The research is about the tea consuming persons in Malaysia. This survey decide that the effect of tea consumption in our economic growth.This is decided in survey report. It relates to many SDGs like reduction of poverty, women empowerment, and others etc.
the data needs to be collected according to one of the 17 sdgs mentioned here
MST2034 / November 2022
1. Preparation of Experimental Design
Decide on a unique research topic which satisfies a 2k factorial design with at least four
replications related to your personal interest that fit into Sustainable Development Goals
(SDGs). In this section, you are required to give an overview on your research topic, such as
how you decide on the components of the experimental design, how you control the data
collection and etc. Append your dataset on the last page of your report.
2. Model Analysis
Conduct relevant model analysis on a self-determined conventional significance level. You
should consider blocking on a confounding system in your model analysis.
3. Modified Model Analysis
Based on section 2 Model Analysis, consider reasonable adjustment towards the original model
and perform relevant model analysis again on a same/ different self-determined conventional
significance level on the modified model.
4. Conclusion and Reporting
Conclude on your findings. You are required to discuss how this research topic could help in
Marking Rubric for
1. Preparation of Experimental Design
2. Model Analysis
3. Modified Model Analysis
data analysis performed with
4. Conclusion and Reporting
Clear and concise
report with many
with some minor
–END OF PAPER-2
9 – 10
Clear research topic
relates to SDGs. All
9 – 10
Excellent flow of
data analysis with
9 – 10
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
MST2034 DESIGN OF EXPERIMENTS
(TUTORIAL 3) (ANSWERS)
(Chapter 3) (Factorial Experiments)
i = 1, 2,…, 4
yijk = + i + j + ( )ij + ijk , j = 1, 2,…, 4
k = 1, 2
3.9364 > 3.24
24.8858 > 3.24
1.5569 < 2.54 12.5035 8.03126 Hence, the analysis from ANOVA suggested that we could reject null hypotheses for factors A and B. However, we do not have sufficient evidence to conclude that there is existence of interaction between the factors. Question 2 i = 1, 2 yijk = + i + j + ( )ij + ijk , j = 1, 2,3, 4 k = 1, 2 Source of Variation Drill speed (D) Feed rate (F) DF Error Total df SS MS F0 1 3 3 8 15 0.1444 0.0925 0.0417 0.0234 0.302 0.1444 0.03083 0.0139 0.02925 49.3675 > 5.32
10.5402 > 4.07
4.7521 > 4.07
Hence, the analysis from ANOVA suggested that we can reject all the null hypotheses. Hence, we
conclude that there is any difference in drill speed and feed rate in affecting the thrust force. There
is existence of interaction too.
MST2034 Design of Experiments
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
21.1426 > 3.89
1.09523 < 3.49 1.9633 < 3.00 Hence, the ANOVA indicates that operators influenced the breaking strength of a synthetic fiber significantly. However, no evidence shows that production machines affecting the braking strength and existence of interaction between the two factors. Question 4 i = 1, 2,3, 4 yijk = + i + j + k + ( )ij + ijk j = 1, 2,3, 4 k = 1, 2 Source of Variation Blocks df SS MS F value 1 2.5312 2.5312 0.3014 Temperature (T) 3 y2 1 2 y − i abn = 94.8437 bn i 31.6146 3.7646 > 3.29
y2j − = 599.5937
23.7993 > 3.29
ij abn − SSA − SSB = 112.5313
1.4889 < 2.59 Error Total 15 31 125.9689 8.3979 2 y y − abn = 935.4688 2 ijk i j k With block effects introduced, the error sum of squares has been reduced, however, block effect does not seem to contribute a lot to the error reducing effect in this experiment, the treatments, temperature and copper content still affect the warping of copper plates. The interaction effect is again does not exist. MST2034 Design of Experiments BinDS, BAS AUGUST 2022 SEMESTER SCHOOL OF MATHEMATICAL SCIENCES Question 5 i = 1, 2,3 yijk = + i + j + k + ( )ij + ijk j = 1, 2,3, 4 k = 1, 2 Source of Variation Blocks df 1 SS 2.04163 MS 2.04163 F value 0.5144 Operator (O) Machine (M) OM Error Total 2 3 6 11 23 160.3333 12.4583 44.6667 43.6584 262.9583 80.1667 4.1528 7.4444 3.9689 20.1987 > 3.98
1.04633 < 3.59 1.8757 < 3.09 With block effects introduced, the error sum of squares has been reduced, however, it does not change the conclusion which we have concluded earlier. Only operators affect the breaking strength of a synthetic fiber. Again, there is no sufficient evidence to show the existence of the interaction effect. Question 6 i = 1, 2,3 j = 1, 2 yijkl = + i + j + k + ( )ij + ( )ik + ( ) jk + ( )ijk + ijkl k = 1, 2,3 l = 1, 2,3 Source Cycle time (A) Temperature (B) Operator (C) AB AC BC ABC Error Total df 2 1 2 2 4 2 4 36 53 SS 426.3333 44.4629 279 68.9260 349.6667 8.0307 39.01273 123.401 1338.8333 MS 213.1667 44.4629 139.5 34.463 87.4167 4.01535 9.7532 3.4278 F value 62.1876 > 3.23
13.1012 > 4.08
40.6967 > 3.23
10.0540 > 3.23
25.5023 > 2.61
1.1714 < 3.23 2.8453 > 2.61
From the ANOVA table, we see that the cycle time, temperature, operators significantly affect the
fill volume. All interaction effects are tested to be exist in this experiment affecting the dyeing of
cotton-synthetic cloth, except for the interaction effect between temperature and operators.
