ANALYSIS OF JAYALAKSHMI AGRO TECH CASES USING THE DATA SHARED

IIMK ADSM 2020-21 Batch-2, Group 6: Ramana, Venugopal, Siju, Vikesh, Abdul

Set the working directory, load the needed libraries and the data

setwd("C:/_MyData_/IIMK/Assignment 1")

library (readxl)
library (dplyr)

## 
## Attaching package: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

agriData_org <- read_excel ("IMB733-XLS-ENG Spreadsheet 3.xlsx", sheet = "Data Sheet") #Load the data
agriData <- agriData_org #Backup the data frame just in case we need it later
str (agriData) #See the structure of the data frame

## tibble [123 x 26] (S3: tbl_df/tbl/data.frame)
##  $ Month-Year   : POSIXct[1:123], format: "2015-06-01" "2015-07-01" ...
##  $ Week         : chr [1:123] "Week4" "Week1" "Week2" "Week3" ...
##  $ No of users  : num [1:123] 2 1 1 4 6 12 13 10 7 12 ...
##  $ Usage        : num [1:123] 4 1 25 70 100 291 225 141 148 215 ...
##  $ D1           : num [1:123] 0 0 0 4 1 12 7 4 1 5 ...
##  $ D2           : num [1:123] 0 0 1 2 1 6 5 4 0 3 ...
##  $ D3           : num [1:123] 1 0 2 3 0 11 6 8 1 6 ...
##  $ D4           : num [1:123] 0 0 2 4 2 4 2 4 5 3 ...
##  $ D5           : num [1:123] 0 0 0 2 0 5 5 3 4 6 ...
##  $ D6           : num [1:123] 0 0 0 4 2 15 6 5 5 12 ...
##  $ D7           : num [1:123] 0 0 2 1 7 7 4 8 3 7 ...
##  $ D8           : num [1:123] 0 0 3 7 9 7 5 7 3 33 ...
##  $ D9           : num [1:123] 0 0 2 3 2 6 6 6 3 11 ...
##  $ D10          : num [1:123] 0 0 1 0 1 10 1 4 2 3 ...
##  $ D11          : num [1:123] 0 0 1 3 4 12 6 3 5 8 ...
##  $ V1           : num [1:123] 0 0 0 5 11 26 28 18 18 20 ...
##  $ V2           : num [1:123] 0 1 0 4 8 20 19 11 16 15 ...
##  $ V3           : num [1:123] 0 0 1 2 5 12 13 8 7 10 ...
##  $ V4           : num [1:123] 0 0 1 2 5 13 8 7 6 6 ...
##  $ V5           : num [1:123] 2 0 1 1 9 16 9 4 15 6 ...
##  $ V6           : num [1:123] 0 0 1 3 3 22 14 4 10 10 ...
##  $ V7           : num [1:123] 0 0 0 3 7 21 13 5 4 9 ...
##  $ V8           : num [1:123] 0 0 0 2 6 14 9 5 10 7 ...
##  $ V9           : num [1:123] 0 0 0 4 7 16 20 10 9 8 ...
##  $ V10          : num [1:123] 0 0 1 4 7 23 17 5 14 10 ...
##  $ Micronutrient: num [1:123] 1 0 6 7 3 13 22 8 7 17 ...

as.Date (agriData$`Month-Year`, "%Y-%m-%d") #Convert the Month-Year field into Date data type

