JAT CaseStudy-Assignment
Ramen
2022-09-14
Hypotheseis Testing- Claims Made by JAT and insights found in Data
setwd(dir="C:\\Users\\Admin\\Dropbox\\PC\\Desktop\\IIMK-BA\\Hypothesis Testing-Jayalaxmi Agro Tech")
library(readxl)
library(janitor)## Warning: package 'janitor' was built under R version 4.2.1##
## Attaching package: 'janitor'## The following objects are masked from 'package:stats':
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## chisq.test, fisher.testlibrary(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, unionlibrary(data.table)##
## Attaching package: 'data.table'## The following objects are masked from 'package:dplyr':
##
## between, first, lastlibrary(ggplot2)
Data <- read_xlsx("IMB733-XLS-ENG.xlsx",sheet = "DataSheet",col_names = T)
Belagavi <- read_xlsx("IMB733-XLS-ENG.xlsx",sheet = "Belagavi_weather",col_names = T)
Dharwad <- read_xlsx("IMB733-XLS-ENG.xlsx",sheet = "Dharwad_weather",col_names = T)
Belagavi <- clean_names(Belagavi)
Dharwad <- clean_names(Dharwad)
str(Data)## tibble [123 × 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 ...Q1. 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
Here We will test the claim using one sample t.test
Hypothesis > Ha : Mu > 60, H0 : Mu <= 60
lets subset the data first, we will use subset function
Data<- clean_names(Data)
Data1 <- subset(Data,Data$month_year>="2017-10-01")
Res1 <- t.test(Data1$d6,mu =60,alternative = "greater")
Res1##
## One Sample t-test
##
## data: Data1$d6
## t = 2.341, df = 28, p-value = 0.01329
## alternative hypothesis: true mean is greater than 60
## 95 percent confidence interval:
## 62.29976 Inf
## sample estimates:
## mean of x
## 68.41379we applied one sample t-test.The results reported that, the sample mean is 68.41379 , sample SE is 3.657714, and a t value of t.test = 2.341, with 28 degrees of freedom
the associated pvalue of test reported p= 0.01329 whihc is less than of alpha (0.05) hence we will be able to reject the null hypothesis, and support the alternative hypothesis
hence there is statistical evidence to support his claim that D6 information was accessed more then 60 times every week on average.
Q3. JAT believes that over the years, the average number of app users have increased significantly. Is there statistical evidence to support that the average number of users in year 2017-2018 is more than average number of users in year 2015-2016 at a=0.05? Support your answer with all necessary tests
We will use two sample t-test to test the Hypothesis.
Ha: Avg user per week for 2017-18 is greater then 2015-16
H0: Avg user per week for 2017-18 is less then or equal to 2015-16
Data$group <- factor(ifelse(Data$month_year >="2017-01-01","2017-18","2015-16"))
str(Data)## tibble [123 × 27] (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 ...
## $ group : Factor w/ 2 levels "2015-16","2017-18": 1 1 1 1 1 1 1 1 1 1 ...Data$group <- relevel(Data$group,ref = "2017-18")
t_test <- t.test(Data$no_of_users~Data$group,alternative = "greater",var.eq=T)
t_test##
## Two Sample t-test
##
## data: Data$no_of_users by Data$group
## t = 9.2567, df = 121, p-value = 4.753e-16
## alternative hypothesis: true difference in means between group 2017-18 and group 2015-16 is greater than 0
## 95 percent confidence interval:
## 107.5685 Inf
## sample estimates:
## mean in group 2017-18 mean in group 2015-16
## 181.10000 50.06849AS p-value is less then that of alpha, hence reject the Null.
inorder to examine the stated hypothesis, the study applied an indpendent sample t-test In this test, we compared 2017-2018 average weekly estimates against 2015-2016 average weekly users. The test resutls supported t = 9.2567, df = 121, p-value = 4.753e-16. This indicates that the associated p value of the test is less than that of alpha (0.05), hence we reject the null hypothesis, and infer that average weekly users during 2017-2018 is higher than that of 2015-2016
Q4. 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. Anand claims that app usage picked up after January 2016; so, test this hypothesis using data from January-2016 – May 2018
to test the claim we are going to use ANOVA
Ha : App Usage is same across the4 weeks of month
H0: App Usage is different across the 4 weeks of month
Data4 <- filter(Data,Data$month_year >= "2016-01-01")
Data5<- select(Data4,week,usage)
Data5$week<- as.factor(Data5$week)
levels(Data5$week)## [1] "Week1" "Week2" "Week3" "Week4"anova <- aov(usage~week,data=Data5)
summary(anova)## Df Sum Sq Mean Sq F value Pr(>F)
