Question 2.32)

Inspector <- c (1:12)
Inspector <- as.character(Inspector)
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)
dat1 <- cbind(Inspector,Caliper1,Caliper2)
dat1 <- as.data.frame(dat1)
str(dat1)

## 'data.frame':    12 obs. of  3 variables:
##  $ Inspector: chr  "1" "2" "3" "4" ...
##  $ Caliper1 : chr  "0.265" "0.265" "0.266" "0.267" ...
##  $ Caliper2 : chr  "0.264" "0.265" "0.264" "0.266" ...

dat1$Caliper1 <- as.numeric(dat1$Caliper1)
dat1$Caliper2 <- as.numeric(dat1$Caliper2)
cor(Caliper1,Caliper2)

## [1] 0.1276307

As we can see that both samples have correlation of 0.127 , we can still go with Paired T Test but as correlation value is weak and likely unimportant , hence we reject Paired t Test .

Hence , Lets check if we can go with Two Sample T Test.

qqnorm(Caliper1,main="NPP for Caliper 1")
qqline(Caliper1)

qqnorm(Caliper2,main="NPP for Caliper 2")
qqline(Caliper2)

boxplot(Caliper1,Caliper2)

As we can see that both the samples are approximately normally distributed , as points fall around the straight line.

We can see that variances does not have significant differneces , hence we can go with the Two Sample T Test

Question a) Is there a significant difference between the means of the population of measurements represented by the two samples? Use α = 0.05

Here we have to test hypotheses that ,

Null Hypthese : Ho : u1-u2= 0

Alternative Hypotheses : Ha : u1-u2=!0

library(dplyr)

t.test(dat1$Caliper1,dat1$Caliper2,alternative="two.sided",var.equal = TRUE)

## 
##  Two Sample t-test
## 
## data:  dat1$Caliper1 and dat1$Caliper2
## t = 0.40519, df = 22, p-value = 0.6893
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.001029568  0.001529568
## sample estimates:
## mean of x mean of y 
##   0.26625   0.26600

Answer :- As from above statistics we can clearly see that p value=0.6893>0.05.Hence we fail to reject NULL Hypotheses . That means there is no significant different in means of two sample collected (u1=u2 : u1-u2=0)

Question b)P value ?

Answer:- P value from above statistics came out to be 0.6893

Question c)Construct 95% confidence interval

Answer :- -0.001029568 < u1 - u2 < 0.001529568

Question 2.34

grider <- c("S1/1","S2/1","S3/1","S4/1","S5/1","S2/1","S2/2","S2/3","S2/4")
K <- c(1.186,1.151,1.322,1.339,1.200,1.402,1.365,1.537,1.559)
L <- c(1.061,0.992,1.063,1.062,1.065,1.178,1.037,1.086,1.052)
mat2 <- cbind(grider,K,L)
dat2 <- as.data.frame(mat2)
dat2$grider <- as.factor(dat2$grider)
dat2$K <- as.numeric(dat2$K)
dat2$L <- as.numeric(dat2$L)
str(dat2)

Q2.34)a) Any evidence to support for difference in mean performance

Null Hypotheses : Ho : u1=u2 : u1-u2 =0 : d = 0

Alternative Hypotheses : Ha : u1=!u2 : u1-u2=!0 : d=!0

library(dplyr)

grider <- c("S1/1","S2/1","S3/1","S4/1","S5/1","S2/1","S2/2","S2/3","S2/4")
K <- c(1.186,1.151,1.322,1.339,1.200,1.402,1.365,1.537,1.559)
L <- c(1.061,0.992,1.063,1.062,1.065,1.178,1.037,1.086,1.052)
mat2 <- cbind(grider,K,L)
dat2 <- as.data.frame(mat2)
dat2$grider <- as.factor(dat2$grider)
dat2$K <- as.numeric(dat2$K)
dat2$L <- as.numeric(dat2$L)
str(dat2)

## 'data.frame':    9 obs. of  3 variables:
##  $ grider: Factor w/ 8 levels "S1/1","S2/1",..: 1 2 6 7 8 2 3 4 5
##  $ K     : num  1.19 1.15 1.32 1.34 1.2 ...
##  $ L     : num  1.061 0.992 1.063 1.062 1.065 ...

cor(dat2$K,dat2$L)

## [1] 0.3821669

As we can see both samples have positive correlation , hence we can leverage this to use Paired T Test.

t.test(dat2$K,dat2$L,alternative="two.sided",paired = TRUE)

## 
##  Paired t-test
## 
## data:  dat2$K and dat2$L
## 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

Answer: As we can see pvalue of test is less than 0.05 . Hence we reject Null Hypothese and hence there is no such evidence to prove that mean of both sample are equal

Q2.34) b) What is the pvalue?