## Warning in as.POSIXlt.POSIXct(x, tz = tz): unknown timezone '%Y-%m-%d'

##   [1] "2015-06-01" "2015-07-01" "2015-07-01" "2015-07-01" "2015-07-01"
##   [6] "2015-08-01" "2015-08-01" "2015-08-01" "2015-08-01" "2015-09-01"
##  [11] "2015-09-01" "2015-09-01" "2015-09-01" "2015-10-01" "2015-10-01"
##  [16] "2015-10-01" "2015-10-01" "2015-11-01" "2015-11-01" "2015-11-01"
##  [21] "2015-11-01" "2015-12-01" "2015-12-01" "2015-12-01" "2015-12-01"
##  [26] "2016-01-01" "2016-01-01" "2016-01-01" "2016-01-01" "2016-02-01"
##  [31] "2016-02-01" "2016-02-01" "2016-02-01" "2016-03-01" "2016-03-01"
##  [36] "2016-03-01" "2016-03-01" "2016-04-01" "2016-04-01" "2016-04-01"
##  [41] "2016-04-01" "2016-05-01" "2016-05-01" "2016-05-01" "2016-05-01"
##  [46] "2016-06-01" "2016-06-01" "2016-06-01" "2016-06-01" "2016-07-01"
##  [51] "2016-07-01" "2016-07-01" "2016-07-01" "2016-08-01" "2016-08-01"
##  [56] "2016-08-01" "2016-08-01" "2016-09-01" "2016-09-01" "2016-09-01"
##  [61] "2016-09-01" "2016-10-01" "2016-10-01" "2016-10-01" "2016-10-01"
##  [66] "2016-11-01" "2016-11-01" "2016-11-01" "2016-11-01" "2016-12-01"
##  [71] "2016-12-01" "2016-12-01" "2016-12-01" "2017-01-01" "2017-01-01"
##  [76] "2017-01-01" "2017-01-01" "2017-02-01" "2017-02-01" "2017-02-01"
##  [81] "2017-02-01" "2017-03-01" "2017-03-01" "2017-03-01" "2017-03-01"
##  [86] "2017-04-01" "2017-04-01" "2017-04-01" "2017-04-01" "2017-05-01"
##  [91] "2017-05-01" "2017-05-01" "2017-05-01" "2017-09-01" "2017-10-01"
##  [96] "2017-10-01" "2017-10-01" "2017-10-01" "2017-11-01" "2017-11-01"
## [101] "2017-11-01" "2017-11-01" "2017-12-01" "2017-12-01" "2017-12-01"
## [106] "2017-12-01" "2018-01-01" "2018-01-01" "2018-01-01" "2018-01-01"
## [111] "2018-02-01" "2018-02-01" "2018-02-01" "2018-02-01" "2018-03-01"
## [116] "2018-03-01" "2018-03-01" "2018-03-01" "2018-04-01" "2018-04-01"
## [121] "2018-04-01" "2018-04-01" "2018-05-01"

Case - 1

Anand, the cofounder of JAT, claims that disease 6 (leaf curl) information was accessed at least 60 times every week on average since October 2017 due to this disease outbreak. Test this claim at a significance level of 0.05 using an appropriate hypothesis test.

Hypothesis

u = Weekly average of accessing disease 6 since Oct 2017
H0: u < 60
Ha: u >= 60

Test: One sample t-test

case1Data <- subset (agriData, agriData$`Month-Year` >= "2017-10-01") #Get the data from Oct 2017
case1Test <- t.test (case1Data$D6, mu = 60, alernative = "greater") #Do the test
case1Test #Display test result

## 
##  One Sample t-test
## 
## data:  case1Data$D6
## t = 2.341, df = 28, p-value = 0.02658
## alternative hypothesis: true mean is not equal to 60
## 95 percent confidence interval:
##  61.05162 75.77597
## sample estimates:
## mean of x 
##  68.41379

Findings

p-value = 0.02658 which is less than 0.05, which means we reject H0 and accept Ha
Disease 6 information is accessed on an average at 61.05 to 75.77 times per week

Conclusion

With 95% confidence we can conclude that information pertaining to disease 6 is accessed at least 60 times every week since Oct 2017

Case - 2

Hypothesis

p = Proportion of users accessing D6 among all app users for disease information
H0: p < 0.15
Ha: p >= 0.15

Test: One proportion Z-test

case2Data <- agriData
case2Data$totalDiseaseAccess <- case2Data$D1 + case2Data$D2 + case2Data$D3 + case2Data$D4 + case2Data$D5 + case2Data$D6 + case2Data$D7 + case2Data$D8 + case2Data$D9 + case2Data$D10 + case2Data$D11 #Get the total of disease access information

case2Test <- prop.test(x = sum(case2Data$D6), n = sum(case2Data$totalDiseaseAccess), p = 0.15, alternative = "greater")
case2Test

## 
##  1-sample proportions test with continuity correction
## 
## data:  sum(case2Data$D6) out of sum(case2Data$totalDiseaseAccess), null probability 0.15
## X-squared = 21.311, df = 1, p-value = 1.953e-06
## alternative hypothesis: true p is greater than 0.15
## 95 percent confidence interval:
##  0.1564156 1.0000000
## sample estimates:
##        p 
## 0.160082

Findings

p-value for proportion of D6 is 1.953e-06 which is < 0.05, which means we reject H0 and accept Ha
Proportion shown in the result is 0.160082

Conclusion

With 95% confidence we can conclude that information related to disease 6 is accessed by at least 15% among all the app users for disease information.

