Question 2.32

h0: mu1 = mu2; Means of the samples are equal;

h1: mu1 != mu2; Means of the samples are different;

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)
#CHECKING CORRELATION
cor(Caliper1,Caliper2)
## [1] 0.1276307
#T-TEST
t.test(Caliper1,Caliper2, paired=TRUE, alpha=0.05)
## 
##  Paired t-test
## 
## data:  Caliper1 and Caliper2
## t = 0.43179, df = 11, p-value = 0.6742
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.001024344  0.001524344
## sample estimates:
## mean of the differences 
##                 0.00025

a)

The means are: mean of x mean of y 0.26625 0.26600 Hence, the difference between the means are smaller than 0.05, thus the difference is not significant.

b)

The p-value of Two sample t-test is 0.6742 which is higher than alpha 0.05. Hence, we fail to reject the Null hypothesis which means that the mean of Caliper1 and Caliper2 are not the same.

c)

The 95 percent confidence interval is: -0.001024344 0.001524344

Question 2.34

μD = μ1 − μ2 = 0

H0: μD = 0;

H1: μD != 0;

#CHECKING CORRELATION
cor(Karlsruhe,Lehigh )
## [1] 0.3821669
#T.TEST PAIRED
t.test(Karlsruhe,Lehigh, paired=TRUE, alpha=0.05)
## 
##  Paired t-test
## 
## data:  Karlsruhe and Lehigh
## 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

a)

We reject the null hypothesis. Thus, yes there is a difference.

b)

P = 0.0002953

c)

95 percent confidence interval: 0.1700423 0.3777355

d)

Samples do not appear normally distributed.

qqnorm(Karlsruhe)
qqline(Karlsruhe)

qqnorm(Lehigh)
qqline(Lehigh)

e)

The difference between the samples looks normally distributed.

dif <- setdiff(Karlsruhe, Lehigh)
qqnorm(dif)
qqline(dif)

f)

The assumption of normality have a moderate importance. In the paired t-test, the assumption of normality applies to the distribution of the differences. Which means that the normality of the samples is not important.

Question 2.29 e,f

e)

There are no significant deviations from the normality assumptions.

qqnorm(data100c, main="100c")
qqline(data100c)

qqnorm(data95c, main="95c")
qqline(data95c)

f)

library(pwr)
?power.t.test
## starting httpd help server ... done
a <- mean(data95c)
b <- mean(data100c)
power.t.test(n = 8,d=2.5, type = "two.sample")
## 
##      Two-sample t test power calculation 
## 
##               n = 8
##           delta = 2.5
##              sd = 1
##       sig.level = 0.05
##           power = 0.9962561
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

power = 0.9962561

Question 2.27 a (non-parametric)

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)

wilcox.test(a,b, alpha = 0.05)
## Warning in wilcox.test.default(a, b, alpha = 0.05): 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

a)

No, C2F6 flow rate does not affect average etch uniformity.