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)

Calculating Mean, Median

summary(temp95)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   6.467   7.845   9.537   9.367  11.205  11.739
summary(temp100)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
##   3.772   6.377   7.216   6.847   7.629   8.963

Plotting normal distribution lines for 95°C & 100°C

qqnorm(temp95,main ="Normal distribution plot for 95°C",ylab= "Thickness of photoresist")
qqline(temp95)

qqnorm(temp100,main = "Normal distribution plot for 100°C",ylab="Thickness of photoresist")
qqline(temp100)

From the normal distributed curve we can see that it is normally distributed

Doing t-test

t.test(log(temp95),log(temp100),var.equal=TRUE,alternative="greater")
## 
##  Two Sample t-test
## 
## data:  log(temp95) and log(temp100)
## t = 2.5046, df = 14, p-value = 0.01262
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  0.09507682        Inf
## sample estimates:
## mean of x mean of y 
##  2.213906  1.893530

As the value p is 0.01262 which is less than 0.05 hence we reject the null hypothesis

We can concluded that the higher banking temperatures result in wafers with a lower mean photoresist thickness as the null Hypothesis is rejected

At 95% the confidence interval is 0.09507682 to ∞

From the normal probability plot we can conclude that at both 95°C & 100°C it is normally Distributed

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)
summary(temp95)
summary(temp100)
qqnorm(temp95,main ="Normal distribution plot for 95°C",ylab= "Thickness of photoresist")
qqline(temp95)
qqnorm(temp100,main = "Normal distribution plot for 100°C",ylab="Thickness of photoresist")
qqline(temp100)
t.test(temp95,temp100,var.equal=TRUE)