Design And Analysis Of Experiments 3

Design and Analysis of Experiments

Take Home Work Exam 1 Student: Nguyen Thi Hang ID: R9605005 Comment:

Mark:

Problem.1. Conductivity measurements (µmho/cm) were taken at four different locations in the aerated lagoon of a pulp and paper mill. The lagoon is supposed to be mixed by aerators so the contents are homogeneous. The conductivity results were shown in the following table: (1) Is the lagoon homogeneously mixed? Conduct an ANOVA for the data and draw your conclusions. (2) Please check if the assumption of equal variances is satisfied. (3) Test all pairs of means using both the Turkey’s test and Fisher LSD test. Please also compare the results from these two multiple comparison methods. What are your conclusions? (4) Use the Kruskal-Wallis test for the experiment. And also compare the conclusions obtained with those from the usual ANOVA. Table1 Conductivity measurements (µmho/cm) at the different locations on the aerated lagoon -------------------------------------------------------------------------------------Location A Location B Location C Location D -----------------------------------------------------------------------------------620 630 680 560 600 670 660 620 630 710 710 600 590 640 670 610 650 680 630 660 680 640 630 590

Solution. Conducting an ANOVA for the data we get below procedure With hypothesis: H0 : µ1 = µ2 = µ3 = µ4 H1 : µi ≠ µ j Using Minitab software (chose α = 0.05) with the following steps: 1. Input and prepare the data for the test

2. Select One-Way of ANOVA method

3. Set up the suitable parameters

we can use some options such as Graphs, Compatisons

Then, we can see the results from using the Minitab

Results and Discussion (1) Is the lagoon homogeneously mixed? Conduct an ANOVA for the data

and draw your conclusions. F0 = 13.57 > F0.05, 3, 20 = 3.10 or P = 0.000 < 0.05 thus we can conclude that H0 can be rejected or the lagoon is not homogeneously mixed. (2) Please check if the assumption of equal variances is satisfied.

H0 : δ 1 = δ 2 = δ 3 = δ 4 H1: δι ≠ δj Use test for equal variances we get these results

To fix suitable parameters:

Both of Pvalue obtained (Pvalue = 0.644 and Pvalue = 0.593) are larger than 0.05 thus H0 can not rejected. It means all variances are the same. (3) Test all pairs of means using both the Turkey’s test and Fisher LSD test. Please also compare the results from these two multiple comparison methods. What are your conclusions?

From

Above results indicate pairs of means that are significantly different for: A vs B, A vs C, B vs D, C vs D And there are not significantly different for: A vs D, B vs C.

Above results indicate pairs of means that are significantly different for: A vs B, A vs C, B vs D, C vs D And there are not significantly different for: A vs D, B vs C. To compare the results from these two multipule comparison methods: These two method give the same results when we compare all pairs of means using both the Tukey’s test and Fisher ‘s test. (4) Use the Kruskal-Wallis test for the experiment. And also compare the

conclusions obtained with those from the usual ANOVA. We have hypotheses: H0 : µ1 = µ2 = µ3 = µ4 H1 : µi ≠ µ j

Above results indicate both of H = 16. 86 and H = 17.02 larger than χ0.05, 3 = 7.8 1 or P = 0.001 < 0.05 we conclude to reject H0 or the lagoon is homogeneously mixed

Problem 2. The data below are from an experiment that attempted to examine the factors affecting the reaeration rate (y) in a laboratory model stream channel. The three experimental factors are stream velocity (V, in m/sec), stream depth (D, in cm), and channel roughness (R).

Table 2 Results of the reaeration rate Run V D R 1 0.25 10 smooth 2 0.5 10 Smooth 3 0.25 15 smooth 4 0.5 15 smooth 5 0.25 10 smooth 6 0.5 10 smooth 7 0.25 15 smooth 8 0.5 15 smooth

107 190 119 188 119 187 140 164

y 117 178 116 191 132 173 133 145

117 179 133 195 126 166 132 144

(1) Analyze the data and draw conclusions. (2) Analyze the residuals.

Solution There are three factors V, D and R, each at two levels, are of interest. The design is called a 23 factorial design. Using DOE method for 23 design to analyze the data: 1. Input and prepare the data on Minitab

2. Set up suitable parameters.

3. Output the data

3. Results and Discussion (1) Analyze the data and draw conclusions. The ANOVA in the above Table is used to confirm the magnitude of these effects. From these Table we note that the main effect of V is highly significant (because has very small P-value). The V*D, V*R, D*R, and V*D*R are also highly significant; thus there are strong interaction between V and D, V and R; V, D, and R on the reaeration rate (y). (2) Analyze the residuals.

From the normal probability plot of these residuals show that doesn’t reveal anything particularly troublesome. Although the largest positive residual (13 at V = 0.05, D = 15, R = coarse) does stand out some what from others.The standardized value of this residual is 13 / 59.4 = 1.69, and this is only residual whose absolute value is smaller than 2.

Residuals Versus the Order of the Data (response is y) 15

Residual

10

5

0

-5

-10 2

4

6

8

10 12 14 16 Observation Order

18

20

22

24

Residuals Versus the Fitted Values (response is y) 15

Residual

10

5

0

-5

-10 110

120

130

140

150 160 Fitted Value

170

180

190

200

From plots the residuals versus the fitted values. This plots indicates that a mild tendency for the variance of the residuals to increase.