Problem 2.32

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

the null Hypothesis is that the measurements for caliper 1 are the same as caliper 2.

The alternative Hypothesis is the measurements are differnet for caliper 1 and caliper 2.

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

    I concluded that there was not any significant differcence in the means not to be significant. I found the p value to be 0.6742. There was not enough evidence to reject the null hypothesis that both calipers were differnt.

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

    p value was 0.6742

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

    The 95 % confidence interval would be ## 95 percent confidence interval is (-0.001024344, 0.001524344)

Caliper_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)
Caliper_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)
t.test(Caliper_1,Caliper_2,paired = TRUE)
## 
##  Paired t-test
## 
## data:  Caliper_1 and Caliper_2
## 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
cor(Caliper_1,Caliper_2)
## [1] 0.1276307

Problem 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:

The null Hypothesis is that the ratios for Karsruhe and Methods that the preformance between the two methods are the same.

The alternate Hypothesis is that the ratios for Karsruhe and Methods that the preformance between the two methods are different.

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

    I concluded that there is enough evidence that there is a difference in the preformance.

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

    p-value = 0.0002953

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

    95 percent confidence interval: 0.1700423 0.3777355

## 
##  Paired t-test
## 
## data:  Karlsruhe_M and Lehigh_M
## 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

  1. Investigate the normality assumption for both samples.

    Both methods look to follow into a straight line and I would say the normaillity assumption holds

    The lehigh data looks like does suffer form long tailedness , but it appears to follow in a straight line

qqnorm(Karlsruhe_M, main = "Karlsruhe QQ plot")

qqnorm(Lehigh_M, main = "Lehigh QQ plot")

  1. Investigate the normality assumption for the difference in ratios for the two methods.

    The differences seem to follow in a straight line and I would say that normaillity assumption holds

qqnorm(Karlsruhe_M-Lehigh_M, main = "Difference of means QQ plot")

  1. Discuss the role of the normality assumption in the paired t-test.

    The normality assumptions is improtant for preforming a paired t- test but the same varation is a really improntant. If the samples diverge from the a normal distribution, The t-test still predicts the results with high accuracy.

Problem 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.

  1. Check the assumption of normality of the photoresist thickness.

    The normaility assumptions appear to follow in a straight line and I would say the normality assumption holds

Degrees_95_C <- c(11.176,7.089,8.097,11.769,11.291,10.759,6.467,8.315)
Degrees_100_C <- c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)

qqnorm(Degrees_100_C, main = "100 degrees C QQ plot")

qqnorm(Degrees_95_C, main = "95 degrees C QQ plot")

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

    power = 0.9865906. The beta would be 0.0134094

library(pwr)
pwr.t.test(n = 8 ,d = 2.25 , sig.level = .05 )
## 
##      Two-sample t test power calculation 
## 
##               n = 8
##               d = 2.25
##       sig.level = 0.05
##           power = 0.9865906
##     alternative = two.sided
## 
## NOTE: n is number in *each* group

Problem 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.

Null hypotheise: The flow rates are not different from each other
Alternatate hypothsise: The flow rates are differnt from each other

  1. Does the C2F6 flow rate affect average etch uniformity? Use alpha = 0.05.

    I found that there is no difference in the means, and the flow rates are the same. The p-value was 0.1994 which is much greater than .05.

C_F_flow_125 <- c(2.7, 4.6, 2.6, 3.0, 3.2 ,3.8)
C_F_flow_200 <- c(4.6, 3.4, 2.9, 3.5, 4.1 ,5.1)


wilcox.test(C_F_flow_125,C_F_flow_200)
## Warning in wilcox.test.default(C_F_flow_125, C_F_flow_200): cannot compute exact
## p-value with ties
## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  C_F_flow_125 and C_F_flow_200
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0