Assignment 30

x=c(2,1,1,2,3,1,2,3,4,1,2)
y=c(3,1,1,4,0,7,1,1,4,6,1)

                   
boxplot(x,y,col=c("blue","red"),horizontal=TRUE,
        names=c("New Brand","Competitior's"),
        xlab="Number of Cavities")

In this boxplot the comparison of the number of cavities found in children using different tooth paste is shown. Children using the compeitor’s brand seem to have a lot more cavities.

y=c(3,1,1,4,0,7,1,1,4,6,1)
x=c(2,1,1,2,3,1,2,3,4,1,2)

qqnorm(x)

qqline(y)

qqnorm(y)

library(psych)

## Warning: package 'psych' was built under R version 3.1.2

describe(x)

##   vars  n mean sd median trimmed  mad min max range skew kurtosis  se
## 1    1 11    2  1      2    1.89 1.48   1   4     3 0.55       -1 0.3

describe(y)

##   vars  n mean   sd median trimmed  mad min max range skew kurtosis  se
## 1    1 11 2.64 2.34      1    2.44 1.48   0   7     7 0.61    -1.22 0.7

pt(-.74852616,10)

## [1] 0.2357

t.test(x,y = NULL,
       alternative="less",var.equal=TRUE)

## 
##  One Sample t-test
## 
## data:  x
## t = 6.633, df = 10, p-value = 1
## alternative hypothesis: true mean is less than 0
## 95 percent confidence interval:
##   -Inf 2.546
## sample estimates:
## mean of x 
##         2

In order to perfrom the Welch’s T-test we must assume that if two samples n1 and n2 are drawn from normal distributions, then the distribution of the n1 and n2 mean differences will be normal. It seems that we could be wrong to assume that the distributions are nromal although they do both just some form of normality. I would say they look normal enough to try the Welch’s T-test.