## week 3 1675404 558468 2.319 0.0804 .
## Residuals 94 22633380 240781
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1model.tables(anova,type = "mean")## Tables of means
## Grand mean
##
## 657.7041
##
## week
## Week1 Week2 Week3 Week4
## 622.2 583.8 541.5 875.8
## rep 25.0 24.0 24.0 25.0Based on the test result pvalue of test is .0804 which suggest that test is insignificant and can not reject the NULL and will go with null Hypothesis and We conclude that Avg App usage is different across the four weeks of a month for data from 01-01-2016 onward
Now we are going to test the claim madeby Mr Anand. Anand claims that app usage picked up after January 2016; so, test this hypothesis using data from January-2016 – May 2018.
We will two sample ttest
Avg App Usage till 31-12-2015 = Mu1
Avg App Usage after 01-01-2016 = Mu2
H0: usage didnt pick up after January 2016 (Mu1 >= Mu2) Ha: usage picked up after January 2016 (Mu1 < Mu2)
Data$group <- factor(ifelse(Data$month_year >="2015-12-31","2016Onward","Before2016"))
Data$group <- relevel(Data$group,ref ="2016Onward" )
Data7<- select(Data,usage,group)
t.test(Data7$usage~Data7$group,alternative = "greater",var.eq=TRUE)##
## Two Sample t-test
##
## data: Data7$usage by Data7$group
## t = 3.5721, df = 121, p-value = 0.0002547
## alternative hypothesis: true difference in means between group 2016Onward and group Before2016 is greater than 0
## 95 percent confidence interval:
## 198.3213 Inf
## sample estimates:
## mean in group 2016Onward mean in group Before2016
## 657.7041 287.6800As per the test result we found that t = 3.5721, df = 121, p-value = 0.0002547, hence we can reject the NULL and Infer that claimmade by Anand is true. which claims app usage picked up after January 2016 at alpha=0.05.
Q4 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.
Data8 <- Data %>% filter(month_year >= "2016-08-01")
Data8$month <- month(as.IDate(Data8$month_year, '%d/%m/%Y'))## Warning in as.POSIXlt.POSIXct(x, tz = tz): unknown timezone '%d/%m/%Y'Data9 <- Data8 %>% select(month_year,usage) %>% group_by(month_year)%>% summarise(Mean_usage = mean(usage))
UsagePlot <- ggplot(data= Data9, aes(x=month_year, y=Mean_usage, group=1)) +
geom_line(color="blue", size=1.5)+
geom_point(color="red", size=3)
UsagePlot
Summary Based on the graphical representation, mean usage shows significant growth in Oct’16 after the launch but didn’t last for long and afterwards shows downward trend and again there a huge spike in Sep’17 followed by a downward slope.. This is quite evident from the graphical visualization shown below.
Note- we do not have 3 months data (June through August 2017)
Q5 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: µ2 <= µ1 (Avg disease access when conditions are not favorable to disease = µ1, µ2 when favorable) ### Ha : µ2 > µ1
test conducted : ttest(two sample)
D1 For Belagavi and Dharwad
Q5BelD1 <- Belagavi
Q5BelD1$Fcond<- factor(ifelse(Q5BelD1$temperature >=20 & Q5BelD1$temperature <=24 & Q5BelD1$relative_humidity >80,"Yes","No"))
Q5BelD1$Fcond<-relevel(Q5BelD1$Fcond,ref = "Yes")
Q5DharD1<- Dharwad
Q5DharD1$Fcond <- factor(ifelse(Q5DharD1$temperature >=20 & Q5DharD1$temperature <= 24 & Q5DharD1$relative_humidity >80,"Yes","No"))
Q5DharD1$Fcond <- relevel(Q5DharD1$Fcond,ref = "Yes")
TestD1Bel <- t.test(d1~Fcond, data = Q5BelD1,alternative="greater",var.eq=TRUE )
TestD1Bel##
## Two Sample t-test
##
## data: d1 by Fcond
## t = 2.7605, df = 22, p-value = 0.005707
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 9.704442 Inf