MST2034 Design of Experiments
AUGUST 2022 SEMESTER
SCHOOL OF MATHEMATICAL SCIENCES
yijk = + i + j + ( )ij + k + ijk , i = 1, 2,3; j = 1, 2,3; k = 1, 2
= overall mean
i = effect of the ith level of the row factor (Temperature)
j = effect of the jth level of the column factor (Pressure)
( )ij = effect of interaction between i and j (interaction effect)
k = effect of the kth block
ijk = random error component
Temp : Pressure
Sum of Squares
H 0 : ( )ij = 0
H1 : At least one ( )ij 0 .
Test statistic: F0 = 2.0952 < F0.05,4,8 = 3.84 We do not reject H0 and conclude that there is no statistically significant in (temperature x pressure) interaction. (d) t0.025, N −a =15 = 2.131 ; LSD = ( 2.131) ( 0.5312555) ( 3(12) + 3(12) ) = 0.8968 y1 − y2 = 0.2167 LSD y1 − y3 = 1.05 LSD * y2 − y3 = 1.2667 LSD * Pressure of 250 & 270, 260 & 270 are significantly different. MST2034 Design of Experiments BinDS, BAS AUGUST 2022 SEMESTER SCHOOL OF MATHEMATICAL SCIENCES MST2034 DESIGNS OF EXPERIMENTS (TUTORIAL 3) (Chapter 3) (Factorial Experiments) Question 1 Johnson describe an experiment to investigate warping of copper plates. The two factors studied were the temperature and the copper content of the plates. The response variable was a measure of the amount of warping. The data were as follows. Is there any indication that either factor affects the amount of warping? Is there any interaction between the factors? Use = 0.05 . Temperature (oC) 50 75 100 125 Copper Content (%) 40 60 80 100 17, 20 16, 21 24, 22 28, 27 12, 19 18, 13 17, 12 27, 31 16, 12 18, 21 25, 23 30, 23 21, 17 23, 21 23, 22 29, 31 Question 2 A mechanical engineer is studying the thrust force developed by a drill press. He suspects that the drilling speed and the feed rate of the material are the most important factors. He selects four feed rates and uses a high and low drill speed chosen to represent the extreme operating conditions. He obtains the following results. Analyze the data and draw conclusions. Use = 0.05 . Drill Speed 125 200 0.015 2.70 2.78 2.83 2.86 Feed rate 0.030 0.045 2.45 2.60 2.49 2.74 2.85 2.86 2.80 2.87 0.060 2.75 2.86 2.94 2.88 Question 3 The factors that influence the breaking strength of a synthetic fiber are being studied. Four production machines and three operators are chosen and a factorial experiment is run using fiber from the same production batch. The results are as follows. Use = 0.05 . Operator 1 2 3 1 109 110 110 112 116 114 Machine 2 3 110 108 115 109 110 111 111 109 112 114 115 119 4 110 108 114 112 120 117 Analyze the data and draw conclusions. MST2034 Design of Experiments BinDS, BAS AUGUST 2022 SEMESTER SCHOOL OF MATHEMATICAL SCIENCES Question 4 Referring to Question 1 again, analyze the data assuming that the replicates are blocks. Conclude whether there are any changes on the conclusions drawn earlier. Question 5 Referring to Question 3 again, analyze the data assuming that the replicates are blocks. Conclude whether there are any changes on the conclusions drawn earlier. Question 6 The quality control department of a fabric finishing plant is studying the effect of several factors on the dyeing of cotton-synthetic cloth used to manufacture men’s shirts. Three operators, three cycle times and two temperatures were selected and three small specimens of cloth were dyed under each set of conditions. The finished cloth was compared to a standard and a numerical score was assigned. The results are as follows. Analyze the data and draw conclusions. Temperature 350oC Operator Operator 1 2 3 1 2 3 23 30 31 24 38 34 24 28 32 23 36 36 25 26 29 28 35 39 36 34 33 37 34 34 35 38 34 39 38 36 36 39 35 35 36 31 28 35 26 26 36 28 24 35 27 29 37 26 27 34 25 25 34 24 300oC Cycle Time 40 50 60 Question 7 The yield of a chemical process is being studied. The two factors of interest are temperature and pressure. Three levels of each factor are selected; however, only 9 runs can be made in one day. The experimenter runs a complete replicate of the design on each day. The data are shown in the following table. Assuming that the days are blocks. Day 1 Pressure Day 2 Pressure Temperature 250 260 270 250 260 270 Low 86.3 84.0 85.8 86.1 85.2 87.3 Medium 88.5 87.3 89.0 89.4 89.9 90.3 High 89.1 90.2 91.3 91.7 93.2 93.7 (a) (b) (c) (d) Write a linear statistical model for the experiment, explain the terms. Compute the analysis of variance for the data. Test the null hypothesis of no Temperature x Pressure interaction. Use LSD test to determine which levels of the pressure factor are significantly different. MST2034 Design of Experiments BinDS, BAS MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies CHAPTER 3 FACTORIAL EXPERIMENTS 3.1) Introduction Many experiments involve the study of effects of two or more factors. In general, factorial