Question 2.32

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

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

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

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

Answer: Null Hypothesis: Ho: mu1 = mu2 or Ho: mu1-mu2 = 0 or D = 0

Alternative Hypothesis: Ha: mu1 ≠ mu2 or Ha: mu1-mu2 ≠ 0 or D ≠ 0

Reading Data

Pop1 <- 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)
Pop2 <- 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)
Diff <- Pop1-Pop2

Plot

qqnorm(Diff,main="Normal Probability Plot of the Difference")

Paired T-Test

t.test(Pop1,Pop2,paired=TRUE)

## 
##  Paired t-test
## 
## data:  Pop1 and Pop2
## 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

Answer Part a: Since our p-value is much greater than a=0.05, so we fail to reject Ho and thus we conclude that there’s no difference between the means of the population of measurements from which the two samples were selected.

Answer Part b: Our p-value = 0.6742

Answer Part c: Lower Bound = -0.001024 and Upper Bound = 0.001524

So confidence interval is -0.001024 <= D (mu1-mu2) <= 0.001524

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.

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

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

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

Answer: Null Hypothesis: Ho: mu3 = mu4 or Ho: mu3-mu4 = 0 or D = 0

Alternative Hypothesis: Ha: mu3 ≠ mu4 or Ha: mu3-mu4 ≠ 0 or D ≠ 0

Reading Data

Pop3 <- c(1.186, 1.151, 1.322, 1.339, 1.200, 1.402, 1.365, 1.537, 1.559)
Pop4 <- c(1.061, 0.992, 1.063, 1.062, 1.065, 1.178, 1.037, 1.086, 1.052)
D <- Pop3-Pop4

Plots

qqnorm(Pop3,main="Karlsruhe Method Normal Probability Plot")

qqnorm(Pop4,main="Lehigh Method Normal Probability Plot")

qqnorm(D,main="Difference Normal Probability Plot")

Paired T-Test

t.test(Pop3,Pop4,paired=TRUE)

## 
##  Paired t-test
## 
## data:  Pop3 and Pop4
## 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

Answer Part a: Since our p-value is very small as compared to a=0.05, so we reject Ho and thus we conclude that there is a difference in mean performance between the two methods.

Answer Part b: Our p-value = 0.0002953

Answer Part c: Lower Bound = 0.17004 and Upper Bound = 0.37773

So confidence interval is 0.17004 <= D (mu1-mu2) <= 0.37773

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.

Null Hypothesis: Ho: muC = muD or Ho: muC - muD = 0

Alternative Hypothesis: Ha: muC > muD or Ha: muC-muD > 0

Reading Data

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

Equality of Variance

boxplot(PopC,PopD, main= "Box plot of datasets PopC & PopD", ylab= "Thickness",names=c("95 C","100 C"))

Two Sample T-Test

t.test(PopC,PopD,alternative = "greater", var.equal=TRUE)

## 
##  Two Sample t-test
## 
## data:  PopC and PopD
## t = 2.6751, df = 14, p-value = 0.009059
## alternative hypothesis: true difference in means is greater than 0
## 95 percent confidence interval:
##  0.8608158       Inf
## sample estimates:
## mean of x mean of y 
##  9.366625  6.846625

Part e. Check the assumption of normality for the data from this experiment.

Answer: To check the assumption of normality for the populations C and D, we’ll draw their normal probability plots.

Normal Probability Plots

qqnorm(PopC,main="Normal Probability Plot for PopC at 95 C", ylab = "Photoresist Thickness")
qqline(PopC)

qqnorm(PopD, main="Normal Probability Plot for PopD at 100 C", ylab = "Photoresist Thickness")
qqline(PopD)

Although normal probability plots aren’t perfect, they still are very close to normality and thus can be assumed normal.

Part f. Find the power of this test for detecting an actual difference in means of 2.5 kA.

Actual difference in means = 2.5

Sp is pooled standard deviation for the two sample T-Test.

library(pwr)
Sp <- sqrt(((8-1)*sd(PopC)^2+(8-1)*sd(PopD)^2)/(8+8-2)) 
pwr.t.test(n=8,d=(2.5/Sp),sig.level=0.05,power=NULL,type="two.sample")

## 
##      Two-sample t test power calculation 
## 
##               n = 8
##               d = 1.32694
##       sig.level = 0.05
##           power = 0.6945872
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

So power of this test is 0.6946.