## sample estimates:
## mean in group Yes mean in group No
## 37.59305 11.91669pvalue of test is 0.005707 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Belagavi, D1 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
TestD1Dhar <- t.test(d1~Fcond, data = Q5DharD1,alternative="greater",var.eq=TRUE )
TestD1Dhar##
## Two Sample t-test
##
## data: d1 by Fcond
## t = 4.5934, df = 20, p-value = 8.801e-05
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 15.66022 Inf
## sample estimates:
## mean in group Yes mean in group No
## 31.590651 6.515126pvalue of test is 8.801e-05 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Dharwad, D1 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
D2 For Belagavi and Dharwad
Q5BelD2 <- Belagavi
Q5BelD2$Fcond<- factor(ifelse(Q5BelD2$temperature >=21.5 & Q5BelD2$temperature <=24.5 & Q5BelD2$relative_humidity >83,"Yes","No"))
Q5BelD2$Fcond<-relevel(Q5BelD2$Fcond,ref = "Yes")
Q5DharD2<- Dharwad
Q5DharD2$Fcond <- factor(ifelse(Q5DharD2$temperature >=21.5 & Q5DharD2$temperature <= 24.5 & Q5DharD2$relative_humidity >83,"Yes","No"))
Q5DharD2$Fcond <- relevel(Q5DharD2$Fcond,ref = "Yes")
TestD2Bel <- t.test(d2~Fcond, data = Q5BelD2,alternative="greater",var.eq=TRUE )
TestD2Bel##
## Two Sample t-test
##
## data: d2 by Fcond
## t = 3.7247, df = 22, p-value = 0.0005887
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 10.89113 Inf
## sample estimates:
## mean in group Yes mean in group No
## 29.380223 9.173547pvalue of test is 0.0005887 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Belagavi, D2 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
TestD2Dhar <- t.test(d2~Fcond, data = Q5DharD2,alternative="greater",var.eq=TRUE )
TestD2Dhar##
## Two Sample t-test
##
## data: d2 by Fcond
## t = 4.0726, df = 20, p-value = 0.0002968
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 19.62338 Inf
## sample estimates:
## mean in group Yes mean in group No
## 40.134921 6.096486pvalue of test is 0.0002968 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Dharwad, D2 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
D3 For Belagavi and Dharwad
Q5BelD3 <- Belagavi
Q5BelD3$Fcond<- factor(ifelse(Q5BelD3$temperature >=22 & Q5BelD3$temperature <=24 ,"Yes","No"))
Q5BelD3$Fcond<-relevel(Q5BelD3$Fcond,ref = "Yes")
Q5DharD3<- Dharwad
Q5DharD3$Fcond <- factor(ifelse(Q5DharD3$temperature >=22 & Q5DharD3$temperature <= 24 ,"Yes","No"))
Q5DharD3$Fcond <- relevel(Q5DharD3$Fcond,ref = "Yes")
TestD3Bel <- t.test(d3~Fcond, data = Q5BelD3,alternative="greater",var.eq=TRUE )
TestD3Bel##
## Two Sample t-test
##
## data: d3 by Fcond
## t = 2.2224, df = 22, p-value = 0.01843
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 4.39784 Inf
## sample estimates:
## mean in group Yes mean in group No
## 30.95773 11.61233pvalue of test is 0.01843 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Belagavi, D3 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
TestD3Dhar <- t.test(d3~Fcond, data = Q5DharD3,alternative="greater",var.eq=TRUE )
TestD3Dhar##
## Two Sample t-test
##
## data: d3 by Fcond
## t = 1.5057, df = 20, p-value = 0.07389
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## -4.118138 Inf
## sample estimates:
## mean in group Yes mean in group No
## 40.26971 11.96166pvalue of test is 0.07389 which is greater then that of alpha(.05) hence can not reject the null means µ2 < =µ1
Inference: based on the test result we can say that for Dharwad, D3 disease is accessed more when conditions are not favorable to it than when the conditions are favorable.