designs are most efficient for this type of experiment. By a factorial design, we mean that in each complete trial or replication of the experiment all possible combinations of the levels of the factors are investigated. For example, if there are a levels of factor A and b levels of factor B, each replicate contains all ab treatment combinations. When factors are arranged in a factorial design, they are often said to be crossed. The effect of a factor is defined to be the change in response produced by a change in the level of the factor. This is frequently called a main effect because it refers to the primary factors of interest in the experiment. For example, consider the simple experiment in the figure below. This is a two-factor factorial experiment with both design factors at two levels. We have called these levels “low” and “high” and denoted them “ –” and “+”, respectively. The main effect of factor A in this two-level design can be thought of as the difference between the average response at the low level of A and the average response at the high level of A. Numerically, this is A = 40+252 − 20+230 = 21 . That is, increasing factor A from the low level to the high level causes an average response increase of 21 units. Similarly, the main effect B is B = 30+252 − 20+2 40 = 11 . If the factors appear at more than two levels, the above procedure must be modified because there are other ways to define the effect of a factor. In some experiments, we may find that the difference in response between the levels of one factor is not the same at all levels of the other factors. When this occurs, there is an interaction between the factors. For example, consider the two-factor factorial experiment shown in figure below. At the low level of factor B (or B − ), the A effect is A = 50 − 20 = 30 and at the high level of factor B (or B + ), the A effect is A = 12 − 40 = −28 . School of Mathematical Sciences (SMS) page 1 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies Because the effect of A depends on the level chosen for factor B, we see that there is interaction between A and B. The magnitude of the interaction effect is the average −28 −30 difference in these two A effects or AB = ( 2 ) = −29 . Clearly, the interaction is large in this experiment. These ideas may be illustrated graphically, shown below. Two-factor interaction graphs such as these are frequently very useful in interpreting significant interactions and in reporting results to nonstatistically trained personnel. However, they should not be utilized as the sole technique of data analysis because their interpretation is subjective and their appearance is often misleading. Left-hand figure: Note that the B − and B + lines are parallel, indicating a lack of interaction between factors A and B. Right-hand figure: Here, we see that the B − and B + lines are not parallel, indicating an interaction between factors A and B. 3.2) The Two-Factor Factorial Design The simplest types of factorial designs involve only two factors or sets of treatments. There are a levels of factor A and b levels of factor B and these are arranged in a factorial design; that is, each replicate of the experiment contains all ab treatment combinations. In general, there are n replicates. Example 1 (The Battery Design Experiment) As an example of a factorial design involving two factors, an engineer is designing a battery for use in a device that will subjected to some extreme variations in temperature. The only design parameter that he can select at this point is the plate material for the battery and he has three possible choices. When the device is manufactured and is shipped to the field, the engineer has no control over the temperature extremes that the device will encounter and he knows from experience that temperature will probably affect the effective battery life. However, temperature can be controlled in the product development laboratory for the purposes of a test. The engineer decides to test all three plate materials at three temperature levels – 15, 70 and 125oF – because these temperature levels are consistent with the product enduse environment. Because there are two factors at three levels, this design is sometimes called a 32 factorial design. Four batteries are tested at each combination of plate material and temperature and all 36 tests are run in random order. The experiment and the resulting observed battery life data are given below. School of Mathematical Sciences (SMS) page 2 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies Material Type 1 2 3 Temperature 15oF 70oF 130 155 34 40 74 180 80 75 150 188 136 122 159 126 106 115 138 110 174 120 168 160 150 139 125oF 20 70 82 58 25 70 58 45 96 104 82 60 In this problem the engineer wants to answer the following questions: 1. What effects do material type and temperature have on the life of the battery? 