Question 2.32

cA<-c(0.265,0.265,0.266,0.267,0.267,0.265,0.267,0.267,0.265,0.268,0.268,0.265)
cB<-c(0.264,0.265,0.264,0.266,0.267,0.268,0.264,0.265,0.265,0.267,0.268,0.269)
cor(cA,cB)
## [1] 0.1276307
sd(cA)
## [1] 0.001215431
sd(cB)
## [1] 0.001758098
boxplot(cA,cB)

qqnorm(cA)
qqline(cA)

qqnorm(cB)
qqline(cB)

wilcox.test(cA,cB)
## Warning in wilcox.test.default(cA, cB): cannot compute exact p-value with ties
## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  cA and cB
## W = 81, p-value = 0.6131
## alternative hypothesis: true location shift is not equal to 0
t.test(cA,cB)
## 
##  Welch Two Sample t-test
## 
## data:  cA and cB
## t = 0.40519, df = 19.559, p-value = 0.6897
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.001038888  0.001538888
## sample estimates:
## mean of x mean of y 
##   0.26625   0.26600

COMMENTS

##cor = 0.12 ## Null Hypotheis: H0 : mu1 - mu2 = 0 ## Alternate Hypothesis:Ha: mu1 - mu2 != 0 ## p-value: 0.6131 ## 95% Confidence Interval : -0.0010 to 0.0015 ## the stadard deivaition for Ca,Cb=0.001215431, 0.001758098

Question 2.34

kmethod<-c(1.186,1.151,1.322,1.339,1.200,1.402,1.365,1.537,1.559)
lmethod<-c(1.061,0.992,1.063,1.062,1.065,1.178,1.037,1.086,1.052)
cor(kmethod,lmethod)
## [1] 0.3821669
qqnorm(kmethod)

qqnorm(lmethod)

boxplot(kmethod,lmethod)

wilcox.test(kmethod,lmethod)
## 
##  Wilcoxon rank sum exact test
## 
## data:  kmethod and lmethod
## W = 80, p-value = 8.227e-05
## alternative hypothesis: true location shift is not equal to 0
t.test(kmethod,lmethod,paired = TRUE)
## 
##  Paired t-test
## 
## data:  kmethod and lmethod
## t = 6.0819, df = 8, p-value = 0.0002953
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.1700423 0.3777355
## sample estimates:
## mean of the differences 
##               0.2738889
t.test(kmethod,lmethod)
## 
##  Welch Two Sample t-test
## 
## data:  kmethod and lmethod
## t = 5.3302, df = 9.8059, p-value = 0.0003557
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.1590886 0.3886892
## sample estimates:
## mean of x mean of y 
##  1.340111  1.066222
qqnorm(kmethod)

qqnorm(lmethod)

d<-kmethod-lmethod
qqnorm(d)

COMMENTS

Null Hypotheis: H0 : mu1 - mu2 = 0, Alternate Hypothesis:Ha: mu1 - mu2 != 0

p-value:0.001 The p-value is 0.001 and it is very less than the alpha value = 0.05 so that we could reject the null hypothesis.So THe evidence to support a claim that there is a difference in mean performance between the two methods

95% confidence interval:0.16 to 0.38

by looking at the test be can say that the difference it is apprx. normal

The normality assumption shows that violated

normallity asumption ion paired t test the value of p-value = 0.0002953

Question 2.29

thickness1<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
thickness2<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
A1<-sd(thickness1)
A2<-sd(thickness2)
A3<-mean(A1,A2)
thickness1<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
thickness2<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
cor(thickness1,thickness2)
## [1] 0.2878289
qqnorm(thickness1)

qqnorm(thickness2)

boxplot(thickness1,thickness2)

power.t.test(n=8,delta = 2.5,sd=1.64,sig.level = 0.05,alternative = "two.sided")
## 
##      Two-sample t test power calculation 
## 
##               n = 8
##           delta = 2.5
##              sd = 1.64
##       sig.level = 0.05
##           power = 0.8090705
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

comments

cor-0.28

we cannot pool the variances for running the test As size of the boxes differ by a considerable magnitude and so we can say that the variances are not equal

Assumption of normality we see that the data points finds to be approx. normally. There is no significant deviation from normality assumption.

Question 2.27

Flow1<-c(2.7,4.6,2.6,3.0,3.2,3.8)
Flow2<-c(4.6,3.4,2.9,3.5,4.1,5.1)
cor(Flow1,Flow2)
## [1] 0.104653
var(Flow1)
## [1] 0.5776667
var(Flow2)
## [1] 0.6746667
wilcox.test(Flow1,Flow2)
## Warning in wilcox.test.default(Flow1, Flow2): cannot compute exact p-value with
## ties
## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  Flow1 and Flow2
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0

Results

#cor:0.10 ### Null Hypothesis: mu1-mu2 = 0 ### Alternate Hypothesis: mu1-mu2 != 0 ### The p-value = 0.1994 and which is > aplha=0.05 and so that failed to reject the null hypothesis. ### So, the conclusion that the C2F6 flow rate does not affect average of etching uniformity.

source code

# Question 2.32

cA<-c(0.265,0.265,0.266,0.267,0.267,0.265,0.267,0.267,0.265,0.268,0.268,0.265)
cB<-c(0.264,0.265,0.264,0.266,0.267,0.268,0.264,0.265,0.265,0.267,0.268,0.269)
cor(cA,cB)
sd(cA)
sd(cB)


boxplot(cA,cB)
qqnorm(cA)
qqline(cA)
qqnorm(cB)
qqline(cB)

wilcox.test(cA,cB)

t.test(cA,cB)

# Question 2.34

kmethod<-c(1.186,1.151,1.322,1.339,1.200,1.402,1.365,1.537,1.559)
lmethod<-c(1.061,0.992,1.063,1.062,1.065,1.178,1.037,1.086,1.052)

cor(kmethod,lmethod)
qqnorm(kmethod)
qqnorm(lmethod)
boxplot(kmethod,lmethod)


wilcox.test(kmethod,lmethod)
t.test(kmethod,lmethod,paired = TRUE)


t.test(kmethod,lmethod)

qqnorm(kmethod)
qqnorm(lmethod)

d<-kmethod-lmethod
qqnorm(d)

# Question 2.29


thickness1<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
thickness2<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
A1<-sd(thickness1)
A2<-sd(thickness2)
A3<-mean(A1,A2)

thickness1<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
thickness2<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
cor(thickness1,thickness2)


qqnorm(thickness1)
qqnorm(thickness2)


boxplot(thickness1,thickness2)


power.t.test(n=8,delta = 2.5,sd=1.64,sig.level = 0.05,alternative = "two.sided")

## question 2.27

Flow1<-c(2.7,4.6,2.6,3.0,3.2,3.8)
Flow2<-c(4.6,3.4,2.9,3.5,4.1,5.1)
cor(A1,A2)


var(A1)
var(A2)

wilcox.test(A1,A2)