Problem 2.32
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
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
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
Hypothesis that we are testing, H0 : µ1 = µ2, H1: µ1 ≠ µ2
As the corelation between the two data of caliper 1 and caliper 2 is less (around 12%), so here we will be doing two sample t-test instead of paired t-test.
a)There no significant difference between the means of the population of measurements.
b)The p-value is 0.6897 which is greater than 0.05, so failed to reject null hypothesis.
c)95 percent confidence interval on the difference in mean diameter measurements for the two types of calipers is -0.001024344 to 0.001524344
Problem 2.34
kmethod<-c(1.186,1.151,1.322,1.339,1.200,1.402,1.365,1.537,1.559)
lmethod<-c(1.061,0.992,1.063,1.062,1.065,1.178,1.037,1.086,1.052)
cor(kmethod,lmethod)
## [1] 0.3821669
d<-kmethod-lmethod
summary(kmethod)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.151 1.200 1.339 1.340 1.402 1.559
summary(lmethod)
## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.992 1.052 1.062 1.066 1.065 1.178
qqnorm(kmethod,main = "NPP for Karlsruthe Method")
qqline(kmethod)
qqnorm(lmethod,,main = "NPP for Lehigh Method")
qqline(lmethod)
qqnorm(d, main = "NPP for two methods")
qqline(d)
t.test(kmethod,lmethod,paired = TRUE)
##
## Paired t-test
##
## data: kmethod and lmethod
## 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 that we are testing, H0 : µ1 = µ2, H1: µ1 ≠ µ2
a)Yes, there is a difference in mean performance between the two methods.
b)The p-value is 0.0002953 which is less than 0.05,so we reject the null hypothesis.
c)95 percent confidence interval on the difference for the two types of methods is 0.1700423 to 0.3777355
d) From the normal probability plot we can say that data for Karlsruthe method is apporximatey normally distributed.
From the normal probability plot we can say that data for Lehigh method is apporximatey normally distributed.
e)The normality assumption for the difference in ratios for the two methods, we can say that data is apporximatey normally distributed.
f)The normality assumption for paired t-test considers the difference between the paired groups.
Problem 2.29
a95<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
b100<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
qqnorm(a95)
qqline(a95)
qqnorm(b100)
qqline(b100)
d<-sd(a95)
e<-sd(b100)
f<-sqrt((d^2+e^2)/2)
f
## [1] 1.884034
power.t.test(n=8,delta = 2.5,sd=1.884034,sig.level = 0.05,power = NULL,alternative = "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
Hypothesis that we are testing, H0 : µ1 = µ2, H1: µ1 ≠ µ2
e)The assumption of normality of the photoresist thickness, we can say that the data is appoximately normally distributed.
f)The power of this test for detecting an actual difference in means of 2.5 kA is 0.809887
Problem 2.27
a<-c(2.7,4.6,2.6,3.0,3.2,3.8)
b<-c(4.6,3.4,2.9,3.5,4.1,5.1)
qqnorm(a)
qqline(a)
qqnorm(b)
qqline(b)
wilcox.test(a,b)
## Warning in wilcox.test.default(a, b): cannot compute exact p-value with ties
##
## Wilcoxon rank sum test with continuity correction
##
## data: a and b
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0
Hypothesis that we are testing, H0 : µ1 = µ2, H1: µ1 ≠ µ2
As the p-value of 0.1994 is greater than 0.05, we fail to reject null hypothesis.
a)No, C2F6 flow rate does not affect average etch uniformity.