2. Is there a choice of material that would give uniformly long life regardless of temperature? This last question is particularly important. It may be possible to find a material alternative that is not greatly affected by temperature. If this so, the engineer can make the battery robust to temperature variation in the field. This is an example of using statistical experimental design for robust product design, a very important engineering problem. This design is a specific example of the general case of a two-factor factorial. 3.3) Statistical Analysis of Two-Factor Factorial Design To pass to the general case, let yijk be the observed response when factor A is at the ith level ( i = 1, 2,..., a ) and factor B is at the jth level ( j = 1, 2,..., b ) for the kth replicate ( k = 1, 2,..., n ) . In general, a two-factor factorial experiment will appear as follows. The order in which the abn observations are taken is selected at random so that this design is a completely randomized design. Factor B 1 2 b 1 y1b1 , y1b 2 , y111 , y112 , y121 , y122 , 2 Factor A a , y11n , y12 n , y1bn y211 , y212 , y221 , y222 , y2b1 , y2b 2 , , y21n , y22 n , y2bn ya11 , ya12 , ya 21 , ya 22 , yab1 , yab 2 , , ya1n , ya 2 n , yabn The observations in a factorial experiment can be described by a model. There are several ways to write the model for a factorial experiment. The effects model is School of Mathematical Sciences (SMS) page 3 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies i = 1, 2,..., a yijk = + i + j + ( )ij + ijk , j = 1, 2,..., b k = 1, 2,..., n where is the overall mean effect, i is the effect of ith level of the row factor A, j is the effect of the jth level of the column factor B, ( )ij is the interaction between i and j , ijk is a random error component. Both factors are assumed to be fixed and the treatment effects are defined as deviations from the overall mean, so a b = 0 ; = 0 ; i =1 i j =1 j Similarly, the interaction effects are fixed and are defined such that a b ( ) = ( ) = 0 . i =1 ij j =1 ij Because there are n replicates of the experiment, there are abn total observations. Another possible model for a factorial experiment is the means model i = 1, 2,..., a yijk = ij + ijk , j = 1, 2,..., b k = 1, 2,..., n where the mean of the ijth cell is ij = + i + j + ( )ij . In the two-factor factorial, both row and column factors, A and B are of equal interest. Specifically, we are interested in testing hypotheses about the equality of row treatment effects, say H 0 : 1 = 2 = ... = a = 0 H1 : at least one i 0 and the equality of column treatment effects, say H 0 : 1 = 2 = = b = 0 H1 : at least one j 0 . We are also interested in determining whether row and column treatments interact. Thus, we also wish to test H 0 : ( )ij = 0 for all i, j H1 : at least one ( )ij 0 . We now discuss how these hypotheses are tested using a two-factor analysis of variance. School of Mathematical Sciences (SMS) page 4 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies Let yi denote the total of all observations under the ith level of factor A, y j denote the total of all observations under the jth level of factor B, yij denote the total of all observations in the ijth cell and y denote the grand total of all the observations. Define yi , y j , yij and y as the corresponding row, column, cell and grand averages. Express mathematically, b n yi , i = 1, 2,..., a bn j =1 k =1 a n y y j = yijk ; y j = j , j = 1, 2,..., b an i =1 k =1 n y yij = yijk ; yij = ij , i = 1, 2,..., a n j = 1, 2,..., b k =1 b n y y = yijk ; y = . abn j =1 k =1 yi = yijk ; yi = The total corrected sum of squares may be written as ( y − y ) = ( y − y ) + ( y − y − y + y ) + ( y − y ) a b n i =1 j =1 k =1 2 ijk ... a b n i =1 j =1 k =1 i ij i j ijk 2 ij = bn ( yi.. − y... ) + an ( y. j . − y... ) + n ( yij . − yi.. − y. j . + y... ) a b 2 i =1 j =1 + ( yijk − yij. ) a 2 b n a b 2 i =1 j =1 2 i =1 j =1 k =1 because the six cross products on the right-hand side are zero. Notice that the total sum of squares has been partitioned into a sum of squares due to “rows”, or factor A, (SSA); a sum of squares due to “columns”, or factor B, (SSB); a sum of squares due to the interaction between A and B, (SSAB); and a sum of squares due to error, (SSE). This is the fundamental ANOVA equation for the two-factor factorial. Symbolically, SSTO = SSA + SSB + SSAB+SSE The number of degrees of freedom associated with each sum of squares is Effect A B AB interaction Degrees of Freedom a–1 b–1 ( a − 1)( b − 1) Error ab ( n − 1) Total abn − 1 We may justify this allocation of the abn − 1 total degrees of freedom to the sum of squares as follows: The main