Answer: As we can see pvalue of test is 0.0002953

Q2.34) c) 95% confidence interval for difference in mean?

Answer: As we can see from statistics of test , 0.1700423<u1-u2:d< 0.3777355

Q2.34) d) Normaliy Assumption?

qqnorm(dat2$K,main="Karlsruhe Method")
qqline(dat2$K)

qqnorm(dat2$L,main = "Lehigh Method")
qqline(dat2$L)

Answer:As we can see if we ignore some outliers specially from Lehigh Method (The column L ) then we can say that both the sample seems to almost looks like normally distributed as almost all points are alligned in almost a straight line

Q2.34) e) Normaliy Assumption for difference in mean?

qqnorm(dat2$K-dat2$L)
qqline(dat2$K-dat2$L)

Answer:As we can see if we ignore some outliers then we can say that the differnece in mean almost looks like normally distributed as almost all points are alligned in almost a straight line

Q2.34) f)Role of Normality assumption in paired t test?

Question 2.29

Getting Data

Thick95<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
Thick100<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
dat3 <-cbind(Thick100,Thick95)
dat3 <- as.data.frame(dat3)

Question 2.29a) claim that the higher baking temperature(Thick100) results in wafers with a lower mean photoresist thickness than higher baking temperature(Thick95)? Use α = 0.05.

Null hypothesis : u1(mean of thick95) = u2 (mean of thick100)

Alternative Hypotheses : u1(mean of thick95) > u2 (mean of thick100)

We can here go with Two sample t test

t.test(Thick95,Thick100,var.equal = TRUE,alternative = "greater")

## 
##  Two Sample t-test
## 
## data:  Thick95 and Thick100
## t = 2.6751, df = 14, p-value = 0.009059
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  0.8608158       Inf
## sample estimates:
## mean of x mean of y 
##  9.366625  6.846625

Answer -: As p value = 0.009 <0.05 , hence we reject Null hypotheses , that means Mean of Thick 100 is lower than mean of Thick95.

Question 2.29b) What is p value?

Answer -: Pvalue from above statistics is 0.009059

Question 2.29c) mean differnece with 95% confidence interval

We can see from above statistics that ,

0.86<u1-u2<infinity

From this we can interpret that ,This lower confidence bound is greater than 0;therefore, there is a difference in the two temperatures on the thickness of the photoresist

Question 2,29e) Check normality

qqnorm(dat3$Thick100,main="NPP for Thick 100")
qqline(dat3$Thick100)

qqnorm(dat3$Thick95, main = "NPP for Thick 95")
qqline(dat3$Thick95)

Answer :- Here we can see the data points on both plot seems to fall almost along a straight line, hence we can say that it is approximately following normal distribution

Question 3f) Find power for detecting actual difference in mean is 2.5KA

library(pwr)

power.t.test(n=8,delta=2.5,sd=1.866,sig.level = 0.05,power=NULL,type = "two.sample",alternative = "one.sided")

## 
##      Two-sample t test power calculation 
## 
##               n = 8
##           delta = 2.5
##              sd = 1.866
##       sig.level = 0.05
##           power = 0.8164169
##     alternative = one.sided
## 
## NOTE: n is number in *each* group

Answer :- Here we got power as 0.816 i.e 81.6%.

Note - Above substituted Standard deviation is calculated using pen and paper using formula of Two Sample t test pooled variance (Sp).

Question 2.27

Getting Data

flow125 <- c(2.7,4.6,2.6,3.0,3.2,3.8)
flow200 <- c(4.6,3.4,2.9,3.5,4.1,5.1)

Question 2.27 a)Does the C2F6flow rate affect average etch uniformity? Use 0.05.

