Question 2.24

2.24A

The null hypothesis is that the means are equal between the machines. While the alternative is that there is a non-zero difference. ## Question 2.24b a t-test is performed to test the hypothesis

mac1 <- c(16.03,16.04,16.05,16.05,16.02,16.01,15.96,15.98,16.02,15.99)

mac2 <- c(16.02,15.97,15.96,16.01,15.99,16.03,16.04,16.02,16.01,16)

t.test(mac1,mac2)
## 
##  Welch Two Sample t-test
## 
## data:  mac1 and mac2
## t = 0.79894, df = 17.493, p-value = 0.435
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.01635123  0.03635123
## sample estimates:
## mean of x mean of y 
##    16.015    16.005

Based on the P-value of 0.435 we reject the null hypothesis

##2.24c

The P value for this experiment is .435, this means that there is a 43.5% chance of the null being incorrect.

##2.24d

the 95% confidence interval of the differences between means is [-0.01635123 , 0.03635123]#meaning that 95% of mean differences are in that range

Problem 2.26

2.26a

t1 <- c(65,81,57,66,82,82,67,59,75,70) # type 1

t2 <- c(64,71,83,59,65,56,69,74,82,79) # type 2


var.test(t1,t2)
## 
##  F test to compare two variances
## 
## data:  t1 and t2
## F = 0.97822, num df = 9, denom df = 9, p-value = 0.9744
## alternative hypothesis: true ratio of variances is not equal to 1
## 95 percent confidence interval:
##  0.2429752 3.9382952
## sample estimates:
## ratio of variances 
##          0.9782168

An F-test is done to analyse if equal variance can be assumed.

The ratio from the F-test is close to one and with a small sample size we can assume equal variance.

##2.26b

We use levene’s t-test to determine if the flare vary with time

t.test(t1,t1,var.equal = T) # Levene's test is performed when var.equal is TRUE
## 
##  Two Sample t-test
## 
## data:  t1 and t1
## t = 0, df = 18, p-value = 1
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -8.704119  8.704119
## sample estimates:
## mean of x mean of y 
##      70.4      70.4

with a P-value of 1 we do not reject the null hypthesis and it can be said with high certainty that the flares do not vary in time.

2.29

Data prep

v95 <- c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
v100 <- c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)

##2.29a

By performing a 1-sided t-test testing for a lower difference in means we can determine if higher tempertures yield thinner layers

t.test(v100,v95,alternative = "l")
## 
##  Welch Two Sample t-test
## 
## data:  v100 and v95
## t = -2.6751, df = 13.226, p-value = 0.009423
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
##        -Inf -0.8539293
## sample estimates:
## mean of x mean of y 
##  6.846625  9.366625

It can be inferred from the t-test that at temperatures of 100 C the thickness is lesser than 95C

2.29b

The P-value for this test is .0094 which means that the results of the t-test are significant and the null is rejected. This means that at higher tempertures thinner layers are formed

##2.29c

The 95% confident interval on the difference in means is [-Inf -0.8539293] With the range being completly negative it can be inferred that all diffrences in means will be negative which proves that baking at 100 C provides as thinner coating.

2.29e

We can test for normality by using a Normal probability plots and histograms

qqnorm(v95)
qqline(v95)

hist(v95)

qqnorm(v100)
qqline(v100)

hist(v100)

Due to small sample the histograms and NPP plots look a little skewed but, they are fairly normal but it appears that there is a skew for the 95 C run.

Complete R code

#Question 2.24

mac1 <- c(16.03,16.04,16.05,16.05,16.02,16.01,15.96,15.98,16.02,15.99)

mac2 <- c(16.02,15.97,15.96,16.01,15.99,16.03,16.04,16.02,16.01,16)
#datq1 <- data.frame(mac1, mac2) # It's not necessary to place the vectors into a data frame

# first a dataframe with the data was created ma1 contains data for
#machine 1 & mc2 has data for machine 2

#Question 2.24a ______________________________________

# the null hypothesis is that the means are equal between the machines. 
#While the alternative is that there is a non-zero difference.


#Question 2.24b_______________________________________________________________
# a t-test is performed to test the hypothesis

t.test(mac1,mac2)

# Based on the P-value of  0.435 we reject the null hypothesis 


#2.24c  The P value for this experiment is .435, this means that there is a 43.5% chance of the null being incorrect.
#is true and there is a non zero difference in the means



#2.24d
# the 95% confidence interval of the differences between means is [-0.01635123 , 0.03635123]
#meaning that 95% of mean differences are in that range 


#________________________________________________________________________________________________
#Question 2.26



# Preparation

t1 <- c(65,81,57,66,82,82,67,59,75,70) # type 1

t2 <- c(64,71,83,59,65,56,69,74,82,79) # type 2

#2.26a
var.test(t1,t2)
# the ratio from the F-test is close to one and with a small sample size we can assume equal variance.


t.test(t1,t1,var.equal = T) # Levene's test is performed when var.equal is TRUE

# 2.26b
# with a p value of 1 we do not reject the null hypthesis and it can be said with high certainty that the machines fill to the same volume

t.test(t1,t1,var.equal = T) # Levene's test is performed when var.equal is TRUE



#2.29_________________________________________________-_____________________________________
v95 <- c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
v100 <- c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)

t.test(v100,v95,alternative = "l")

#by performing a 1-sided t-test testing for a lower difference in means
#It can be inferred from the t-test that at temperatures of 100 C the thickness is lesser than 95C

#2.29b the P-value for this test is .0094 which means that the results of the t-test are significant and the null is rejected

#29.9c the 95% confident interval on the difference in means is   [-Inf -0.8539293]
# with the range being completly negative it can be inferred that all diffrences in means will be negative which proves that baking at 
#100 C provides as thinner coating.







#29.9e
# we can test for normality by using a Normal probability plots and histograms
qqnorm(v95)
hist(v95)
qqline(v95)


qqnorm(v100)
qqline(v100)
hist(v100)
#Due to small sample the histograms and NPP plots look a little skewed but, they are fairly normal but it appears that there is a skew for the 95 C run