effects A and B have a and b levels, respectively; therefore, they have School of Mathematical Sciences (SMS) page 5 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies a – 1 and b – 1 degrees of freedom as shown. The interaction degrees of freedom are simply the number of degrees of freedom for cells (which is ab − 1 ) minus the number of degrees of freedom for the two main effects A and B; that is ab − 1 − ( a − 1) − ( b − 1) = ( a − 1)( b − 1) . Within each of the ab cells, there are n – 1 degrees of freedom between the n replicates; thus, there are ab ( n − 1) degrees of freedom for error. If there are column treatment effects or interaction present, then the corresponding mean squares will be larger than MSE. Therefore, to test the significance of both main effects and their interaction, simply divide the corresponding mean square by the error mean square. Large values of this ratio imply that the data do not support the null hypothesis. Computationally, manual computing of the sum of squares is shown below. a b n SSTO = yijk2 − i =1 j =1 k =1 y2 ; abn SSA = 1 a 2 y2 yi − abn ; bn i =1 SSB = y2 1 b 2 y − j abn . an j =1 It is convenient to obtain the SSAB in two stages. First, we compute the sum of squares between ab cell totals, which is called the sum of squares due to “subtotals”: SSSubtotals = 1 a b 2 y2 yij − abn . n i =1 j =1 This sum of squares also contains SSA and SSB. Therefore, the second step is to compute SSAB as SSAB = SSSubtotals − SSA − SSB . We may compute SSE by subtraction as SSE = SSTO − SSSubtotals . SSE = SSTO − SSAB − SSA − SSB The ANOVA Table Source Factor A a −1 SS SSA MSA F0 MSA MSE Factor B b −1 SSB MSB MSE SSAB MSAB MSAB MSE Error ( a − 1)( b − 1) ab ( n − 1) SSB b−1 AB Interaction SSE MSE Total abn − 1 SSTO School of Mathematical Sciences (SMS) Df page 6 MS August 2022 Semester or MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies Example 2 Perform ANOVA for Example 1. Temperature 150F 700F Material Type 1 130 155 74 180 150 188 159 126 138 110 168 160 144.833 2 3 y. j. 34 40 80 75 136 122 106 115 174 120 150 139 107.583 1 2 3 yi 2. yi1. yi1. 134.75 155.75 144.00 b 20 70 82 58 25 70 58 45 96 104 82 60 64.167 83.167 108.333 125.083 1250F yi 2. n 2 ijk yi 3. yi 3. 57.25 119.75 145.75 y2 2 2 SSTO = y − = (130 ) + (155) + abn i =1 j =1 k =1 a yi.. Temperature 700F 150F Material Type 1250F 57.5 49.5 85.5 + ( 60 ) ( 3799 ) = 77646.97 . − 2 2 36 ( 3799 ) = 10683.72 . y2 1 a 1 2 2 2 SSMaterial = yi2 − = ( 998) + (1300 ) + (1501) − bn i =1 abn ( 3)( 4 ) 36 2 ( 3799 ) = 39118.72 . y2 1 b 2 1 2 2 2 SSTemperature = y j − = (1738) + (1291) + ( 770 ) − an j =1 abn 3 ( 4 ) 36 2 SSInteractions = y2 1 a b 2 y − ij abn − SSMaterial − SSTemperature n i =1 j =1 ( 3799 ) − 10683.72 − 39118.72 = 9613.78 + ( 342 ) − 36 and SSE = SSTO − SSMaterial − SSTemperature − SS Interaction . = 77646.97 − 10683.72 − 39118.72 − 9613.78 = 18230.75 2 1 2 2 = ( 539 ) + ( 229 ) + 4 Source Temperature Material Type Temperature X MT Error Total School of Mathematical Sciences (SMS) 2 Df 2 2 4 27 35 SS 39118.72 10683.72 9613.78 28230.75 77646.97 page 7 MS 19559.36 5341.86 2403.44 675.21 F0 28.97 7.91 3.56 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies Because F0.05,4,27 = 2.73 , we conclude that there is a significant interaction between material types and temperature. Furthermore, F0.05,2,27 = 3.35 , so the main effects of material type and temperature are also significant. - - - - - - - - - - - - - - - - - - - - - -- - - - - - - - - - - - - -- - - - - - - - - - - - - - - - - - - - -- - - To assist in interpreting the results of this experiment, it is helpful to construct a graph of the average responses at each treatment combination. The graph is shown below. The significant interaction is indicated by the lack of parallelism of the lines. In general, longer life is attained at low temperature, regardless of material type. Changing from low to intermediate temperature, battery life with material type 3 may actually increase, whereas it decreases for type 1 and 2. From intermediate to high temperature, battery life decreases for material types 2 and 3 and is essential unchanged for type 1. Material type 3 seems to give the best results if we want less loss of effective life as the temperature changes. 