This report captures work done for the individual homework for Week 3. R code along with the results are provided. The required homework problems were taken from “Design and Analysis of Experiments 8th Edition”:
   1) 2.32
   2) 2.34
   3) 2.29e, 2.29f
   4) 2.27a (using non-Parametric) Answers to the questions are in purple.

Problem 1 (2.32)

QUESTION

The diameter of a ball bearing was measured by 12 inspectors, each using two different kinds of caliper. [See table of values in book.]
  (a) Is there a significant difference between the means of the population of measurements from which the two samplers were selected?
    Use alpha = 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.

PROBLEM SETUP

Statistical Test and Rationale:  Paired t-Test
Hypothesis Test:        Null:     H0:μ1 = μ2
               Alternate: H1:μ1 ≠ μ2
               Where μ1 & μ2 = mean diameters of ball bearings measured using Calipers 1 and 2 (respectively).
One or Two Sided:      Two-sided
Alpha:            0.05



STATISTICAL TEST

t.test(Caliper1,Caliper2, alternative = "two.sided", paired = TRUE, var.equal = TRUE, conf.level = 0.95)
## 
##  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

CONCLUSION

(a & b) The p-value above is 0.6742, and with an alpha of 0.05, we fail to reject the Null hypothesis. We failed to detect a significant difference.
(c)   The statistical output above shows the 95% confidence interval for the difference in mean diameters: -0.001024344 to 0.001524344.

Problem 2 (2.34)

QUESTION

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 shown in a table of values in the book.
  (a) Is there any evidence to support a claim that there is a difference in mean performance between the two methods? Use alpha = 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.

PROBLEM SETUP

Statistical Test and Rationale:  Paired t-test
Hypothesis Test:        Null:     H0:nK - nL = 0
               Alternate: H1:nK - nL ≠ 0
               Where K=Karlsuruhe and L=Lehigh methods.
One or Two Sided:      Two-sided
Alpha:            0.05

STATISTICAL TEST

t.test(Karlsruhe,Lehigh, alternative = "two.sided", paired = TRUE, var.equal = TRUE, conf.level = 0.95)
## 
##  Paired t-test
## 
## data:  Karlsruhe and Lehigh
## t = 6.085, df = 8, p-value = 0.0002943
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  0.1701636 0.3778364
## sample estimates:
## mean of the differences 
##                   0.274

CONCLUSION

(a & b) The p-value from the paired t-test is 0.0002943 while the p-value for the Mann-Whitney is 0.003906, and in either case with an alpha of 0.05, we reject the Null hypothesis. There is evidence to support the claim of a significant difference.
(c)   The statistical output above shows the 95% confidence interval for the difference in mean diameters: 0.1701636 to 0.3778364.
(d)   In order to investigate the normality assumption for both samples, the normality probability plots for both the Karlsruhe and Lehigh methods are shown below. The plot for Karlsruhe indicates that it is normal while the Lehigh method does not look normal. The correct test to use in this situation is a Mann-Whitney U test that is run below.

qqnorm(Karlsruhe,
       main = "Normal Probability Plot for Karlsruhe Method",
       pch = 21,
       cex = 2,
       col = "black",
       bg = "black",
       lwd = 2)
qqline(Karlsruhe, lwd = 2)

qqnorm(Lehigh,
       main = "Normal Probability Plot for Lehigh Method",
       pch = 21,
       cex = 2,
       col = "black",
       bg = "black",
       lwd = 2)
qqline(Lehigh, lwd = 2)

Mann-Whitney U test should be ran because of the non-normal nature of the Lehigh sample. It is provided below:

wilcox.test(Karlsruhe,Lehigh, alternative = "two.sided", paired = TRUE, var.equal = TRUE, conf.level = 0.95)
## 
##  Wilcoxon signed rank exact test
## 
## data:  Karlsruhe and Lehigh
## V = 45, p-value = 0.003906
## alternative hypothesis: true location shift is not equal to 0


(e) The normal probability plot for the difference in methods can be seen below, with the difference not normally distributed. 

KarlsruheLehigh <- c(.125,.159,.259,.277,.135,.224,.328,.451,.507)
qqnorm(KarlsruheLehigh,
       main = "Normal Probability Plot for the Difference between Methods",
       pch = 21,
       cex = 2,
       col = "black",
       bg = "black",
       lwd = 2)
qqline(Lehigh, lwd = 2)

(f)   The normality assumption in the paired t-test is one of normality in the difference of the paired values.

Problem 3 (2.29e and 2.29f)

QUESTION

Photoresist is a light-sensitive material applied to semiconductors… [See table of values in book.]
  (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.

CHECK FOR NORMALITY

We can check for normality by looking at the normal probability plots for both the 95°C and 100°C plots, as shown below.
(e) Both plots show that the data is non-normal for each baking temperature.

qqnorm(BakingAt95,
       main = "Normal Probability Plot for Photoresist Thickness for Baking at 95°C", pch = 21, cex = 2, col = "black", bg = "Red", lwd = 2)
qqline(BakingAt95, lwd = 2)

qqnorm(BakingAt100,
       main = "Normal Probability Plot for Photoresist Thickness for Baking at 100°C", pch = 21, cex = 2, col = "black", bg = "Blue", lwd = 2)
qqline(BakingAt100, lwd = 2)

PROBLEM SETUP


Alpha:            0.05
Beta:            TBD
Difference:          1.88 / 2.5 = .79616
Standard Deviation:      1.88 (pooled std. dev. from the entries in the data table)

STATISTICAL TEST

pwr.t.test(n=8, d= 0.79616, sig.level = 0.05,power = NULL, type="two.sample", alternative="two.sided")
## 
##      Two-sample t test power calculation 
## 
##               n = 8
##               d = 0.79616
##       sig.level = 0.05
##           power = 0.3172597
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

CONCLUSION

(f) The power to detect a difference of 2.5 kA with a pooled standard deviation of 1.88 and an effect size of .79616 is low, at 0.317.

Problem 4 (2.27a)

QUESTION

An article in Solid State Technology, “Orthogonal Plasma Etching” … [See table of values in book.]
  (a) Does the C2F6 flow rate affect average etch uniformity? Use alpha = .05.

PROBLEM SETUP

Statistical Test and Rationale:  Mann-Whitney
Hypothesis Test:        Null:     H0:n125 - n200 = 0
               Alternate: H1:n125 - n250 ≠ 0
One or Two Sided:      Two-sided
Alpha:            0.05

STATISTICAL TEST

wilcox.test(Flow125,Flow200, alternative="two.sided", paired = FALSE, conf.level = .95)
## Warning in wilcox.test.default(Flow125, Flow200, alternative = "two.sided", :
## cannot compute exact p-value with ties
## 
##  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 0

CONCLUSION

(a) With a p-value of 0.1994 which is greater than the significance level of 0.05, we fail to reject the null hypothesis and with the information given, we can only conclude that the flow rate doesn’t impact the etch uniformity.