Case - 3

JAT believes that over the years, the average number of app users have increased significantly. Is there statistical evidence to support that the average number of users in year 2017-2018 is more than average number of users in year 2015-2016 at a=0.05? Support your answer with all necessary tests.

Hypothesis

u1 = 2015-16 [average number of app users per week]
u2 = 2017-18
H0: u1 >= u2
Ha: u1 < u2

Test: Two sample t-test

case3Data <- agriData
case3Data$yearGroup = "Year" #Add a column to segregate the data into two groups
case3Data$yearGroup[case3Data$`Month-Year` >= "2015-01-01" & case3Data$`Month-Year` <= "2016-12-31"] = "2015-16" #Set the flag for 2015-16 observations
case3Data$yearGroup[case3Data$`Month-Year` >= "2017-01-01" & case3Data$`Month-Year` <= "2018-12-31"] = "2017-18" #Set the flag for 2017-18 observations

case3Test <- t.test (case3Data$`No of users` ~ case3Data$yearGroup, alternative = "less", var.equal = TRUE) #Do the test
case3Test #Display test result

## 
##  Two Sample t-test
## 
## data:  case3Data$`No of users` by case3Data$yearGroup
## t = -9.2567, df = 121, p-value = 4.753e-16
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
##       -Inf -107.5685
## sample estimates:
## mean in group 2015-16 mean in group 2017-18 
##              50.06849             181.10000

Findings

p-value is 4.7536e-16 which is less than 0.05, which means we reject H0 and accept Ha
Mean users during 2015-16 is 50 and during 2017-18 is 181

Conclusion

With 95% confidence we can conclude that average number of app users is higher in 2017-18 than in 2015-16.

Case - 4a

Farmers use apps to access information throughout the month. Using the data, check whether app usage is same or different across the four weeks of a month.

Hypothesis

u1 = Average usage in 1st week of every month
u2 = Average usage in 2nd week of every month
u3 = Average usage in 3rd week of every month
u4 = Average usage in 4th week of every month
H0: u1 = u2 = u3 = u4
Ha: Not all u are equal (At least two u are different)

Test: One-way Anova

case4aData <- data.frame(Usage = agriData$Usage, Week = agriData$Week) #Create a data frame only with Usage and Week data
case4aData$Week <- factor (case4aData$Week) #Convert Week to categorical variable
agriAnova <- aov(Usage~Week, data = case4aData) #Do the test
summary (agriAnova) #Display summary of the test

##              Df   Sum Sq Mean Sq F value Pr(>F)  
## Week          3  1515178  505059    2.22 0.0894 .
## Residuals   119 27074553  227517                 
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

model.tables (agriAnova, type = "means") #Check the means by week

## Tables of means
## Grand mean
##          
## 582.4959 
## 
##  Week 
##     Week1 Week2 Week3 Week4
##     551.9 522.2   480 764.7
## rep  31.0  30.0    30  32.0

Findings

Average between weeks variance is 505059
Average within weeks variance is 227517
F-Value the ratio of above two averages is close to 2 which is not that high.
In other words the F-value is statistically insignificant at 95% confidence level (5% error)
p-value is 0.0894 which is > 0.05. So, we accept H0 and reject Ha
The average usage per week is like this. Week1 - 551.9, Week2 - 522.2, Week3 - 480, Week4 - 764.7 which are close enough

Conclusion

With 5% error we can conclude that app usage is almost same across the four weeks of a month.

Case - 4b

Anand claims that app usage picked up after January 2016; so, test this hypothesis using data from January-2016 - May 2018.

Hypothesis

u1 = Average usage per week since Jan 2016
u2 = Average usage per week until Dec 2015
H0: u1 <= u2
Ha: u1 > u2

Test: Two sample t-test

case4bData <- agriData
case4bData$yearGroup = "Year" #Add a column to segregate the data into two year groups
case4bData$yearGroup[case4bData$`Month-Year` <= "2015-12-31"] = "Till_2015"
case4bData$yearGroup[case4bData$`Month-Year` >= "2016-01-01"] = "From_2016"

case4Test <- t.test (case4bData$Usage ~ case4bData$yearGroup, alternative = "greater", var.equal = TRUE) #Do the test
case4Test #Display test result

## 
##  Two Sample t-test
## 
## data:  case4bData$Usage by case4bData$yearGroup
## t = 3.5721, df = 121, p-value = 0.0002547
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  198.3213      Inf
## sample estimates:
## mean in group From_2016 mean in group Till_2015 
##                657.7041                287.6800

Findings

p-value is 0.0002547 which is < 0.05. So we reject H0 and accept Ha
Mean usage until Dec 2015 is 287 and mean usage from Jan 2016 is 657

Conclusion

With 95% confidence we can conclude that app usage picked up after Jan 2016.

