Homework 1 Week 2
Saad Mirza
Last compiled on September 11, 2022 at 11:44 PM - CDT
Homework problems are from "Design and Analysis of Experiments 8th Edition":
2.24
2.26a (using Levene's Test) and 2.26b
2.29a, 2.29b, 2.29c, 2.29e
1 Question 2.24:
Two machines are used for filling plastic bottles with a net volume of 16.0 ounces. The filling processes can be assumed to be normal, with standard deviation of σ1 = 0.015 and σ2 = 0.018. The quality engineering department suspects that both machines fill to the same net volume, whether or not this volume is 16.0 ounces. An experiment is performed by taking a random sample from the output of each machine.
| 16.03 | 16.01 | 16.02 | 16.03 |
| 16.04 | 15.96 | 15.97 | 16.04 |
| 16.05 | 15.98 | 15.96 | 16.02 |
| 16.05 | 16.02 | 16.01 | 16.01 |
| 16.02 | 15.99 | 15.99 | 16.00 |
State the hypotheses that should be tested in this experiment.
Test these hypotheses using \(\alpha=\) 0.05. What are your conclusions?
Find the P-value for this test.
Find a 95 percent confidence interval on the difference in mean fill volume for the two machines.
1.1 Solution 2.24:
a) Stating the Hypothesis:
The null and alternative hyptheses are tested with an α = 0.05:
\[H_o: \mu_{1} = \mu_{2}\]
\[
H_a: \mu_{1}\neq \mu_{2}
\]
where μ1 = Machine 1 Net Fill Volume and μ2 =
Machine 2 Net Fill Volume
b) Testing the Hypothesis:
We will perform Two Sample T-Test with pooled variance. As given in problem description the filling process can be assumed to be normal. The standard deviation for Machine 1 = 0.015 (Variance = 0.000225) and standard deviation for Machine 2 = 0.018 (Variance = 0.000324), which are quite close.
So we have satisfied both the assumptions and we can proceed to perform the Two Sample T-Test with pooled variance.
Reading the Data:
Machine1 <- c(16.03,16.04,16.05,16.05,16.02,16.01,15.96,15.98,16.02,15.99)
Machine2 <- c(16.02,15.97,15.96,16.01,15.99,16.03,16.04,16.02,16.01,16.00)Two Sample T-Test:
t.test(Machine1,Machine2, var.equal = TRUE, alternative = "two.sided")##
## Two Sample t-test
##
## data: Machine1 and Machine2
## t = 0.79894, df = 18, p-value = 0.4347
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -0.01629652 0.03629652
## sample estimates:
## mean of x mean of y
## 16.015 16.005Conclusion:
Sample averages for Machine 1 and Machine 2 are 16.015 and 16.005 respectively.
T statistic = 0.79894
With a p-value = .4347, i.e. much greater than \(\alpha=0.05\) t/f we fail to reject the Null hypothesis and conclude that there is no significant difference in the two fill volumes of the machines.
c) P-Value for the Test:
P-value for the test = 0.4347
d) 95% Confidence Interval for difference in the mean fill volumes of two machines:
Lower bound for the confidence interval is -0.016296 and upper bound is 0.036296
i.e.
\[ -0.016296\leq \mu_{1} - \mu_{2} \leq 0.036296 \]
2 Question 2.26:
The following are the burning times (in minutes) of chemical flares of two different formulations. The design engineers are interested in both the mean and variance of the burning times.
| 65 | 82 | 64 | 56 |
| 81 | 67 | 71 | 69 |
| 57 | 59 | 83 | 74 |
| 66 | 75 | 59 | 82 |
| 82 | 70 | 65 | 79 |
Test the hypothesis that the two variances are equal. Use \(\alpha = 0.05\).
Using the results of (a), test the hypothesis that the mean burning times are equal. Use \(\alpha = 0.05\). What is the P-value for this test?
2.1 Solution 2.26:
a) Test for Equal Variance using Levene's Test:
The null and alternative hypotheses are tested with an α = 0.05:
\[H_O: \sigma_{1}^{2} = \sigma_{2}^{2} \]
\[H_a: \sigma_{1}^{2} \neq
\sigma_{2}^{2}\]
Where σ12 is the variance for Flare Type 1 Burning
Times and σ22 is the variance for Flare Type 2
Burning Times.
Reading the Data:
Pop <- c(65, 81, 57, 66, 82, 82, 67, 59, 75, 70, 64, 71, 83, 59, 65, 56, 69, 74, 82, 79)
B <- c("A", "A", "A", "A", "A", "A", "A", "A", "A", "A", "B", "B", "B", "B", "B", "B", "B", "B", "B", "B")
dat <- cbind(Pop, B)
dat1 <- data.frame(dat)
str(dat1)## 'data.frame': 20 obs. of 2 variables:
## $ Pop: chr "65" "81" "57" "66" ...
## $ B : chr "A" "A" "A" "A" ...dat1$B <- as.factor(dat1$B)
dat1$Pop <- as.numeric(dat1$Pop)
str(dat1)## 'data.frame': 20 obs. of 2 variables:
## $ Pop: num 65 81 57 66 82 82 67 59 75 70 ...
## $ B : Factor w/ 2 levels "A","B": 1 1 1 1 1 1 1 1 1 1 ...Now Using Levene’s Test to test the Hypothesis:
library(lawstat)
levene.test(dat1$Pop,dat1$B, location = "mean")##
## Classical Levene's test based on the absolute deviations from the mean
## ( none not applied because the location is not set to median )
##
## data: dat1$Pop
## Test Statistic = 0.0014598, p-value = 0.9699Conclusion:
So as per Levene's test, our test statistic: 0.0014598
Our P-value: 0.9699, which is greater then \(\alpha=0.05\)
Therefore, we fail to reject the Null hypothesis and thus variances of the burning times of the two flare types are equal.
b) Test for Equal Mean Burning Times:
Based on equal variance and normality of the data we will perform a two sample T-test to test the hypothesis that whether the mean burning times are equal and we will also obtain P-Value:
The null and alternative hypotheses are tested with an α = 0.05:
\[H_o: \mu_{1} = \mu_{2}\]
\[
H_a: \mu_{1}\neq \mu_{2}
\]
where μ1 = Flare Type 1 Burning Time and μ2 =
Flare Type 1 Burning Time. All times are in minutes.