3.3.1 Multiple Comparisons When the ANOVA indicates that row or column means differ. It is usually of interest to make comparisons between the individual row or column means to discover the specific differences. Referring to Example 1, it indicates that interaction effect is significant. When interaction is significant, comparisons between the means of one factor (e.g. A) may be obscured by the AB interaction. One approach to this situation is to fix factor B at a specific level and apply Tukey’s test to the means of factor A at the level. To illustrate, let us refer to the following example. Example 3 From Example 1, we are interested in detecting differences among the means of three material types. Because the interaction is significant, we make this comparison at just one level of temperature, say level 2 (70 o F). We assume that the best estimate of the error variance is the MSE from the ANOVA table, utilizing the assumption that the experimental error variance is the same over all treatment combinations. The three material type averages at 70 oF arranged in ascending order are y12 = 57.25 (material type 1) y22 = 119.75 (material type 2) y32 = 145.75 (material type 3) and T0.05 = q0.05 ( 3, 27 ) School of Mathematical Sciences (SMS) MSE n = ( 3.50 ) page 8 675.21 4 = 45.47 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies where we obtained q0.05 ( 3, 27 ) 3.50 by interpolation. The pairwise comparisons yield 3 vs 1: 145.75 − 57.25 = 88.50 T0.05 = 45.47 3 vs 2: 145.75 − 119.75 = 26.00 T0.05 = 45.47 2 vs 1: 119.75 − 57.25 = 62.50 T0.05 = 45.47 This analysis indicates that at the temperature level 70 oF, the mean battery life is the same for material types 2 and 3 and that the mean battery life for material type 1 is significantly lower in comparison to both types 2 and 3. 3.4) The General Factorial Design The results for the two-factor factorial design may be extended to the general case where there are a levels of factor A, b levels of factor B, c levels of factor C and so on, arranged in a factorial experiment. In general, there will be abc…n total observations if there are n replicates to determine a sum of squares due to error if all possible interactions are included in the model. If all factors in the experiment are fixed, we may easily formulate and test hypotheses about the main effects and interactions using the ANOVA. For a fixed effects model, test statistics for each main effect and interaction by the mean square error. All of these F tests will be upper-tail, one-tail tests. The number of degrees of freedom for any main effect is the number of levels of the factor minus one and the number of degrees of freedom for an interaction is the product of the number of degrees of freedom associated with the individual components of the interaction. For example, consider the three-factor analysis of variance model: i = 1, 2,..., a j = 1, 2,..., b yijkl = + i + j + k + ( )ij + ( )ik + ( ) jk + ( )ijk + ijkl k = 1, 2,..., c l = 1, 2,..., n Assuming that A, B, C are fixed, the ANOVA table is shown below. The F tests on main effects and interactions follow directly from the expected mean squares. Source A a −1 df SS SSA MS MSA B b −1 SSB MSB C c −1 SSC MSC AB ( a − 1)( b − 1) SSAB MSAB School of Mathematical Sciences (SMS) page 9 F0 MSA MSE MSB MSE MSC MSE MSAB MSE August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies AC ( a − 1)( c − 1) SSAC MSAC BC ( b − 1)( c − 1) SSBC MSBC ABC ( a − 1)( b − 1)( c − 1) SSABC MSABC Error abc ( n − 1) SSE MSE Total abcn − 1 SSTO MSAC MSE MSBC MSE MSABC MSE Manual computing formulas for the sum of squares are as follows. a b c n y2 2 SSTO = yijkl − ; abcn i =1 j =1 k =1 l =1 y2 1 a 2 yi − ; bcn i =1 abcn b y2 1 2 SSB = y − j abcn ; acn j =1 SSA = SSC = y2 1 c 2 y − k abcn ; abn k =1 To compute the two-factor interaction sums of squares, the total for the A B , A C and B C cells are needed. It is frequently helpful to collapse the original data table into three two-way tables to compute these quantities. The sums of squares are found from SSAB = y2 1 a b 2 y − ij abcn − SSA − SSB = SSSubtotals( AB) − SSA − SSB ; cn i =1 j =1 y2 1 a b 2 y − ik abcn − SSA − SSC = SSSubtotals( AC ) − SSA − SSC ; bn i =1 k =1 y2 1 b c 2 SSBC = y − jk abcn − SSB − SSC = SSSubtotals( BC ) − SSB − SSC . an j =1 k =1 SSAC = Note that the sums of squares for the two-factor subtotals are found from the totals at each two-way table. The three-factor interaction sum of squares is computed from the threeway cell totals yijk as SSABC = y2 1 a b c 2 y − ijk abcn − SSA − SSB − SSC − SSAB − SSAC − SSBC . n i =1 j =1 k =1 = SSSubtotals( ABC ) − SSA − SSB − SSC − SSAB − SSAC − SSBC The error sum of squares may be found by subtracting the sum of squares for each main effect and interaction from the total sum of squares or by SSE = SSTO − SSSubtotals( ABC ) . School of Mathematical Sciences (SMS) page 10 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies Example 4 The Soft Drink Bottling Problem A soft drink bottler is interested in obtaining more uniform fill heights in the bottles produced by his manufacturing process. The filling machine theoretically fills each bottle to the correct target height, but in practice, there is variation around this target and the