Case - 5

Anand claims that number of users have increased over a period of two years. He wants to understand if app usage (number of times his app is accessed in a month by various users) has increased with the increased number of users. Prove this claim statistically. Also suggest a suitable statistical test to prove that the correlation between users and usage is non-zero.

Hypothesis

H0: The app usage has not increased with the number of users (in last 2 years)
Ha: The app usage has increased with the number of users (in last 2 years)

Test: Pearson’s Correlation test

case5Data <- subset (agriData, agriData$`Month-Year` >= "2016-05-01") #We have data till May 2018. So select data from May 2016
groupByMonthCase5 <- group_by (case5Data, `Month-Year`) #Group the data by month
case5Summary <- summarise(groupByMonthCase5, Users = sum(`No of users`), Usage = sum(Usage)) #Summarize the data by total users and total usage in the month

## `summarise()` ungrouping output (override with `.groups` argument)

cor.test (case5Summary$Users, case5Summary$Usage) #Do the test

## 
##  Pearson's product-moment correlation
## 
## data:  case5Summary$Users and case5Summary$Usage
## t = 9.1887, df = 20, p-value = 1.287e-08
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.7691185 0.9577074
## sample estimates:
##       cor 
## 0.8991594

Findings

p-value is less than 0.05 which means, we reject H0 and accept Ha
The estimated correlation coefficient value is 0.899 which is non-zero and is statistically significant.
The correlation coefficient interval is 0.769 to 0.957

Conclusion

With 95% confidence we can conclude that the app usage has increased with the number of users in last 2 years.

Case - 6

A new version of the app was released in August 2016. Anand wants to understand which month in the given time frame after the launch of the new version, the mean usage pattern would start to show a statistically significant shift.

Test: Draw a line graph

case6Data <- subset (agriData, agriData$`Month-Year` >= "2016-09-01") #Extract data from Sep 2016
groupByMonthCase6 <- group_by (case6Data, `Month-Year`) #Group the data by month
case6Summary <- summarise (groupByMonthCase6, Usage = mean(Usage)) #Summarize the data by mean usage per month

## `summarise()` ungrouping output (override with `.groups` argument)

case6Summary #Display summary

## # A tibble: 18 x 2
##    `Month-Year`        Usage
##    <dttm>              <dbl>
##  1 2016-09-01 00:00:00  396.
##  2 2016-10-01 00:00:00 1215 
##  3 2016-11-01 00:00:00  751 
##  4 2016-12-01 00:00:00  496.
##  5 2017-01-01 00:00:00  442.
##  6 2017-02-01 00:00:00  429.
##  7 2017-03-01 00:00:00  681.
##  8 2017-04-01 00:00:00  356.
##  9 2017-05-01 00:00:00  203 
## 10 2017-09-01 00:00:00 1923 
## 11 2017-10-01 00:00:00 1398.
## 12 2017-11-01 00:00:00 1300.
## 13 2017-12-01 00:00:00 1446.
## 14 2018-01-01 00:00:00 1123.
## 15 2018-02-01 00:00:00  718 
## 16 2018-03-01 00:00:00  865.
## 17 2018-04-01 00:00:00  623.
## 18 2018-05-01 00:00:00  380

plot (case6Summary$`Month-Year`, case6Summary$Usage, type="o", col = "red", xlab = "Year", ylab = "Average Usage",
      main = "Line Graph for Average Usage by Month") #Draw line graph

Findings

As we can see from the data and the graph, usage Pattern started to show statistically significant shift from Sep 2017.

Case - 7

If a disease is likely to spread in particular weather condition (data given in the disease index sheet), then the access of that disease should be more in the months having suitable weather conditions. Help the analyst in coming up with a statistical test to support the claim for two districts for which the sample of weather and disease access data is provided in the data sheet. Identify the diseases for which you can support this claim. Test this claim both for temperature and relative humidity at 95% confidence.

