Question 1 (2.32)

Given Data

Caliper1<-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)
Caliper2<-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(Caliper1,Caliper2)

## [1] 0.1276307

Calculating the Means of the samples

summary(Caliper1)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.2650  0.2650  0.2665  0.2662  0.2670  0.2680

summary(Caliper2)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##  0.2640  0.2647  0.2655  0.2660  0.2672  0.2690

a) H0 : µ1 = µ2 Hence there is no significant difference in the Mean

we can see that the co-relation is too less hence iPerforming two sample t.test

t.test(Caliper1,Caliper2)

## 
##  Welch Two Sample t-test
## 
## data:  Caliper1 and Caliper2
## 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

a) H0 : µ1 = µ2 Hence there is no significant difference in the Mean

b)p-value = 0.6897, which is greater than 0.05 Hence we failed to reject the null Hypothesis

c)95 percent confidence interval for the two calipers are:-0.001038888 - 0.001538888

Question 2(2.34)

Given Data

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

## [1] 0.3821669

Finding the means and median

summary(Karlsruhe_Method)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   1.151   1.200   1.339   1.340   1.402   1.559

summary(Lehigh_Method)

##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   0.992   1.052   1.062   1.066   1.065   1.178

Plottig normal probability plot

qqnorm(Karlsruhe_Method,main = "Normal Probability plot for Karlsruhe Method")
qqline(Karlsruhe_Method)

qqnorm(Lehigh_Method,main = "Normal Probability plot for Lehigh Method")
qqline(Lehigh_Method)

#### We can see that both are normally distributed

Calculating the difference in ratios of the two methods

difference<-Karlsruhe_Method-Lehigh_Method
qqnorm(difference,main = "Normal Probability plot for difference")
qqline(difference)

As the data of both methods is correlated positively so we perform the paired t-test to test the hypothesis

Performing T Test

t.test(Karlsruhe_Method,Lehigh_Method,paired = TRUE)

## 
##  Paired t-test
## 
## data:  Karlsruhe_Method and Lehigh_Method
## 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

Hypothesis H0 : µ1 = µ2, H1: µ1 ≠ µ2

a)Yes there is difference in the mean for both the methods

b)The p value is 0.0002953 which is less than 0.05 hence we reject the null hypothesis

c)95 percent confidence interval for both the methods is: 0.1700423 0.3777355

d)From the above normal probability plot for karlsruhe method, Lehigh method we can see that the data is approximately normally distributed.

e)from the normal probability plot for the difference between the two method we can see it is normally distributed

f)Since the sample size is below 30 so we must assume the normality by visualizing the data from normal probability plot or box plots. The normal assumption for paired t-test determines the difference between the paired groups

Question 3(2.29)

Given Data

temp95<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
temp100<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)

Plotting Normal Probability Plot

qqnorm(temp95)
qqline(temp95)

qqnorm(temp100)
qqline(temp100)

a<-sd(temp95)
b<-sd(temp100)
z<-sqrt((a^2+b^2)/2)
power.t.test(n=8,delta = 2.5,sd=1.884034,sig.level = 0.05,power = NULL,alternative = c("one.sided"))

## 
##      Two-sample t test power calculation 
## 
##               n = 8
##           delta = 2.5
##              sd = 1.884034
##       sig.level = 0.05
##           power = 0.809887
##     alternative = one.sided
## 
## NOTE: n is number in *each* group

e)From assumption of normality of the photoresist thickness, we can say that the data is appoximately normally distributed.

f)The power of t-test is 0.7320891

Hypothesis H0 : µ1 = µ2, H1: µ1 ≠ µ2

Question 4(2.27)

Given Data

UO125<-c(2.7,4.6,2.6,3.0,3.2,3.8)
UO200<-c(4.6,3.4,2.9,3.5,4.1,5.1)

Plotting Normal Probability Plot

qqnorm(UO125)
qqline(UO125)

qqnorm(UO200)
qqline(UO200)

wilcox.test(UO125,UO200)

## Warning in wilcox.test.default(UO125, UO200): cannot compute exact p-value with
## ties

## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  UO125 and UO200
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0

Homework 2

Jestin Job Sibychan

18/09/2021

Question 1 (2.32)

Given Data

Calculating the Means of the samples

a) H0 : µ1 = µ2 Hence there is no significant difference in the Mean

we can see that the co-relation is too less hence iPerforming two sample t.test

a) H0 : µ1 = µ2 Hence there is no significant difference in the Mean

b)p-value = 0.6897, which is greater than 0.05 Hence we failed to reject the null Hypothesis

c)95 percent confidence interval for the two calipers are:-0.001038888 - 0.001538888

Question 2(2.34)

Given Data

Finding the means and median

Plottig normal probability plot

Calculating the difference in ratios of the two methods

As the data of both methods is correlated positively so we perform the paired t-test to test the hypothesis

Performing T Test

Hypothesis H0 : µ1 = µ2, H1: µ1 ≠ µ2

a)Yes there is difference in the mean for both the methods

b)The p value is 0.0002953 which is less than 0.05 hence we reject the null hypothesis

c)95 percent confidence interval for both the methods is: 0.1700423 0.3777355

d)From the above normal probability plot for karlsruhe method, Lehigh method we can see that the data is approximately normally distributed.

e)from the normal probability plot for the difference between the two method we can see it is normally distributed

f)Since the sample size is below 30 so we must assume the normality by visualizing the data from normal probability plot or box plots. The normal assumption for paired t-test determines the difference between the paired groups

Question 3(2.29)

Given Data

Plotting Normal Probability Plot

e)From assumption of normality of the photoresist thickness, we can say that the data is appoximately normally distributed.

f)The power of t-test is 0.7320891

Hypothesis H0 : µ1 = µ2, H1: µ1 ≠ µ2

Question 4(2.27)

Given Data

Plotting Normal Probability Plot

from the above test we can conclude that No, C2F6 flow rate does not affect average etch uniformity.

as value of p is 0.1994 which is greater than 0.05 hence we fail to reject the null hopothesis