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A. Use the default data set InsectSprays: The counts of insects in agricultural experimental units treated with di???erent insecticides. Column 1: Insect count Column 2: The type of spray H0 : Mean of insect counts are same corresponding to every spray.

  1. How many different types of spray categories are in the data set? Using proper R-command ???nd if every spray category has the same number of rows. For example, if spray A has 10 rows, then other categories should have 10 rows.
attach(InsectSprays)
names(InsectSprays)
## [1] "count" "spray"
View(InsectSprays)
str(InsectSprays)
## 'data.frame':    72 obs. of  2 variables:
##  $ count: num  10 7 20 14 14 12 10 23 17 20 ...
##  $ spray: Factor w/ 6 levels "A","B","C","D",..: 1 1 1 1 1 1 1 1 1 1 ...
summary(InsectSprays)
##      count       spray 
##  Min.   : 0.00   A:12  
##  1st Qu.: 3.00   B:12  
##  Median : 7.00   C:12  
##  Mean   : 9.50   D:12  
##  3rd Qu.:14.25   E:12  
##  Max.   :26.00   F:12
is.factor(InsectSprays$spray)
## [1] TRUE

  1. Find the mean of insect count for the spray category A. Results in #3

  2. Show all the means corresponding to every category in a row (as output). You may use the function tapply().

tapply(InsectSprays$count, InsectSprays$spray, mean)
##         A         B         C         D         E         F 
## 14.500000 15.333333  2.083333  4.916667  3.500000 16.666667
  1. Find the variance of insect count corresponding to every category. Hint: You can use the tapply() here as well.
tapply(InsectSprays$count, InsectSprays$spray, var)
##         A         B         C         D         E         F 
## 22.272727 18.242424  3.901515  6.265152  3.000000 38.606061
  1. Check if all the A-F spray category versus insect counts are normally distributed. Print the graphs as a matrix graph with 2 rows and 3 columns.
par(mfrow = c(2,3))
qqnorm(InsectSprays$count[InsectSprays$spray=="A"])
qqline(InsectSprays$count[InsectSprays$spray=="A"], col="red")
qqnorm(InsectSprays$count[InsectSprays$spray=="B"])
qqline(InsectSprays$count[InsectSprays$spray=="B"], col="orange")
qqnorm(InsectSprays$count[InsectSprays$spray=="C"])
qqline(InsectSprays$count[InsectSprays$spray=="C"], col="brown")
qqnorm(InsectSprays$count[InsectSprays$spray=="D"])
qqline(InsectSprays$count[InsectSprays$spray=="D"], col="purple")
qqnorm(InsectSprays$count[InsectSprays$spray=="E"])
qqline(InsectSprays$count[InsectSprays$spray=="E"], col="yellow")
qqnorm(InsectSprays$count[InsectSprays$spray=="F"])
qqline(InsectSprays$count[InsectSprays$spray=="F"], col="green")

  1. What is your conclusion regarding the homogeneity of variances.
bartlett.test(count~spray, InsectSprays)
## 
##  Bartlett test of homogeneity of variances
## 
## data:  count by spray
## Bartlett's K-squared = 25.96, df = 5, p-value = 9.085e-05

The variances are not homogenic.

  1. Do the one way ANOVA test using aov().
Anova<- aov(count~spray, data=InsectSprays)
aov(count~spray, data=InsectSprays)
## Call:
##    aov(formula = count ~ spray, data = InsectSprays)
## 
## Terms:
##                    spray Residuals
## Sum of Squares  2668.833  1015.167
## Deg. of Freedom        5        66
## 
## Residual standard error: 3.921902
## Estimated effects may be unbalanced
  1. Print the summary of the anova result. Write the numerical values of the followings:
  1. Degrees of Freedom: dfwithin, dfbetween (b) Sum of Squares: SSwithin, SSbetween (c) Mean of Squares (Variances): MSwithin, MSbetween (d) F value (e) p value. How do you interpret the p-value with respect to the null hypothesis?
summary(Anova)
##             Df Sum Sq Mean Sq F value Pr(>F)    
## spray        5   2669   533.8    34.7 <2e-16 ***
## Residuals   66   1015    15.4                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

