Question no 2

Getting Data

Type <- c("Type1","Type1","Type1","Type1","Type1","Type1","Type1","Type1","Type1","Type1","Type2","Type2","Type2","Type2","Type2","Type2","Type2","Type2","Type2","Type2")
dataq2 <- c(65,81,57,66,82,82,67,59,75,70,64,71,83,59,65,56,69,74,82,79)
datamat <- cbind(Type,dataq2)
data2 <- as.data.frame(datamat)
data2$Type <- as.factor(data2$Type)
data2$dataq2 <- as.numeric(data2$dataq2)

Question 2a) Using Levene Test Prove equality of variance

library(lawstat)
levene.test(data2$dataq2,data2$Type,location="mean")
## 
##  Classical Levene's test based on the absolute deviations from the mean
##  ( none not applied because the location is not set to median )
## 
## data:  data2$dataq2
## Test Statistic = 0.0014598, p-value = 0.9699

Answer - Here are P value is > 0.05 , Hence accept we can say that accept Null Hypotheses that variance are equal

Question 2b)

As from above test we came to know that varinaces are equal , hence we can use Two Sample t test with pooled variance to test the hypotheses

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

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

library(dplyr)
datt1 <- data2 %>% filter(Type=="Type1") %>% select(dataq2)
datt2 <- data2 %>% filter(Type=="Type2") %>% select(dataq2)
t.test(datt1,datt2,var.equal = TRUE)

Answer :- As p value=0.96 > 0.05, hence accept Null Hypotheses that mean burning time is equal . P value for above test came out to be 0.9622.

Question no 3

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)
data <-cbind(Thick100,Thick95)
data <- as.data.frame(data)

Question 3a) 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(Thick100,Thick95,var.equal = TRUE,alternative = "less")
## 
##  Two Sample t-test
## 
## data:  Thick100 and Thick95
## t = -2.6751, df = 14, p-value = 0.009059
## alternative hypothesis: true difference in means is less than 0
## 95 percent confidence interval:
##        -Inf -0.8608158
## sample estimates:
## mean of x mean of y 
##  6.846625  9.366625

Answer -: As p value = 0.009 <0.5 , hence we reject Null hypotheses , that means Mean of Thick 100 is lower than mean of Thick95.

Question 3b) What is p value?

Answer -: Pvalue from above statistics is 0.009059

Question 3c) 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 3e) Check normality

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

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