Homework 3
Author Kininge, Rucha
Last compiled on September 17, 2023 at 4:14 PM - PDT
1 Homework
1.1 - Question 2.32
The diameter of a ball bearing was measured by 12 inspectors, each using two different kinds of calipers. The results were:
| Inspector | Caliper 1 | Caliper 2 |
|---|---|---|
| 1 | 0.265 | 0.264 |
| 2 | 0.265 | 0.265 |
| 3 | 0.266 | 0.264 |
| 4 | 0.267 | 0.266 |
| 5 | 0.267 | 0.267 |
| 6 | 0.265 | 0.268 |
| 7 | 0.267 | 0.264 |
| 8 | 0.267 | 0.265 |
| 9 | 0.265 | 0.265 |
| 10 | 0.268 | 0.267 |
| 11 | 0.268 | 0.268 |
| 12 | 0.265 | 0.269 |
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)
b. Find the P-value for the test in part (a).
c. Construct a 95 percent confidence interval on the difference in mean diameter measurements for the two types of calipers.
1.2 Solution 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)
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)
Inspector <- c("1 : 12")# paired t-test
t.test(caliper1, caliper2, paired = TRUE)##
## Paired t-test
##
## data: caliper1 and caliper2
## 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#NPP'S
qqnorm(caliper1, main = "NPP Caliper 1" )
qqnorm(caliper2, main = "NPP Caliper 2")
Difference <- caliper1-caliper2
qqnorm(Difference,main = "NPP of the Difference", cex =3, bg = "blue", lwd = 2, pch = 21)
cor(caliper1, caliper2)## [1] 0.1276307t.test(caliper1,caliper2,paired=TRUE)##
## Paired t-test
##
## data: caliper1 and caliper2
## 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
## 0.00025Part B
P-value = 0.6742
C Construct 95% confidence level
-0.001024344 , 0.001524344
1.3 - 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:
| Girder | Karlsruhe Method | Lehigh Method |
|---|---|---|
| S1/1 | 1.186 | 1.061 |
| S2/1 | 1.151 | 0.992 |
| S3/1 | 1.322 | 1.063 |
| S4/1 | 1.339 | 1.062 |
| S5/1 | 1.200 | 1.065 |
| S2/1 | 1.402 | 1.178 |
| S2/2 | 1.365 | 1.037 |
| S2/3 | 1.537 | 1.086 |
| S2/4 | 1.559 | 1.052 |
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)
b. What is the P-value for the test in part (a)?
c. Construct a 95 percent confidence interval for the difference in mean predicted to observed load.
d. Investigate the normality assumption for both samples.
e. Investigate the normality assumption for the difference in ratios for the two methods.
f. Discuss the role of the normality assumption in the paired t-test.
# The data is assigned as dataset2
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)
Difference <- (karlsruhe-lehigh)a. Paired t-test
t.test(karlsruhe,lehigh, paired = TRUE)##
## 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#d. normality assumption for both samples
qqnorm(karlsruhe, main = "karlsruhe Method", bg = "black")
qqnorm(lehigh, main = "lehigh Method", bg = "black")
e. Normality assumption for the difference in ratio for the two methods
qqnorm(Difference, main="NPP of Ratio Differences", cex=2, bg="red", lwd=2, pch=21)
f. Role of the normality assumption in the paired t-test
Normality assumption in paired t-test refers to the difference between paired observations. Confidence intervals provide a range of plausible values for the population parameter.
1.4 - Question 2.29
Photoresist is a light-sensitive material applied to semiconductor wafers so that the circuit pattern can be imaged on to the wafer. After application, the coated wafers are 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.
| 95 C | 100 C |
|---|---|
| 11.176 | 5.263 |
| 7.089 | 6.748 |
| 8.097 | 7.461 |
| 11.739 | 7.015 |
| 11.291 | 8.133 |
| 10.759 | 7.418 |
| 6.476 | 3.772 |
| 8.315 | 8.963 |
e. Check the assumption of normality of the photoresist thickness.
f. Find the power of this test for detecting an actual difference in means of 2.5 kA.
##Solution
e. Check the assumption of normality of the photoresist thickness.