Null Hypotheses = Ho : u1 = u2 : u1-u2 = 0

Alternative Hypotheses = Ha : u1 =!u2 : u1-u2=!0

wilcox.test(flow125,flow200)

## Warning in wilcox.test.default(flow125, flow200): cannot compute exact p-value
## with ties

## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  flow125 and flow200
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0

Answer : As p value is 0.1994 > 0.05 , hence we fail to reject NULL hypotheses , i.e. C2F6 flow rate does not affect average etch uniformity . i.e. mean for flow 125 and mean of flow 200 is same

Homework - 03

Sujit Tilakraj Thakur

9/14/2021

Question 2.32)

As we can see that both samples have correlation of 0.127 , we can still go with Paired T Test but as correlation value is weak and likely unimportant , hence we reject Paired t Test .

As we can see that both the samples are approximately normally distributed , as points fall around the straight line.

We can see that variances does not have significant differneces , hence we can go with the Two Sample T Test

Question a) Is there a significant difference between the means of the population of measurements represented by the two samples? Use α = 0.05

Here we have to test hypotheses that ,

Null Hypthese : Ho : u1-u2= 0

Alternative Hypotheses : Ha : u1-u2=!0

Answer :- As from above statistics we can clearly see that p value=0.6893>0.05.Hence we fail to reject NULL Hypotheses . That means there is no significant different in means of two sample collected (u1=u2 : u1-u2=0)

Question b)P value ?

Answer:- P value from above statistics came out to be 0.6893

Question c)Construct 95% confidence interval

Answer :- -0.001029568 < u1 - u2 < 0.001529568

Question 2.34

Q2.34)a) Any evidence to support for difference in mean performance

Null Hypotheses : Ho : u1=u2 : u1-u2 =0 : d = 0

Alternative Hypotheses : Ha : u1=!u2 : u1-u2=!0 : d=!0

As we can see both samples have positive correlation , hence we can leverage this to use Paired T Test.

Answer: As we can see pvalue of test is less than 0.05 . Hence we reject Null Hypothese and hence there is no such evidence to prove that mean of both sample are equal

Q2.34) b) What is the pvalue?

Answer: As we can see pvalue of test is 0.0002953

Q2.34) c) 95% confidence interval for difference in mean?

Answer: As we can see from statistics of test , 0.1700423<u1-u2:d< 0.3777355

Q2.34) d) Normaliy Assumption?

Answer:As we can see if we ignore some outliers specially from Lehigh Method (The column L ) then we can say that both the sample seems to almost looks like normally distributed as almost all points are alligned in almost a straight line

Q2.34) e) Normaliy Assumption for difference in mean?

Answer:As we can see if we ignore some outliers then we can say that the differnece in mean almost looks like normally distributed as almost all points are alligned in almost a straight line

Q2.34) f)Role of Normality assumption in paired t test?

Question 2.29

Getting Data

Question 2.29a) claim that the higher baking temperature(Thick100) results in wafers with a lower mean photoresist thickness than higher baking temperature(Thick95)? Use α = 0.05.

Null hypothesis : u1(mean of thick95) = u2 (mean of thick100)

Alternative Hypotheses : u1(mean of thick95) > u2 (mean of thick100)

We can here go with Two sample t test

Answer -: As p value = 0.009 <0.05 , hence we reject Null hypotheses , that means Mean of Thick 100 is lower than mean of Thick95.

Question 2.29b) What is p value?

Answer -: Pvalue from above statistics is 0.009059

Question 2.29c) mean differnece with 95% confidence interval

We can see from above statistics that ,

0.86<u1-u2<infinity

From this we can interpret that ,This lower confidence bound is greater than 0;therefore, there is a difference in the two temperatures on the thickness of the photoresist

Question 2,29e) Check normality

Answer :- Here we can see the data points on both plot seems to fall almost along a straight line, hence we can say that it is approximately following normal distribution

Question 3f) Find power for detecting actual difference in mean is 2.5KA

Answer :- Here we got power as 0.816 i.e 81.6%.

Note - Above substituted Standard deviation is calculated using pen and paper using formula of Two Sample t test pooled variance (Sp).

Question 2.27

Getting Data

Question 2.27 a)Does the C2F6flow rate affect average etch uniformity? Use 0.05.

Null Hypotheses = Ho : u1 = u2 : u1-u2 = 0

Alternative Hypotheses = Ha : u1 =!u2 : u1-u2=!0

Answer : As p value is 0.1994 > 0.05 , hence we fail to reject NULL hypotheses , i.e. C2F6 flow rate does not affect average etch uniformity . i.e. mean for flow 125 and mean of flow 200 is same