Hypothesis

H0: A disease information should be accessed similarly irrespective of weather condition being favorable or unfavorable to that disease
Ha: A disease information should be accessed more when the weather condition is favorable to that disease
Weather condition is determined based on temperature and humidity
u1 = Mean access of the disease when the weather condition is favorable to that disease
u2 = Mean access of the disease when the weather condition is unfavorable to that disease
H0: u1 = u2
Ha: u1 > u2 or u1 < u2

Test: Two sample t-test (two tailed)

Load district wise data

belagaviAgriData <- read_excel ("IMB733-XLS-ENG Spreadsheet 3.xlsx", sheet = "Belagavi_weather") #Load the data of Belagavi district
dharwadAgriData <- read_excel ("IMB733-XLS-ENG Spreadsheet 3.xlsx", sheet = "Dharwad_weather") #Load the data of Dharwad district

Analyze for D1 in Belagavi

belagaviD1Data <- belagaviAgriData
belagaviD1Data$D1Favorable = "No" #Initialize the D1 Favorable flag with No
belagaviD1Data$D1Favorable[belagaviD1Data$Temperature >= 20 & belagaviD1Data$Temperature <= 24 & belagaviD1Data$`Relative Humidity` > 80] = "Yes" #Set the flag to Yes for the observations where weather is favorable

belagaviD1Test <- t.test (D1 ~ D1Favorable, data = belagaviD1Data, alternative = "two.sided", var.equal = TRUE) #Do the test

belagaviD1Test #Display test result

## 
##  Two Sample t-test
## 
## data:  D1 by D1Favorable
## t = -2.7605, df = 22, p-value = 0.01141
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -44.966364  -6.386351
## sample estimates:
##  mean in group No mean in group Yes 
##          11.91669          37.59305

Findings

p-value is 0.011 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 37.5 and in unfavorable weather condition is 11.91

Conclusion

With 95% confidence we can conclude that, in Belagavi district, D1 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D1 in Dharwad

dharwadD1Data <- dharwadAgriData
dharwadD1Data$D1Favorable = "No" #Initialize the D1 Favorable flag with No
dharwadD1Data$D1Favorable[dharwadD1Data$Temperature >= 20 & dharwadD1Data$Temperature <= 24 & dharwadD1Data$`Relative Humidity` > 80] = "Yes" #Set the flag to Yes for the observations where weather is favorable

dharwadD1Test <- t.test (D1 ~ D1Favorable, data = dharwadD1Data, alternative = "two.sided", var.equal = TRUE)
dharwadD1Test

## 
##  Two Sample t-test
## 
## data:  D1 by D1Favorable
## t = -4.5934, df = 20, p-value = 0.000176
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -36.46288 -13.68817
## sample estimates:
##  mean in group No mean in group Yes 
##          6.515126         31.590651

Findings

p-value is 0.000176 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 31.5 and in unfavorable weather condition is 6.5

Conclusion

With 95% confidence we can conclude that, in Dharwad district, D1 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D2 in Belagavi

belagaviD2Data <- belagaviAgriData
belagaviD2Data$D2Favorable = "No"
belagaviD2Data$D2Favorable[belagaviD2Data$Temperature >= 21.5 & belagaviD2Data$Temperature <= 24.5 & belagaviD2Data$`Relative Humidity` > 83] = "Yes"

belagaviD2Test <- t.test (D2 ~ D2Favorable, data = belagaviD2Data, alternative = "two.sided", var.equal = TRUE)
belagaviD2Test

## 
##  Two Sample t-test
## 
## data:  D2 by D2Favorable
## t = -3.7247, df = 22, p-value = 0.001177
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -31.457485  -8.955867
## sample estimates:
##  mean in group No mean in group Yes 
##          9.173547         29.380223

Findings

p-value is 0.001177 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 29.4 and in unfavorable weather condition is 9.1

Conclusion

With 95% confidence we can conclude that, in Belagavi district, D2 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D2 in Dharwad

dharwadD2Data <- dharwadAgriData
dharwadD2Data$D2Favorable = "No"
dharwadD2Data$D2Favorable[dharwadD2Data$Temperature >= 21.5 & dharwadD2Data$Temperature <= 24.5 & dharwadD2Data$`Relative Humidity` > 83] = "Yes"

dharwadD2Test <- t.test (D2 ~ D2Favorable, data = dharwadD2Data, alternative = "two.sided", var.equal = TRUE)
dharwadD2Test