ThickAt95 <- c(11.176, 7.089, 8.097, 11.739, 11.291, 10.759, 6.467, 8.315)
ThichAt100 <- c(5.263, 6.748, 7.461, 7.015, 8.133, 7.418, 3.772, 8.963)#Normal Probability Plot for 95 degree C
library(stats)
qqnorm(ThickAt95,main="NPP for Thick at 95 C", ylab = "Photoresist Thickness", cex=2, lwd=2, pch=21)
#Normal Probability Plot for 100 degree C
qqnorm(ThickAt95,main="NPP for Thick at 100 C", ylab = "Photoresist Thickness", cex=2, lwd=2, pch=21)
f. Power for detectiong an actual difference in means of 2.5 kA
Given data
n = 8
sig level = 0.05
d = 2.5/sd
Type = 2 sample t-test
2 sample t-test
sd95<- c(sd(ThickAt95))
print(sd95)## [1] 2.099564library(pwr)
Power1<- c(pwr.t.test(n=8,d=1.327,sig.level=0.05,power=NULL,type="two.sample"))
print(Power1)## $n
## [1] 8
##
## $d
## [1] 1.327
##
## $sig.level
## [1] 0.05
##
## $power
## [1] 0.6946262
##
## $alternative
## [1] "two.sided"
##
## $note
## [1] "n is number in *each* group"
##
## $method
## [1] "Two-sample t test power calculation"1.5 - 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:
| C2F6 Flow (SCCM) |
1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 125 | 2.7 | 4.6 | 2.6 | 3.0 | 3.2 | 3.8 |
| 200 | 4.6 | 3.4 | 2.9 | 3.5 | 4.1 | 5.1 |
Does the C2F6 flow rate affect average etch uniformity? Use α�= 0.05. And use Non - Parametric Method:
Solution
#The data assign as data3 flow125 <- c(2.7,4.6,2.6,3.0,3.2,3.8) flow200 <- c(4.6,3.4,2.9,3.5,4.1,5.1)
wilcox.test(flow125,flow200)##
## Wilcoxon rank sum test with continuity correction
##
## data: flow125 and flow200
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 02 Complete R Code
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)
Inspector <- c("1 : 12")
# paired t-test
t.test(caliper1, caliper2, paired = TRUE)
#NPP'S
qqnorm(caliper1, main = "NPP Caliper 1" )
qqnorm(caliper2, main = "NPP Caliper 2")
qqnorm(Difference,main = "NPP of the Difference", cex =3, bg = "blue", lwd = 2, pch = 21)
cor(caliper1, caliper2)
t.test(caliper1, caliper2, paired = TRUE)
karlsrushe <-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)
Difference <- (karlsruhe-lehigh)
t.test(karlsruhe,lehigh, paired = TRUE)
qqnorm(karlsruhe, main = "karlsruhe Method", bg = "black")
qqnorm(lehigh, main = "lehigh Method", bg = "black")
qqnorm(Difference, main="NPP Ratio Differences", cex=2, bg="red", lwd=2, pch=21)
qqline(Difference, lwd=2)
ThickAt95 <- c(11.176, 7.089, 8.097, 11.739, 11.291, 10.759, 6.467, 8.315)
ThichAt100 <- c(5.263, 6.748, 7.461, 7.015, 8.133, 7.418, 3.772, 8.963)
#Normal Probability Plot for 95 degree C
qqnorm(thickAt95,main="NPP Thick at 95 C", ylab = "Photoresist Thickness", cex=2, lwd=2, pch=21)
#Normal Probability Plot for 100 degree C
qqnorm(ThickAt95,main="NPP Thick 100 C", ylab = "Photoresist Thickness", cex=2, lwd=2, pch=21)
sd95<- c(sd(ThickAt95))
print(sd95)
Power1<- c(pwr.t.test(n=8,d=1.327,sig.level=0.05,power=NULL,type="two.sample"))
print(Power1)
flow125 <- c(2.7,4.6,2.6,3.0,3.2,3.8)
flow200 <- c(4.6,3.4,2.9,3.5,4.1,5.1)
wilcox.test(flow125,flow200)```