D4 For Belagavi and Dharwad
Q5BelD4 <- Belagavi
Q5BelD4$Fcond<- factor(ifelse(Q5BelD4$temperature >=22 & Q5BelD4$temperature <=26 & Q5BelD4$relative_humidity>85 ,"Yes","No"))
Q5BelD4$Fcond<-relevel(Q5BelD4$Fcond,ref = "Yes")
Q5DharD4<- Dharwad
Q5DharD4$Fcond <- factor(ifelse(Q5DharD4$temperature >=22 & Q5DharD4$temperature <= 26 & Q5DharD4$relative_humidity>85,"Yes","No"))
Q5DharD4$Fcond <- relevel(Q5DharD4$Fcond,ref = "Yes")
TestD4Bel <- t.test(d4~Fcond, data = Q5BelD4,alternative="greater",var.eq=TRUE )
TestD4Bel##
## Two Sample t-test
##
## data: d4 by Fcond
## t = 1.793, df = 22, p-value = 0.04337
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 0.4785112 Inf
## sample estimates:
## mean in group Yes mean in group No
## 24.28984 12.97384pvalue of test is 0.04337 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Belagavi, D4 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
TestD4Dhar <- t.test(d4~Fcond, data = Q5DharD4,alternative="greater",var.eq=TRUE )
TestD4Dhar##
## Two Sample t-test
##
## data: d4 by Fcond
## t = 2.3147, df = 20, p-value = 0.01569
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 6.896259 Inf
## sample estimates:
## mean in group Yes mean in group No
## 39.16667 12.10875pvalue of test is 0.01569 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Dharwad, D4 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
D5 For Belagavi and Dharwad
Q5BelD5 <- Belagavi
Q5BelD5$Fcond<- factor(ifelse(Q5BelD5$temperature >=22 & Q5BelD5$temperature <=24.5 & Q5BelD5$relative_humidity>77 & Q5BelD5$relative_humidity <85,"Yes","No"))
Q5BelD5$Fcond<-relevel(Q5BelD5$Fcond,ref = "Yes")
Q5DharD5<- Dharwad
Q5DharD5$Fcond <- factor(ifelse(Q5DharD5$temperature >=22 & Q5DharD5$temperature <= 24.5 & Q5DharD5$relative_humidity>77 & Q5DharD5$relative_humidity<85,"Yes","No"))
Q5DharD5$Fcond <- relevel(Q5DharD5$Fcond,ref = "Yes")
TestD5Bel <- t.test(d5~Fcond, data = Q5BelD5,alternative="greater",var.eq=TRUE )
TestD5Bel##
## Two Sample t-test
##
## data: d5 by Fcond
## t = 3.6675, df = 22, p-value = 0.0006761
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 13.85781 Inf
## sample estimates:
## mean in group Yes mean in group No
## 36.57407 10.51547pvalue of test is 0.0006761 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Belagavi, D5 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
TestD5Dhar <- t.test(d5~Fcond, data = Q5DharD5,alternative="greater",var.eq=TRUE )
TestD5Dhar##
## Two Sample t-test
##
## data: d5 by Fcond
## t = 0.10853, df = 20, p-value = 0.4573
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## -16.53381 Inf
## sample estimates:
## mean in group Yes mean in group No
## 14.17749 13.06725pvalue of test is 0.4573 which is greater then that of alpha(.05) hence can not reject the null means µ2<=µ1
Inference: based on the test result we can say that for Dharwad, D5 disease is accessed more when conditions are not favorable to it than when the conditions are favorable.
D7 For Belagavi and Dharwad
Q5BelD7 <- Belagavi
Q5BelD7$Fcond<- factor(ifelse(Q5BelD7$temperature >25 & Q5BelD7$relative_humidity >80 ,"Yes","No"))
Q5BelD7$Fcond<-relevel(Q5BelD7$Fcond,ref = "Yes")
Q5DharD7<- Dharwad
Q5DharD7$Fcond <- factor(ifelse(Q5DharD7$temperature >25 & Q5DharD7$relative_humidity >80 ,"Yes","No"))
Q5DharD7$Fcond <- relevel(Q5DharD7$Fcond,ref = "Yes")
TestD7Bel <- t.test(d7~Fcond, data = Q5BelD7,alternative="greater",var.eq=TRUE )
TestD7Bel##
## Two Sample t-test
##
## data: d7 by Fcond
## t = 3.4275, df = 22, p-value = 0.001204
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## 25.65723 Inf
## sample estimates:
## mean in group Yes mean in group No
## 72.42328 21.00642pvalue of test is 0.001204 which is less then that of alpha(.05) hence reject the null means µ2<=µ1
Inference: based on the test result we can say that for Belagavi, D7 disease is accessed more when conditions are favorable to it than when the conditions are not favorable.
TestD7Dhar <- t.test(d7~Fcond, data = Q5DharD7,alternative="greater",var.eq=TRUE )
TestD7Dhar##
## Two Sample t-test
##
## data: d7 by Fcond
## t = 0.72663, df = 20, p-value = 0.2379
## alternative hypothesis: true difference in means between group Yes and group No is greater than 0
## 95 percent confidence interval:
## -20.73009 Inf
## sample estimates:
## mean in group Yes mean in group No
## 35.00000 19.90822