Homework 2 Solutions
Sahil Gaikwad
9/19/2021
Question No.2.32
a
Testing Hypothesis:-
H0 : µ1 = µ2 Null Hypothesis is that two means are equal and there is no difference between two means
H1: µ1 ≠ µ2 Alternative Hypothesis is that two means are not equal and there is difference between two means
μ1 is the mean of measurements taken by caliper1
μ2 is the mean of measurements taken by caliper2
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)
cor(caliper1,caliper2)## [1] 0.1276307summary(caliper1)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2650 0.2650 0.2665 0.2662 0.2670 0.2680summary(caliper2)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.2640 0.2647 0.2655 0.2660 0.2672 0.2690ANSWER a)- The mean of measurements of diameter of a ball bearing measured by Caliper1 and Caliper2 are 0.2662 and 0.2660 respectively. There is no significance difference between the mean of two population
b
Now as the correlation between the ball bearing diameter measurements taken by caliper1 and caliper2 is too small to conduct paired t-test. Hence to test the hypothesis we will perform two sample t-test
t.test(caliper1,caliper2)##
## Welch Two Sample t-test
##
## data: caliper1 and caliper2
## t = 0.40519, df = 19.559, p-value = 0.6897
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.001038888 0.001538888
## sample estimates:
## mean of x mean of y
## 0.26625 0.26600ANSWER b) The p-value after performing the paired t-test is 0.6897 The value of p is greater than 0.05. Hence we failed to reject the null hypothesis
ANSWER c) The 95% confidence interval on the difference in mean diameter measurements for the two types of calipers is between -0.001038888 to 0.001538888.
Question No.2.34
a
Testing Hypothesis:-
H0 : µ1 = µ2 Null Hypothesis is that two means are equal and there is no difference between two means
H1: µ1 ≠ µ2 Alternative Hypothesis is that two means are not equal and there is difference between two means
μ1 is the mean of measurements taken by Karlsruhe Method
μ2 is the mean of measurements taken by Lehigh Method
Karlsruhe_Method<-c(1.186,1.151,1.322,1.339,1.200,1.402,1.365,1.537,1.559)
Lehigh_Method<-c(1.061,0.992,1.063,1.062,1.065,1.178,1.037,1.086,1.052)
cor(Karlsruhe_Method,Lehigh_Method)## [1] 0.3821669summary(Karlsruhe_Method)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 1.151 1.200 1.339 1.340 1.402 1.559summary(Lehigh_Method)## Min. 1st Qu. Median Mean 3rd Qu. Max.
## 0.992 1.052 1.062 1.066 1.065 1.178ANSWER a) The mean of shear strength for steel plate girders by Karlsruhe Method and Lehigh Method are 1.340 and 1.066 respectively. The mean difference of the two methods is 0.27388. Yes, there is evidence to support the claim there is a difference in mean performance between the two methods
b
The correlation shows that data from the both the methods is positively correlated, hence to test the hypothesis we will perform the paired t-test
t.test(Karlsruhe_Method,Lehigh_Method, paired=TRUE)##
## Paired t-test
##
## data: Karlsruhe_Method and Lehigh_Method
## 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
## 0.2738889ANSWER b) The p-value after performing the paired t-test is 0.0002953 The value of p is lesser than 0.05. Hence we reject the null hypothesis
ANSWER c) The 95% confidence interval for the difference in mean predicted to observed load measurements for the two types of methods is between 0.1700423 to 0.3777355
d
To investigate the Normality assumption for both samples we need to plot the Normal Probability Plots for both the methods
qqnorm(Karlsruhe_Method, main ="Normal Probability Plot for Karlsruhe Method", ylab="ratio")
qqline(Karlsruhe_Method)
qqnorm(Lehigh_Method, main ="Normal Probability Plot for Lehigh Method", ylab= "ratio" )
qqline(Lehigh_Method)
ANSWER d) From the Normal probability Plots of the Karlsruhe Method and Lehigh Method, we can state that the data of both the methods is normally distributed