bottler would like to understand the sources of this variability better and eventually reduce it. The process engineer can control three variables during the filling process: the percent carbonation (A), the operating pressure in the filler (B) and the bottles produced per minute or the line speed (C). The pressure and speed are easy to control but the percent carbonation is more difficult to control during actual manufacturing because it varies with product temperature. However, for purposes of an experiment, the engineer can control carbonation at three levels: 10, 12 and 14%. She chooses two levels for pressure (25 and 30 psi) and two levels for line speed (200 and 250 bpm). She decided to run 2 replicates of a factorial design in these three factors, with all 24 runs taken in random order. The response variable observed is the average deviation from the target fill height observed in a production run of bottles at each set of conditions. The data that resulted from this experiment is given below. Positive deviations are fill heights above the target, whereas negative deviations are fill heights below the target. Operating Pressure (B) 25 psi 30 psi Line Speed (C) Line Speed (C) 200 250 200 250 −1 −1 −3 1 0 0 −1 1 0 2 2 6 1 1 3 5 5 7 7 10 4 6 9 11 6 15 20 34 % of Carbonation (A) 10 12 14 B C Totals, y jk 21 y. j .. B 25 −5 4 22 −4 20 59 75 = y.... 54 A C Totals, yi.k . C A 200 250 −5 10 1 12 6 14 14 25 34 A B Totals, yij .. A 10 12 14 yi 30 1 16 37 The total corrected sum of squares is found as ( 75) y2 SSTO = y − = 571 − = 336.625 abcn 24 i =1 j =1 k =1 l =1 and the sum of squares for the main effects are a b c 2 n 2 ijkl School of Mathematical Sciences (SMS) page 11 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies ( 75) y2 1 a 2 1 2 2 2 SSCarbonation = yi − = ( −4 ) + ( 20 ) + ( 59 ) − = 252.75 ; bcn i =1 abcn 8 24 2 ( 75) 1 2 2 y2 1 b 2 = ( 21) + ( 54 ) − = 45.375 ; and SSPressure = y − j 24 acn j =1 abcn 12 2 ( 75) 1 y2 1 c 2 2 2 = ( 26 ) + ( 49 ) − = 22.042 ; y − SSSpeed = k 24 abn k =1 abcn 12 2 To calculate the sums of squares for the two-factor interactions, we must find the two-way cell totals. Hence, the sums of squares are y2 1 a b SSAB = yij2 − − SSA − SSB cn i =1 j =1 abcn ; 2 75 ) ( 1 2 2 2 = ( −5 ) + (1) + + ( 37 ) − − 252.75 − 45.375 = 5.25 4 24 The carbonation-speed or AC interaction uses the A C cell totals yik shown below. SSAC = y2 1 a b 2 y − ik abcn − SSA − SSC bn i =1 k =1 1 2 2 = ( −5) + (1) + 4 ( 75) ; 2 + ( 34 ) − − 252.75 − 22.042 = 0.583 24 2 The pressure-speed or BC interaction is found from the B C cell totals y jk shown below. y2 1 b c 2 SSBC = y jk − abcn − SSB − SSC an j =1 k =1 = ( 75) ; 2 1 2 2 2 2 − 45.375 − 22.042 = 1.042 ( 6 ) + (15) + ( 20 ) + ( 34 ) − 6 24 The three-factor interaction sum of squares is found from the A B C cell totals yijk as follows. y2 1 a b c 2 SSABC = yijk − − SSA − SSB − SSC − SSAB − SSAC − SSBC n i =1 j =1 k =1 abcn 1 2 2 = ( −4 ) + ( −1) + 2 = 1.083. ( 75) + ( 21) − − 252.75 − 45.375 − 22.042 − 5.25 − 0.583 − 1.042 24 2 2 y2 1 a b c 2 y − ijk abcn = 328.125 , we have n i =1 j =1 k =1 SSE = SSTO − SSSubtotals( ABC ) = 336.625 − 328.125 = 8.5 . Finally, noting that SSSubtotals( ABC ) = School of Mathematical Sciences (SMS) page 12 August 2022 Semester MST2034 Design of Experiments Source % of carbonation (A) Pressure (B) Line Speed (C) AB AC BC ABC Error Total B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies df 2 1 1 2 2 1 2 12 23 SS 252.75 45.375 22.042 5.25 0.583 1.042 1.083 8.500 336.625 MS 126.375 45.375 22.042 2.625 0.292 1.042 0.542 0.708 F value 178.412 64.059 31.118 3.706 0.412 1.471 0.765 From the ANOVA table, we see that the percentage of carbonation, operating pressure and line speed significantly affect the fill volume. The carbonation-pressure interaction F ratio indicating some interaction between these factors. 3.5) Blocking in a Factorial Design We have discussed factorial designs in the context of a completely randomized experiment. Sometimes, it is not feasible or practical to completely randomize all of the runs in a factorial. For example, the presence of a nuisance factor may require that the experiment be run in blocks. We discussed the basic concepts of blocking in the context of a singlefactor experiment in previous chapter. We now show how blocking can be incorporated in a factorial. Consider a factorial experiment with two factors (A and B) and n replicates. The linear statistical model for this design is i = 1, 2,..., a yijk = + i + j + ( )ij + ijk j = 1, 2,..., b k = 1, 2,..., n where i , j and ( )ij represent the effects factor A, B and the AB interaction, respectively. Now suppose that to run this experiment a particular raw material is required. This raw material is available in batches