2.27. An article in Solid State Technology, “Orthogonal Design for Process Optimization and Its Application to Plasma Etching” by G. Z. Yin and D. W. Jillie (May 1987) describes an experiment to determine the effect of the C2F6 flow rate on the uniformity of the etch on a silicon wafer used in integrated circuit manufacturing. All of the runs were made in random order. Data for two flow rates are as follows:

Entry of sample data and Formation of data frame.

C2F6_125<-c(2.7,4.6,2.6,3.0,3.2,3.8)
C2F6_200<-c(4.6,3.4,2.9,3.5,4.1,5.1)
 
dat<-data.frame(C2F6_125,C2F6_200)

Normal Probility plot of C2F6-125 and C2F6-200.

qqnorm(C2F6_125, main = "Normal Probility plot of C2F6-125")
qqline(C2F6_125)

qqnorm(C2F6_200, main = "Normal Probility plot of C2F6-200")
qqline(C2F6_200)

##Comments: The normal probability plot of the sample is hard to judge and to take decision on the normality because of small sample size, we can’t use T-Test for the given samples size; hence we will use nonparametric test.

Test Hypothesis

\[ H_o:\mu1=\mu2 \\ H_a:\mu1\neq \mu2 \] Null Hypothesis(Ho), Alternative Hypothesis(Ha).

where; μ1 = mean of C2F6-125, μ2 = mean of C2F6-200.

Non Parametric Test.

wilcox.test(C2F6_125,C2F6_200)
## Warning in wilcox.test.default(C2F6_125, C2F6_200): cannot compute exact p-value
## with ties
## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  C2F6_125 and C2F6_200
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0

2.27(a) Does the C2F6 flow rate affect average etch uniformity? Use α = 0.05.

Answer: From the test we could say the we fail to reject Ho because of P-value = 0.1994 this means that there is not much of affect on the average etch uniformity.

Source Code:

# Entry of sample data and Formation of data frame.

C2F6_125<-c(2.7,4.6,2.6,3.0,3.2,3.8)
C2F6_200<-c(4.6,3.4,2.9,3.5,4.1,5.1)



dat<-data.frame(C2F6_125,C2F6_200)


# Normal Probability plot of C2F6-125 and C2F6-200
qqnorm(C2F6_125, main = "Normal Probility plot of C2F6-125")
qqline(C2F6_125)
qqnorm(C2F6_200, main = "Normal Probility plot of C2F6-200")
qqline(C2F6_200)


# Non Parametric test.
wilcox.test(C2F6_125,C2F6_200)

Entering the sample data…..

C95<-c(11.176,7.089,8.097,11.739,11.291,11.291,6.467,8.315)
C100<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)

2.29(e) Check the assumption of normality of the photoresist thickness.

Normal Probility plot of 95°C and 100°C.

qqnorm(C95, main = "Normal Probility plot of 95°C")
qqline(C95)

qqnorm(C100, main = "Normal Probility plot of 100°C")
qqline(C100)

Comments: The normal probability plots of both 95°C and 100°C has the points scattered around the line as we could see and few are on the line, Hence I think we could assume it to be normally distributed.

2.29(f) Find the power of this test for detecting an actual difference in means of 2.5 kA.

Given: we know that number of sample is n1 and n2 is 8, difference in means (μ1-μ2) = 2.5.

Sp = \(\sqrt{\frac{(n1-1)*var(95°C)+(n2-1)*var(100°C)}{n1+n2-2}}\), d = \(\frac{μ1-μ2}{Sp}\)

Power Calculation.

# Pooled Standard deviation calculation.
Sp<-sqrt(((7*var(C95))+(7*var(C100)))/14)

# Calculating d-effect size.
d<- 2.5/Sp

# Power test.

library(pwr)
## Warning: package 'pwr' was built under R version 4.1.3
pwr.t.test(n = 8, d = 1.304, sig.level = 0.05, power = NULL, type = "two.sample", alternative = "greater")
## 
##      Two-sample t test power calculation 
## 
##               n = 8
##               d = 1.304
##       sig.level = 0.05
##           power = 0.7978548
##     alternative = greater
## 
## NOTE: n is number in *each* group

Answer: power = 0.7978548

Source Code:

# Entering the sample data.....
C95<-c(11.176,7.089,8.097,11.739,11.291,11.291,6.467,8.315)
C100<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)

### Normal Probability plot of 95°C and 100°C.
qqnorm(C95, main = "Normal Probility plot of 95°C")
qqline(C95)
qqnorm(C100, main = "Normal Probility plot of 100°C")
qqline(C100)

# Pooled Standard deviation calculation.
Sp<-sqrt(((7*var(C95))+(7*var(C100)))/14)

# Calculating d-effect size.
d<- 2.5/Sp

# Power test.

library(pwr)
pwr.t.test(n = 8, d = 1.304, sig.level = 0.05, power = NULL, type = "two.sample", alternative = "greater")