A. df=5 B. Sum Squares= 2669, 1015 C. Variances= 534, 15 D. F value= 34.7 E. Reject the null hypothesis.

  1. Conduct the Post Hoc test to compare the mean insect counts for di???erent spray-pairs. Visualize the results with boxplot().
TukeyHSD(Anova)
##   Tukey multiple comparisons of means
##     95% family-wise confidence level
## 
## Fit: aov(formula = count ~ spray, data = InsectSprays)
## 
## $spray
##            diff        lwr       upr     p adj
## B-A   0.8333333  -3.866075  5.532742 0.9951810
## C-A -12.4166667 -17.116075 -7.717258 0.0000000
## D-A  -9.5833333 -14.282742 -4.883925 0.0000014
## E-A -11.0000000 -15.699409 -6.300591 0.0000000
## F-A   2.1666667  -2.532742  6.866075 0.7542147
## C-B -13.2500000 -17.949409 -8.550591 0.0000000
## D-B -10.4166667 -15.116075 -5.717258 0.0000002
## E-B -11.8333333 -16.532742 -7.133925 0.0000000
## F-B   1.3333333  -3.366075  6.032742 0.9603075
## D-C   2.8333333  -1.866075  7.532742 0.4920707
## E-C   1.4166667  -3.282742  6.116075 0.9488669
## F-C  14.5833333   9.883925 19.282742 0.0000000
## E-D  -1.4166667  -6.116075  3.282742 0.9488669
## F-D  11.7500000   7.050591 16.449409 0.0000000
## F-E  13.1666667   8.467258 17.866075 0.0000000
boxplot(count~as.factor(spray), data=InsectSprays)

  1. Use colors in boxplot for di???erent sprays. Also, add ylabel to the ???gure. Are there any outliers?
boxplot(count~spray, ylab="count",col=c("blue", "green", "red", "orange", "purple", "brown"), data=InsectSprays)

  1. Create the same boxplot using ggplot() and geom boxplot(). Fill the boxplots with colors as well.
library(ggplot2) 
ggplot(InsectSprays, aes(x =spray, y=count)) +

geom_boxplot(fill= "orange", colour= "blue") + 
scale_x_discrete() + xlab("Spray") +
ylab("Count")

B. Consider the default data set mtcars for the next two problems.

  1. Print the Pearson’s correlation coe???cients up to 4 decimal places.
attach(mtcars)
## The following object is masked from package:ggplot2:
## 
##     mpg
cor(mtcars, method="pearson") 
##             mpg        cyl       disp         hp        drat         wt
## mpg   1.0000000 -0.8521620 -0.8475514 -0.7761684  0.68117191 -0.8676594
## cyl  -0.8521620  1.0000000  0.9020329  0.8324475 -0.69993811  0.7824958
## disp -0.8475514  0.9020329  1.0000000  0.7909486 -0.71021393  0.8879799
## hp   -0.7761684  0.8324475  0.7909486  1.0000000 -0.44875912  0.6587479
## drat  0.6811719 -0.6999381 -0.7102139 -0.4487591  1.00000000 -0.7124406
## wt   -0.8676594  0.7824958  0.8879799  0.6587479 -0.71244065  1.0000000
## qsec  0.4186840 -0.5912421 -0.4336979 -0.7082234  0.09120476 -0.1747159
## vs    0.6640389 -0.8108118 -0.7104159 -0.7230967  0.44027846 -0.5549157
## am    0.5998324 -0.5226070 -0.5912270 -0.2432043  0.71271113 -0.6924953
## gear  0.4802848 -0.4926866 -0.5555692 -0.1257043  0.69961013 -0.5832870
## carb -0.5509251  0.5269883  0.3949769  0.7498125 -0.09078980  0.4276059
##             qsec         vs          am       gear        carb
## mpg   0.41868403  0.6640389  0.59983243  0.4802848 -0.55092507
## cyl  -0.59124207 -0.8108118 -0.52260705 -0.4926866  0.52698829
## disp -0.43369788 -0.7104159 -0.59122704 -0.5555692  0.39497686
## hp   -0.70822339 -0.7230967 -0.24320426 -0.1257043  0.74981247
## drat  0.09120476  0.4402785  0.71271113  0.6996101 -0.09078980
## wt   -0.17471588 -0.5549157 -0.69249526 -0.5832870  0.42760594
## qsec  1.00000000  0.7445354 -0.22986086 -0.2126822 -0.65624923
## vs    0.74453544  1.0000000  0.16834512  0.2060233 -0.56960714
## am   -0.22986086  0.1683451  1.00000000  0.7940588  0.05753435
## gear -0.21268223  0.2060233  0.79405876  1.0000000  0.27407284
## carb -0.65624923 -0.5696071  0.05753435  0.2740728  1.00000000
  1. Visualize histogram, scatter plot with ???tted curve and Kendall’s correlation coe???cient for mtcars in a same matrix- graph. Make sure the data on the graph is readable, especially the correlation coe???cient (up to two decimal places are enough to show).
library(psych)
## 
## Attaching package: 'psych'
## The following objects are masked from 'package:ggplot2':
## 
##     %+%, alpha
pairs.panels(mtcars, method="kendall")