path <- "C:/Users/91999/OneDrive/Documents/Gradedata.csv"
gradedata <- read.csv(path,header = T, sep=",")
View(gradedata)
names(gradedata)
## [1] "Marks" "Grade"
boxplot(gradedata$Marks ~ gradedata$Grade, las=1, ylab= "Marks", xlab="Grade", main="Grade by marks")

aov(gradedata$Marks ~ gradedata$Grade)
## Call:
## aov(formula = gradedata$Marks ~ gradedata$Grade)
##
## Terms:
## gradedata$Grade Residuals
## Sum of Squares 7494.964 542.000
## Deg. of Freedom 3 24
##
## Residual standard error: 4.752192
## Estimated effects may be unbalanced
Anova1 <- aov(gradedata$Marks ~ gradedata$Grade)
summary(Anova1)
## Df Sum Sq Mean Sq F value Pr(>F)
## gradedata$Grade 3 7495 2498.3 110.6 3.45e-14 ***
## Residuals 24 542 22.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
attributes(Anova1)
## $names
## [1] "coefficients" "residuals" "effects" "rank"
## [5] "fitted.values" "assign" "qr" "df.residual"
## [9] "contrasts" "xlevels" "call" "terms"
## [13] "model"
##
## $class
## [1] "aov" "lm"
Anova1$coefficients
## (Intercept) gradedata$GradeB gradedata$GradeC gradedata$GradeD
## 94.14286 -13.28571 -27.42857 -44.00000
summary(Anova1)
## Df Sum Sq Mean Sq F value Pr(>F)
## gradedata$Grade 3 7495 2498.3 110.6 3.45e-14 ***
## Residuals 24 542 22.6
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
TukeyHSD(Anova1)
## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = gradedata$Marks ~ gradedata$Grade)
##
## $`gradedata$Grade`
## diff lwr upr p adj
## B-A -13.28571 -20.29300 -6.278424 0.0001289
## C-A -27.42857 -34.43586 -20.421281 0.0000000
## D-A -44.00000 -51.00729 -36.992710 0.0000000
## C-B -14.14286 -21.15015 -7.135567 0.0000557
## D-B -30.71429 -37.72158 -23.706995 0.0000000
## D-C -16.57143 -23.57872 -9.564138 0.0000054
plot(TukeyHSD(Anova1))

plot(TukeyHSD(Anova1), las=1)

kruskal.test(gradedata$Marks ~ gradedata$Grade)
##
## Kruskal-Wallis rank sum test
##
## data: gradedata$Marks by gradedata$Grade
## Kruskal-Wallis chi-squared = 25.345, df = 3, p-value = 1.308e-05