## 
##  Two Sample t-test
## 
## data:  D2 by D2Favorable
## t = -4.0726, df = 20, p-value = 0.0005937
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -51.47275 -16.60412
## sample estimates:
##  mean in group No mean in group Yes 
##          6.096486         40.134921

Findings

p-value is 0.0005937 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 40.1 and in unfavorable weather condition is 6.1

Conclusion

With 95% confidence we can conclude that, in Dharwad district, D2 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D3 in Belagavi

belagaviD3Data <- belagaviAgriData
belagaviD3Data$D3Favorable = "No"
belagaviD3Data$D3Favorable[belagaviD3Data$Temperature >= 22 & belagaviD3Data$Temperature <= 24] = "Yes"

belagaviD3Test <- t.test (D3 ~ D3Favorable, data = belagaviD3Data, alternative = "two.sided", var.equal = TRUE)
belagaviD3Test

## 
##  Two Sample t-test
## 
## data:  D3 by D3Favorable
## t = -2.2224, df = 22, p-value = 0.03685
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -37.398250  -1.292554
## sample estimates:
##  mean in group No mean in group Yes 
##          11.61233          30.95773

Findings

p-value is 0.03685 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 30.9 and in unfavorable weather condition is 11.6

Conclusion

With 95% confidence we can conclude that, in Belagavi district, D3 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D3 in Dharwad

dharwadD3Data <- dharwadAgriData
dharwadD3Data$D3Favorable = "No"
dharwadD3Data$D3Favorable[dharwadD3Data$Temperature >= 22 & dharwadD3Data$Temperature <= 24] = "Yes"

dharwadD3Test <- t.test (D3 ~ D3Favorable, data = dharwadD3Data, alternative = "two.sided", var.equal = TRUE)
dharwadD3Test

## 
##  Two Sample t-test
## 
## data:  D3 by D3Favorable
## t = -1.5057, df = 20, p-value = 0.1478
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -67.52597  10.90986
## sample estimates:
##  mean in group No mean in group Yes 
##          11.96166          40.26971

Findings

p-value is 0.1478 which is > 0.05 and hence we accept H0
Mean access in favorable weather condition is 40.3 and in unfavorable weather condition is 12

Conclusion

With 95% confidence we can conclude that, in Dharwad district, D3 disease is accessed similarly irrespective of weather condition is favorable to it or not.
Doubt: The means show that it is accessed more in favorable condition. How to interpret this?

Analyze for D4 in Belagavi

belagaviD4Data <- belagaviAgriData
belagaviD4Data$D4Favorable = "No"
belagaviD4Data$D4Favorable[belagaviD4Data$Temperature >= 22 & belagaviD4Data$Temperature <= 26 & belagaviD4Data$`Relative Humidity` > 85] = "Yes"

belagaviD4Test <- t.test (D4 ~ D4Favorable, data = belagaviD4Data, alternative = "two.sided", var.equal = TRUE)
belagaviD4Test

## 
##  Two Sample t-test
## 
## data:  D4 by D4Favorable
## t = -1.793, df = 22, p-value = 0.08674
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -24.404931   1.772927
## sample estimates:
##  mean in group No mean in group Yes 
##          12.97384          24.28984

Findings

p-value is 0.08674 which is > 0.05 and hence we accept H0
Mean access in favorable weather condition is 24.2 and in unfavorable weather condition is 13

Conclusion

With 95% confidence we can conclude that, in Belagavi district, D4 disease is accessed similarly irrespective of weather condition is favorable to it or not.
Doubt: The means show that it is accessed more in favorable condition. How to interpret this?

Analyze for D4 in Dharwad

dharwadD4Data <- dharwadAgriData
dharwadD4Data$D4Favorable = "No"
dharwadD4Data$D4Favorable[dharwadD4Data$Temperature >= 22 & dharwadD4Data$Temperature <= 26 & dharwadD4Data$`Relative Humidity` > 85] = "Yes"

dharwadD4Test <- t.test (D4 ~ D4Favorable, data = dharwadD4Data, alternative = "two.sided", var.equal = TRUE)
dharwadD4Test

## 
##  Two Sample t-test
## 
## data:  D4 by D4Favorable
## t = -2.3147, df = 20, p-value = 0.03138
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -51.442463  -2.673366
## sample estimates:
##  mean in group No mean in group Yes 
##          12.10875          39.16667