that are not large enough to allow all abn treatment combinations to be run from the same batch. However, if a batch contains enough material for ab observations, then an alternative design is to run each of the n replicates using a separate batch of raw material. Consequently, the batches of raw material represent a randomization restriction or a block and a single replicate of a complete factorial experiment is run within each block. The effects model for this new design is i = 1, 2,..., a yijk = + i + j + ( )ij + k + ijk j = 1, 2,..., b k = 1, 2,..., n where k is the effect of the kth block. Of course, within a block the order in which the treatment combinations are run is completely randomized. School of Mathematical Sciences (SMS) page 13 August 2022 Semester MST2034 Design of Experiments B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies ANOVA Table for a Two-Factor Factorial in a Randomized Complete Block Source of df SS F value Variation Blocks n −1 y2 1 2 y − abn k ab k Factor A a −1 y2 1 yi2 − bn i abn MSA MSE Factor B b −1 y2 1 2 y − j abn an j MSB MSE AB ( a − 1)( b − 1) MSAB MSE Error ( ab − 1)( n − 1) ( abn − 1) y2 1 yij2 − − SSA − SSB n i abn Total Subtraction yijk2 − i j k y2 abn The model assumes that interaction between blocks and treatments is negligible. This was assumed previously in the analysis of randomized block designs. If these interactions do exist, they cannot be separated from the error component. In fact, the error 2 2 2 term in this model really consists of the ( )ik , ( ) jk and ( )ijk interactions. The ANOVA outlined a layout which closely resembles that of a factorial design, with the error sum of squares reduced by the sum of squares for blocks. Computationally, we find the sum of squares for blocks as the sum of squares between the n block totals yk . Example 5 An engineer is studying methods for improving the ability to detect targets on a radar scope. Two factors she considers to be important are the amount of background noise or “ground clutter”, on the scope and the type of filter placed over the screen. An experiment is designed using three levels of ground clutter and two filter types. We will consider these as fixed-type factors. The experiment is performed by randomly selecting a treatment combination (ground clutter level and filter type) and then introducing a signal representing the target into the scope. The intensity of this target is increased until the operator observes it. This intensity level at detection is then measured as the response variable. Because operator availability, it is convenient to select an operator and keep him or her at the scope until all the necessary runs have been made. Furthermore, operators differ in their skill and ability to use the scope. Consequently, it seems logical to use the operators as blocks. Once an operator is chosen, the order in which the six treatment combinations are run is randomly determined. Thus, we have a 3 2 factorial experiment run in a randomized complete block. The data are shown below. School of Mathematical Sciences (SMS) page 14 August 2022 Semester MST2034 Design of Experiments Operators (blocks) Filter Type Low Medium High B Sc (Hons) in Industrial Statistics B Sc (Hons) in Actuarial Studies 1 2 2 1 2 Ground clutter 90 86 96 84 102 87 106 90 114 93 112 91 1 3 4 1 2 1 2 100 105 108 92 97 95 92 96 98 81 80 83 The linear model for this experiment is i = 1, 2,..., a yijk = + i + j + ( )ij + k + ijk j = 1, 2,..., b k = 1, 2,..., n where i represents the ground clutter effect, j represents the filter type effect ( )ij is the interaction, k is the block effect and ijk is the NID ( 0, 2 ) error component. The sum of squares for ground clutter, filter type and their interaction are computed in the usual manner. The sum of squares due to blocks is found from the operator totals yk as follows. SSBlocks = y2 1 n 2 y − k abn ab k =1 ( 2278) 1 2 2 2 2 = 572 ) + ( 579 ) + ( 597 ) + ( 530 ) − = 402.17 ( ( 3)( 2 )( 4 ) ( 3) 2 2 ANOVA Table Source Ground clutter (G) Filter type (F) GF Blocks Error Total Df 2 1 2 3 15 23 SS MS F0 335.58 157.79 15.13 1066.67 1066.67 96.19 77.08 38.54 3.48 402.17 134.06 166.33 11.09 2047.83 The complete ANOVA indicates that all effects are tested by dividing their mean squares by the mean square error. Both ground clutter level and filter type are significant at the 1 percent level, whereas their interaction is significant only at the 10 percent level. Thus, we conclude that both ground clutter level and the type of scope filter used affect the operator’s ability to detect the target and there is some evidence of mild interaction between these factors. School of Mathematical Sciences (SMS) page 15 August 2022 Semester
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