Two Sample T-Test:
FlareType1 <- c(65, 81, 57, 66, 82, 82, 67, 59, 75, 70)
FlareType2 <- c(64, 71, 83, 59, 65, 56, 69, 74, 82, 79)
t.test(FlareType1,FlareType2,var.equal=TRUE, alternative = "two.sided")##
## Two Sample t-test
##
## data: FlareType1 and FlareType2
## t = 0.048008, df = 18, p-value = 0.9622
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## -8.552441 8.952441
## sample estimates:
## mean of x mean of y
## 70.4 70.2Conclusion:
So, our t statistic: 0.048
P-value: 0.9622
Since P-value is much greater than 0.05, thus we fail to reject Ho. In other words, mean burning times are equal.
3 Question 2.29:
Photo resist 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 photo resist 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 |
| 6.467 | 3.772 |
| 8.315 | 8.963 |
- Is there evidence to support the claim that the higher baking temperature results in wafers with a lower mean photo resist thickness? Use \(\alpha=0.05\)
What is the P-value for the test conducted in part (a)?
Find a 95 percent confidence interval on the difference in means. Provide a practical interpretation of this interval.
- Check the assumption of normality of the photo resist thickness.
3.1 Solution 2.29:
a) Do higher baking temps result in wafers with lower thicknesses?
We have to formulate two hypotheses to test whether the mean photo resist thickness is equal for the two populations or does higher baking temperature results in wafers with a lower mean photo resist thickness. We will do a Two sample T-test with pooled variance to test these hypotheses.
The null and alternative hypotheses are tested with an α = 0.05:
\[H_o: \mu_{1} = \mu_{2}\]
\[H_a: \mu_{1} > \mu_{2}\]
where μ1 = Photo resist thickness for 95°C baking temperature and μ2 = Photo resist thickness for 100°C baking temperature. All thicknesses are in kA.
But first to conduct a 2-Sample T-test., we must satisfy the requirements for normality (done in part e) and equal variances.
BakingAt95 <- c(11.176, 7.089, 8.097, 11.739, 11.291, 10.759, 6.467, 8.315)
BakingAt100 <- c(5.263, 6.748, 7.461, 7.015, 8.133, 7.418, 3.772, 8.963)
boxplot(BakingAt95,BakingAt100, main= "Box plot of datasets BakingAt95 & BakingAt100", ylab= "Thickness",names=c("95 C","100 C"))
Since the variance for 95°C looks to be roughly twice the size as the variance for 100°C
We will test the hypotheses for the equal variances (Levene's
Test).
The null and alternative hypotheses are tested with an α = 0.05:
\[H_O: \sigma_{1}^{2} = \sigma_{2}^{2} \]
\[H_a: \sigma_{1}^{2} \neq
\sigma_{2}^{2}\]
Where σ12 is the variance for Thicknesses at 95°C
and σ22 is the variance for Thicknesses at
100°C.
Thickness <- c(11.176,7.089,8.097,11.739,11.291,10.759,6.467,8.315,5.263,6.748,7.461,7.015,8.133,7.418,3.772,8.963)
BakingTemp <- c(95,95,95,95,95,95,95,95,100,100,100,100,100,100,100,100)Levene’s Test:
levene.test(Thickness,BakingTemp, location="mean")##
## Classical Levene's test based on the absolute deviations from the mean
## ( none not applied because the location is not set to median )
##
## data: Thickness
## Test Statistic = 2.5625, p-value = 0.1317Since the p-value is greater than .05, see above, we fail to reject the Null hypothesis. Variances are equal.
Now Performing Test for Equal Mean Thicknesses:
t.test(BakingAt95,BakingAt100,alternative = "greater", var.equal=TRUE)##
## Two Sample t-test
##
## data: BakingAt95 and BakingAt100
## 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 6.846625Conclusion:
T statistic: 2.6751
P-value: 0.009059
Since P-value is much smaller than 0.05, thus we have to reject Ho. In other words, mean photo resist thickness is lower at higher baking temperatures.
b) P-value for the test conducted in part (a):
Answer: P-value is 0.009059
c) 95 percent confidence interval on the difference in means:
Answer: Lower bound for 95% Confidence Interval is 0.8608158
i.e.
\[0.8608158 \leq \mu_{1} - \mu_{2}\]Conclusion:
This means that one can be 95% certain that the wafers baked at the 95°C temperature will have at least 0.8608158 kA greater thickness than those baked at the 100°C temperature.
e) Assumption of normality of the photo resist thickness:
To check the assumption of normality for the Photo resist thickness for 95°C and 100°C baking temperature, we'll draw their normal probability plots:
Normal Probability Plots:
qqnorm(BakingAt95,main="Normal Probability Plot for PopC at 95 C", ylab = "Photoresist Thickness")
qqline(BakingAt95)
qqnorm(BakingAt100, main="Normal Probability Plot for PopD at 100 C", ylab = "Photoresist Thickness")
qqline(BakingAt100)
Comment:
Although normal probability plots aren't perfect, they still are very close to normality and thus can be assumed normal.