Findings

p-value is 0.03138 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 39.1 and in unfavorable weather condition is 12.1

Conclusion

With 95% confidence we can conclude that, in Dharwad district, D4 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D5 in Belagavi

belagaviD5Data <- belagaviAgriData
belagaviD5Data$D5Favorable = "No"
belagaviD5Data$D5Favorable[belagaviD5Data$Temperature >= 22 & belagaviD5Data$Temperature <= 24.5 & belagaviD5Data$`Relative Humidity` >= 77 & belagaviD5Data$`Relative Humidity` <= 85] = "Yes"

belagaviD5Test <- t.test (D5 ~ D5Favorable, data = belagaviD5Data, alternative = "two.sided", var.equal = TRUE)
belagaviD5Test

## 
##  Two Sample t-test
## 
## data:  D5 by D5Favorable
## t = -3.6675, df = 22, p-value = 0.001352
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -40.79405 -11.32315
## sample estimates:
##  mean in group No mean in group Yes 
##          10.51547          36.57407

Findings

p-value is 0.001352 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 36.6 and in unfavorable weather condition is 10.5

Conclusion

With 95% confidence we can conclude that, in Belagavi district, D5 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D5 in Dharwad

dharwadD5Data <- dharwadAgriData
dharwadD5Data$D5Favorable = "No"
dharwadD5Data$D5Favorable[dharwadD5Data$Temperature >= 22 & dharwadD5Data$Temperature <= 24.5 & dharwadD5Data$`Relative Humidity` >= 77 & dharwadD5Data$`Relative Humidity` <= 85] = "Yes"

dharwadD5Test <- t.test (D5 ~ D5Favorable, data = dharwadD5Data, alternative = "two.sided", var.equal = TRUE)
dharwadD5Test

## 
##  Two Sample t-test
## 
## data:  D5 by D5Favorable
## t = -0.10853, df = 20, p-value = 0.9147
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -22.44986  20.22939
## sample estimates:
##  mean in group No mean in group Yes 
##          13.06725          14.17749

Findings

p-value is 0.9147 which is > 0.05 and hence we accept H0
Mean access in favorable weather condition is 14.1 and in unfavorable weather condition is 13

Conclusion

With 95% confidence we can conclude that, in Dharwad district, D5 disease is accessed similarly irrespective of weather condition being favorable to it or not.

Analyze for D7 in Belagavi

belagaviD7Data <- belagaviAgriData
belagaviD7Data$D7Favorable = "No"
belagaviD7Data$D7Favorable[belagaviD7Data$Temperature > 25 & belagaviD7Data$`Relative Humidity` > 80] = "Yes"

belagaviD7Test <- t.test (D7 ~ D7Favorable, data = belagaviD7Data, alternative = "two.sided", var.equal = TRUE)
belagaviD7Test

## 
##  Two Sample t-test
## 
## data:  D7 by D7Favorable
## t = -3.4275, df = 22, p-value = 0.002408
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -82.52794 -20.30579
## sample estimates:
##  mean in group No mean in group Yes 
##          21.00642          72.42328

Findings

p-value is 0.002408 which is < 0.05 and hence we reject H0
Mean access in favorable weather condition is 72.4 and in unfavorable weather condition is 21

Conclusion

With 95% confidence we can conclude that, in Belagavi district, D7 disease is accessed more when the weather condition is favorable to it than when it is unfavorable to it.

Analyze for D7 in Dharwad

dharwadD7Data <- dharwadAgriData
dharwadD7Data$D7Favorable = "No"
dharwadD7Data$D7Favorable[dharwadD7Data$Temperature > 25 & dharwadD7Data$`Relative Humidity` > 80] = "Yes"

dharwadD7Test <- t.test (D7 ~ D7Favorable, data = dharwadD7Data, alternative = "two.sided", var.equal = TRUE)
dharwadD7Test

## 
##  Two Sample t-test
## 
## data:  D7 by D7Favorable
## t = -0.72663, df = 20, p-value = 0.4759
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -58.41659  28.23304
## sample estimates:
##  mean in group No mean in group Yes 
##          19.90822          35.00000

Findings

p-value is 0.4759 which is > 0.05 and hence we accept H0
Mean access in favorable weather condition is 35 and in unfavorable weather condition is 20

Conclusion

With 95% confidence we can conclude that, in Dharwad district, D7 disease is accessed similarly irrespective of the weather condition is favorable to it or not.
Doubt: The means show that it is accessed more in favorable condition. How to interpret this?

