Question 2.32

The diameter of a ball bearing was measured by 12inspectors, each using two different kinds of calipers. The results were

Data Entry of the values of Caliper 1 and Caliper 2

c_1<-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)
c_2<-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)

we are using paired T-Test because the c_1 and c_2 are from the same inspector

t.test(c_1,c_2, paired=TRUE,conf.level = 0.95)
## 
##  Paired t-test
## 
## data:  c_1 and c_2
## t = 0.43179, df = 11, p-value = 0.6742
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  -0.001024344  0.001524344
## sample estimates:
## mean difference 
##         0.00025

2.32(a) Is there a significant difference between the means of the population of measurements from which the two samples were selected? Use α=0.05.

Answer : we got the p-value is 0.6742 so we reject the null hypothesis and c_1 and c_2 are from the same Inspector we are going for paried t-test and there is no difference between them the mean difference is 0.00025

2.32(b)Find the P-value for the test in part (a)

Answer : The p-value is 0.6742 > 0.05 so we fail to reject null hypothesis Ho

2.32(c)Construct a 95 percent confidence interval on the difference in mean diameter measurements for the two types of calipers.

Answer : The 95 percent confidence interval is -0.001024344 and 0.001524344

Question 2.34 An article in the Journal of Strain Analysis (vol. 18, no. 2, 1983) compares several procedures for predicting the shear strength for steel plate girders. Data for nine girders in the form of the ratio of predicted to observed load for two of these procedures, the Karlsruhe and Lehigh methods, are as follows:

Karlsruhe<-c(1.186,1.151,1.322,1.339,1.200,1.402,1.365,1.537,1.559)
Lehigh<-c(1.061,0.992,1.063,1.062,1.065,1.178,1.037,1.086,1.052)

data <-cbind.data.frame("Karlsruhe,Lehigh")

d<-c(Karlsruhe-Lehigh)
t.test(Karlsruhe,Lehigh,paired =TRUE,conf.level = 0.95)
## 
##  Paired t-test
## 
## data:  Karlsruhe and Lehigh
## t = 6.0819, df = 8, p-value = 0.0002953
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  0.1700423 0.3777355
## sample estimates:
## mean difference 
##       0.2738889

2.34(a)Is there any evidence to support a claim that there is a difference in mean performance between the two methods? Use α=0.05.

Answer : There is a difference in the mean performance of the two methods 0.2738889

2.34(b)What is the P-value for the test in part (a)?

Answer : we got the p-value 0.0002953 < level of significance 0.05

2.34(c)Construct a 95 percent confidence interval for the difference in mean predicted to observed load.

Answer : 95 percent confidence interval: 0.1700423 and 0.3777355

2.34(d)Investigate the normality assumption for both samples. Normal probability for karlsruhe

qqnorm(Karlsruhe,main = "Normal Probability of karlsruhe")
qqline(Karlsruhe)

Normal probability of lehigh

qqnorm(Lehigh,main = "Normal probability of Lehigh")
qqline(Lehigh)

2.34(e)Investigate the normality assumption for the difference in ratios for the two methods

qqnorm(d,main = "Normal Probability ratios of Karlsruhe and Lehigh")
qqline(d)

2.34(f)Discuss the role of the normality assumption in the paired t-test Answer : The normal probability of paired t-test we can see that the most of the points fall on the straight line but some of them are out of the line

Question 2.29 Photoresist is a light-sensitive material applied tosemiconductor wafers so that the circuit pattern can be imaged on to the wafer. After application, the coated wafersare baked to remove the solvent in the photoresist mixture and to harden the resist. Here are measurements of photoresist thickness (in kA) for eight wafers baked at two different temperatures. Assume that all of the runs were made in random order.

degree95<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
degree100<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)

Question 2.29(e) Check the assumption of normality of the photoresist thickness. QQ plot for 95 and 100 Degrees

qqnorm(degree95,main = "Normal Probability of 95_degree")
qqline(degree95)

qqnorm(degree100,main = "Normal probability of 100_degree")
qqline(degree100)

Question 2.29(f)Find the power of this test for detecting an actual difference in means of 2.5 kA. Answer: to find the value of d we need to calculate the sp =1.884

Sp<-sqrt(((7*var(degree95))+(7*var(degree100)))/14)

delta = 2.5/sp =1.3289

power.t.test(n = 8, d = 1.3269, sig.level = 0.05, power = NULL, type = "two.sample")
## 
##      Two-sample t test power calculation 
## 
##               n = 8
##           delta = 1.3269
##              sd = 1
##       sig.level = 0.05
##           power = 0.694557
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

power of this test is 0.694557

Question 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:

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

c2f6Flow1<-c(2.7,4.6,2.6,3.0,3.2,3.8)
c2f6Flow2<-c(4.6,3.4,2.9,3.5,4.1,5.1)
?wilcox.test
## starting httpd help server ... done
wilcox.test(c2f6Flow1,c2f6Flow2)
## Warning in wilcox.test.default(c2f6Flow1, c2f6Flow2): cannot compute exact p-
## value with ties
## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  c2f6Flow1 and c2f6Flow2
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0

The p value is 0.1994 which is greater than the level of signifance we cannot reject null hypothesis Ho and There is no difference in the mean