Problem No.2.24

knitr::opts_chunk$set(echo = TRUE)

### As per given data in the problem,

machine1<-c(16.03,16.04,16.05,16.05,16.02,16.01,15.96,15.98,16.02,15.99)
machine2<-c(16.02,15.97,15.96,16.01,15.99,16.03,16.04,16.02,16.01,16.00)
p<-cbind(machine1,machine2)

##Now we will find the summary of machine1 and machine2

summary(machine1)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   15.96   15.99   16.02   16.02   16.04   16.05

##The summary description of machine1 data shows the minimum or lowest amount filled by machine1 is 15.96 ounces and maximum is 16.05 ounces with a mean 16.02 ounces. The first quartile is 15.99 ounces While the third quartile is 16.04 ounces.

summary(machine2)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   15.96   15.99   16.01   16.00   16.02   16.04

##The summary description of machine2 data shows the minimum or lowest amount filled by machine2 is 15.96 ounces and maximum is 16.04 ounces with a mean 16.00 ounces. The first quartile is 15.96 ounces While the third quartile is 16.02 ounces.

Now we will plot the boxplots and Normal Probability distribution plots respectively.

boxplot(machine1,machine2,names=c("machine1","machine2"),main="Boxplot for Machine1 and Machine2")

##By comparing the both boxplot we can state that mean of machine1 and machine2 are almost equal and it appears that the filling data for machine1 is more distributed as compared to filling data machine2 .

qqnorm(machine1,main="Normal Probability plot of machine1",xlab="machine1", ylab = "Net Quantity in Ounces")

qqline(machine1)

##From the Normal distribution probability plot of machine1 we can state that the data is normally distributed and 80% of the Machine1 filling distribution is within 2 standard deviation of the from the mean.

qqnorm(machine2,main="Normal Probability plot of machine2",xlab="machine2", ylab = "Net Quantity in Ounces")
qqline(machine2)

##From the Normal distribution probability plot of machine2 we can state that data is normally distributed and 80% of the Machine2 filling distribution is within 2 standard deviation of the from the mean.

Now performing the t-test

t.test(machine1,machine2,var.equal = TRUE)
## 
##  Two Sample t-test
## 
## data:  machine1 and machine2
## t = 0.79894, df = 18, p-value = 0.4347
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.01629652  0.03629652
## sample estimates:
## mean of x mean of y 
##    16.015    16.005

Answers:-

The value of p from t-test is 0.4347

Conclusion:- As the value of p is greater than 0.05, so we state that we failed to reject the null hypothesis.

The 95% confidence interval is -0.01629 to 0.03629


{"Allcode, include=FALSE"}