Summary of conclusion for all cases

Cases - 1 to 6

Case	Hypothesis Accepted	Conclusion
1	Alternative	With 95% confidence we can conclude that information pertaining to disease 6 is accessed at least 60 times every week since Oct 2017.
2	Alternative	With 95% confidence we can conclude that information related to disease 6 is accessed by at least 15% among all the app users for disease information.
3	Alternative	With 95% confidence we can conclude that average number of app users is higher in 2017-18 than in 2015-16.
4a	Null	With 5% error we can conclude that app usage is almost same across the four weeks of a month.
4b	Alternative	With 95% confidence we can conclude that app usage picked up after Jan 2016.
5	Alternative	With 95% confidence we can conclude that the app usage has increased with the number of users in last 2 years.
6	Not Applicable	As we can see from the data and the graph, usage Pattern started to show statistically significant shift from Sep 2017.

Case - 7 Disease and District wise conclusion

Disease	Belagavi	Dharwad
D1	Alternative. Accessed more in favorabe weather condition.	Alternative. Accessed more in favorabe weather condition.
D2	Alternative. Accessed more in favorabe weather condition.	Alternative. Accessed more in favorabe weather condition.
D3	Alternative. Accessed more in favorabe weather condition.	Null. Accessed similarly irrespective of weather condition being favorable or not.
D4	Null. Accessed similarly irrespective of weather condition being favorable or not.	Alternative. Accessed more in favorabe weather condition.
D5	Alternative. Accessed more in favorabe weather condition.	Null. Accessed similarly irrespective of weather condition being favorable or not.
D7	Alternative. Accessed more in favorabe weather condition.	Null. Accessed similarly irrespective of weather condition being favorable or not.

ANALYSIS OF JAYALAKSHMI AGRO TECH CASES USING THE DATA SHARED

IIMK ADSM 2020-21 Batch-2, Group 6: Ramana, Venugopal, Siju, Vikesh, Abdul

Set the working directory, load the needed libraries and the data

Case - 1

Anand, the cofounder of JAT, claims that disease 6 (leaf curl) information was accessed at least 60 times every week on average since October 2017 due to this disease outbreak. Test this claim at a significance level of 0.05 using an appropriate hypothesis test.

Hypothesis

Test: One sample t-test

Findings

Conclusion

Case - 2

Among the app users for disease information, at least 15% of them access disease information related to disease 6. Use an appropriate hypothesis test to check this claim at a = 0.05

Hypothesis

Test: One proportion Z-test

Findings

Conclusion

Case - 3

JAT believes that over the years, the average number of app users have increased significantly. Is there statistical evidence to support that the average number of users in year 2017-2018 is more than average number of users in year 2015-2016 at a=0.05? Support your answer with all necessary tests.

Hypothesis

Test: Two sample t-test

Findings

Conclusion

Case - 4a

Farmers use apps to access information throughout the month. Using the data, check whether app usage is same or different across the four weeks of a month.

Hypothesis

Test: One-way Anova

Findings

Conclusion

Case - 4b

Anand claims that app usage picked up after January 2016; so, test this hypothesis using data from January-2016 - May 2018.

Hypothesis

Test: Two sample t-test

Findings

Conclusion

Case - 5

Hypothesis

Test: Pearson’s Correlation test

Findings

Conclusion

Case - 6

A new version of the app was released in August 2016. Anand wants to understand which month in the given time frame after the launch of the new version, the mean usage pattern would start to show a statistically significant shift.

Test: Draw a line graph

Findings

Case - 7

Hypothesis

Test: Two sample t-test (two tailed)

Load district wise data

Analyze for D1 in Belagavi

Findings

Conclusion

Analyze for D1 in Dharwad

Findings

Conclusion

Analyze for D2 in Belagavi

Findings

Conclusion

Analyze for D2 in Dharwad

Findings

Conclusion

Analyze for D3 in Belagavi

Findings

Conclusion

Analyze for D3 in Dharwad

Findings

Conclusion

Analyze for D4 in Belagavi

Findings

Conclusion

Analyze for D4 in Dharwad

Findings

Conclusion

Analyze for D5 in Belagavi

Findings

Conclusion

Analyze for D5 in Dharwad

Findings

Conclusion

Analyze for D7 in Belagavi

Findings

Conclusion

Analyze for D7 in Dharwad