e
To investigate the normality assumption for the difference in ratios for the two methods
difference<- Karlsruhe_Method-Lehigh_Method
qqnorm(difference, main = "Normal Probability Plot of difference bewteen both methods ")
qqline(difference)
ANSWER e) From the Normal probability Plot of difference in ratios of the Karlsruhe Method and Lehigh Method, we can state that the normality assumptions is that data is normally distributed
ANSWER f) The role of normality assumption in the paired t-test is that the paired t-test considers the difference between the paired groups. For a sample size below 30, We must assume the normality of the given data and this can be checked by visualizing the data by plotting Boxplots or Normal Probability Plots.In case of sample size above 30 we assume normality, hence it meets the central limit theorem
Question No.2.29
e
Testing Hypothesis:-
H0 : µ1 = µ2 Null Hypothesis is that two means are equal and there is no difference between two means
H1: µ1 ≠ µ2 Alternative Hypothesis is that two means are not equal and there is difference between two means
p95<-c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315)
q100<-c(5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
sd(p95)## [1] 2.099564sd(q100)## [1] 1.640427qqnorm(p95,main ="Normal Probability Plot for p95°C", ylab= "kA", xlab="temp")
qqline(p95)
qqnorm(q100,main ="Normal Probability Plot for q100°C", ylab= "kA", xlab="temp")
qqline(q100)
ANSWER e) From the Normal probability Plot of photoresist thickness that is Normal Probability Plot for p95°C and the Normal Probability Plot for q100°C ,we can state that the normality assumptions is that data is normally distributed with few exceptions
Also smaller sample size can result in non normal distribution as the sample size has a direct effect on the distribution
f
To find the power of this test for detecting actual difference in means of 2.5kA
d<-sd(p95)
e<-sd(q100)
f<-sqrt((d^2+e^2)/2)
f## [1] 1.884034power.t.test(n=8,delta=2.5,sd=1.884034,sig.level = 0.05, power = NULL,alternative = c("one.sided"))##
## Two-sample t test power calculation
##
## n = 8
## delta = 2.5
## sd = 1.884034
## sig.level = 0.05
## power = 0.809887
## alternative = one.sided
##
## NOTE: n is number in *each* groupANSWER f) The value of power of this test is 0.809887
Question No.2.27
Testing Hypothesis:-
H0 : µ1 = µ2 Null Hypothesis is that two means are equal and there is no difference between two means
H1: µ1 ≠ µ2 Alternative Hypothesis is that two means are not equal and there is difference between two means
μ1 is the mean of data of C2F6 p125 flow
μ2 is the mean of data of C2F6 q200 flow
Here as sample size is not large so We will check the normality by plotting the Normal Probability plots
p125<-c(2.7,4.6,2.6,3.0,3.2,3.8)
q200<-c(4.6,3.4,2.9,3.5,4.1,5.1)
qqnorm(p125, main = "Normal Probability Plot for p125")
qqline(p125)
qqnorm(q200, main = "Normal Probability plot for q200")
qqline(q200)
From the Normal probability Plot of photoresist thickness that is Normal Probability Plot for p95°C and the Normal Probability Plot for q100°C ,we can state that the normality assumptions is that data is normally distributed with few exceptions
To test the hypothesis using Non-parametric method, we will test the hypothesis using the mann whitney U test
wilcox.test(p125,q200)## Warning in wilcox.test.default(p125, q200): cannot compute exact p-value with
## ties##
## Wilcoxon rank sum test with continuity correction
##
## data: p125 and q200
## W = 9.5, p-value = 0.1994
## alternative hypothesis: true location shift is not equal to 0ANSWER a) No, the flow rate C2F6 does not affect the average etch uniformity. As the value of p is 0.1994,which is greater than 0.05. Hence we fail to reject the null hypothesis.