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.

Reading data:

PopA <- c(2.7, 4.6, 2.6, 3.0, 3.2, 3.8)
PopB <- c(4.6, 3.4, 2.9, 3.5, 4.1, 5.1)

Part A: Does the C2F6 flow rate affect average etch uniformity? Use a=0.05.

Null Hypothesis: Ho:muA=muB or muA-muB=0

Alternative Hypothesis: muA≠muB or muA-muB≠0

Normal Probability Plots

qqnorm(PopA,main="Flow 125 SCCM Normal Probability Plot")

qqnorm(PopB,main="Flow 200 SCCM Normal Probability Plot")

First distribution is slightly skewed to the right, while second distribution is approximately normal.

Boxplots

boxplot(PopA,PopB,names=c('Flow 125 SCCM','Flow 200 SCCM'),main='Flow 125 SCCM vs Flow 200 SCCM Boxplots')

Mann-Whitney-U test with alpha=0.05:

wilcox.test(PopA,PopB)

## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  PopA and PopB
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0

Answer Part a: Since our p-value is 0.1994, so at a significance level of 0.05 we fail to reject Ho. In other words, we conclude that flow rate doesn’t affect average etch uniformity.

HW 3 R Markdown

Muneeb

9/17/2021

Question 2.32

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

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

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

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

Answer: Null Hypothesis: Ho: mu1 = mu2 or Ho: mu1-mu2 = 0 or D = 0

Alternative Hypothesis: Ha: mu1 ≠ mu2 or Ha: mu1-mu2 ≠ 0 or D ≠ 0

Reading Data

Plot

Paired T-Test

Answer Part a: Since our p-value is much greater than a=0.05, so we fail to reject Ho and thus we conclude that there’s no difference between the means of the population of measurements from which the two samples were selected.

Answer Part b: Our p-value = 0.6742

Answer Part c: Lower Bound = -0.001024 and Upper Bound = 0.001524

So confidence interval is -0.001024 <= D (mu1-mu2) <= 0.001524

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.

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

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

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

Answer: Null Hypothesis: Ho: mu3 = mu4 or Ho: mu3-mu4 = 0 or D = 0

Alternative Hypothesis: Ha: mu3 ≠ mu4 or Ha: mu3-mu4 ≠ 0 or D ≠ 0

Reading Data

Plots

Paired T-Test

Answer Part a: Since our p-value is very small as compared to a=0.05, so we reject Ho and thus we conclude that there is a difference in mean performance between the two methods.

Answer Part b: Our p-value = 0.0002953

Answer Part c: Lower Bound = 0.17004 and Upper Bound = 0.37773

So confidence interval is 0.17004 <= D (mu1-mu2) <= 0.37773

Question 2.29

Null Hypothesis: Ho: muC = muD or Ho: muC - muD = 0

Alternative Hypothesis: Ha: muC > muD or Ha: muC-muD > 0

Reading Data

Equality of Variance

Two Sample T-Test

Part e. Check the assumption of normality for the data from this experiment.

Answer: To check the assumption of normality for the populations C and D, we’ll draw their normal probability plots.

Normal Probability Plots

Although normal probability plots aren’t perfect, they still are very close to normality and thus can be assumed normal.

Part f. Find the power of this test for detecting an actual difference in means of 2.5 kA.

Actual difference in means = 2.5

Sp is pooled standard deviation for the two sample T-Test.

So power of this test is 0.6946.

Question 2.27

Reading data:

Part A: Does the C2F6 flow rate affect average etch uniformity? Use a=0.05.

Null Hypothesis: Ho:muA=muB or muA-muB=0

Alternative Hypothesis: muA≠muB or muA-muB≠0

Normal Probability Plots

First distribution is slightly skewed to the right, while second distribution is approximately normal.

Boxplots

Mann-Whitney-U test with alpha=0.05:

Answer Part a: Since our p-value is 0.1994, so at a significance level of 0.05 we fail to reject Ho. In other words, we conclude that flow rate doesn’t affect average etch uniformity.