use R to construct a Correlation Matrix for all the variables in the dataset.
round(cor(store.df), digits=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
Task 2j
Use R to measure the correlation between Profit and MTenure.
cor(store.df$Profit,store.df$MTenure)
## [1] 0.4388692
Use R to measure the correlation between Profit and CTenure.
cor(store.df$Profit,store.df$CTenure)
## [1] 0.2576789
Task 2k
Corrgram based on all variables in the dataset.
library(corrgram)
corrgram(store.df, order=FALSE, lower.panel=panel.shade,
upper.panel=panel.pie, text.panel=panel.txt,
main="Corrgram of store variables intercorrelations")
Task 2l
Run a Pearson’s Correlation test on the correlation
between Profit and MTenure.
What is the p-value?
cor.test(store.df$Profit,store.df$MTenure)
##
## Pearson's product-moment correlation
##
## data: store.df$Profit and store.df$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
p-value is 8.193e-05
Run a Pearson’s Correlation test on the correlation between
Profit and CTenure.
What is the p-value?
cor.test(store.df$Profit,store.df$CTenure)
##
## Pearson's product-moment correlation
##
## data: store.df$Profit and store.df$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
p-value is 0.02562
Task 2m
Run a regression of Profit on
{MTenure, CTenure Comp, Pop, PedCount, Res, Hours24, Visibility}.
store.df.lm = lm(Profit ~ MTenure+CTenure+Comp+Pop+PedCount+Res+Hours24+Visibility, data= store.df)
summary(store.df.lm)
##
## Call:
## lm(formula = Profit ~ MTenure + CTenure + Comp + Pop + PedCount +
## Res + Hours24 + Visibility, data = store.df)
##
## 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
Adjusted R-squared= 0.594
p-value= 5.382e-12
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(store.df.lm)$coefficients["MTenure",1], digits=0)
## [1] 761
761
What is expected change in the Profit at a store,
if the Crew’s tenure i.e. number of months of experience with Store24,
increases by one month?
round(summary(store.df.lm)$coefficients["CTenure",1], digits=0)
## [1] 945
945
task 2n
List the explanatory variable(s) whose beta-coefficients are statistically significant (p < 0.05).
summary(store.df.lm)$coef[summary(store.df.lm)$coef[,4] <= .05, 4]
## MTenure CTenure Comp Pop PedCount
## 9.715897e-08 2.839955e-02 1.938381e-05 1.489046e-02 3.664408e-04
## Res Hours24
## 2.262320e-02 1.993586e-03
List the explanatory variable(s) whose beta-coefficients are not statistically significant (p > 0.05).
summary(store.df.lm)$coef[summary(store.df.lm)$coef[,4] > 0.05, 4]
## (Intercept) Visibility
## 0.9096745 0.1694106
Task 2n
The visibility of the store does not matter that much in increasing profit as compared to other variables.
If the number of competitors increases by 1 unit then the profit wil directly decrease by 25286.887