LECTURE Notes on Design of Experiments

L E C T U R E N O T E S O N D E S I G N O F E X P E R I M E N T S - D r . V . G N A N A R A J

Page 1

DESIGN OF EXPERIMENTS

INTRODUCTION

Analysis of Variance (ANOVA) is a hypothesis-testing technique used to test the equality of two or

more population (or treatment) means by examining the variances of samples that are taken.

ANOVA allows one to determine whether the differences between the samples are simply due to

random error (sampling errors) or whether there are systematic treatment effects that cause the mean

in one group to differ from the mean in another.

Most of the time ANOVA is used to compare the equality of three or more means, however when the

means from two samples are compared using ANOVA it is equivalent to using a t-test to compare the

means of independent samples.

ANOVA is based on comparing the variance (or variation) between the data samples to variation within

each particular sample. If the between variation is much larger than the within variation, the means of

different samples will not be equal. If the between and within variations are approximately the same

size, then there will be no significant difference between sample means.

ELEMENTS OF A DESIGNED EXPERIMENT

Definition 1

The response variable is the variable of interest to be measured in the experiment. We also refer to

the response as the dependent variable.

Definition 2

Factors are those variables whose effect on the response is of interest to the experimenter.

Quantitative factors are measured on a numerical scale, whereas qualitative factors are not

(naturally) measured on a numerical scale.

Definition 3

Factor levels are the values of the factor utilized in the experiment.

Definition 4

The treatments of an experiment are the factor-level combinations utilized.

Definition 5

An experimental unit is the object on which the response and factors are observed or measured.

Definition 6

A designed experiment is an experiment in which the analyst controls the specification of the

treatments and the method of assigning the experimental units to each treatment. An observational

experiment is an experiment in which the analyst simply observes the treatments and the response on

a sample of experimental units


Page 2

Definition 7

The completely randomized design is a design in which treatments are randomly assigned to the

experimental units or in which independent random samples of experimental units are selected for

each treatment.

Design of Experiments is classified into three types

1) Completely Randomized Design or One-way Analysis of Variance

2) Randomized Design or Two-way Analysis of Variance

3) Latin Square Design or Three-way Analysis of Variance

1) COMPLETELY RANDOMIZED DESIGN or One-way Analysis of Variance

The test procedure compares the variation in observations between samples to the variation within

samples. Completely randomized designs are the simplest in which the treatments are assigned to the

experimental units completely at random. This allows every experimental unit, i.e., plot, animal, soil

sample, etc., to have an equal probability of receiving a treatment.

Suppose we wish to compare k population means ( 2k ). This situation can arise in two ways. If the study is observational, we are obtaining independently drawn samples from k distinct populations and

we wish to compare the population means for some numerical response of interest. If the study is

experimental, then we are using a completely randomized design to obtain our data from k distinct

treatment groups. In a completely randomized design the experimental units are randomly assigned to

one of k treatments and the response value from each unit is obtained. The mean of the numerical

response of interest is then compared across the different treatment groups.

Advantages of Completely Randomized Designs

1. Complete flexibility is allowed - any number of treatments and replicates may be used.

2. Relatively easy statistical analysis, even with variable replicates and variable experimental errors for

different treatments.

3. Analysis remains simple when data are missing.

4. Provides the maximum number of degrees of freedom for error for a given number of experimental

units and treatments.

Disadvantages of Completely Randomized Designs

1. Relatively low accuracy due to lack of restrictions which allows environmental variation to enter

experimental error.


Page 3

2. Not suited for large numbers of treatments because a relatively large amount of experimental

material is needed which increases the variation.

Appropriate Use of Completely Randomized Designs

1. Under conditions where the experimental material is homogeneous, i.e., laboratory, or growth

chamber experiments.

2. Where a fraction of the experimental units is likely to be destroyed or fail to respond.

3. In small experiments where there is a small number of degrees of freedom.

The completely randomized design is seldom used in field experiments where the randomized

complete block design has been consistently more accurate since there are usually recognizable

sources of environmental variation.

In One way Classification

Null Hypothesis koH === ...: 21 Alternative Hypothesis: . somefor i.e. differ, means population least twoat : jiH jia

Assumptions:

1. Samples are drawn independently (completely randomized design)

2. Population variances are equal, i.e. 22221 k === L .

3. Populations are normally distributed.

Notations:

Number of Samples (or levels) = k

Number of observations in i th sample = ni, i = 1,2,3,, k

Total Number of observations = =i

inn

Observation j in the i th sample = xij, j = 1,2,3,.,ni

Sum of ni observations in i th sample = j

ijx= Ti

Sum of all observations ==i ji

i xijTT

The Computational Formulae


Page 4

Total Sum of Squares =i j

ijTn

TxSS

22

Between samples sum of squares, n

Tn

TSSi i

iB

22

=

Within samples sum of squares SSW = SST - SSB

Total mean square 1

=

n

SSMS TT Between samples square 1=

kSSMS BB

Within sample square, kn

SSMS WW

= Number of d.o.f = (k -1) + (n k) = n 1.

ANOVA TABLE

Source of variation Sum ofSquares Degrees of Freedom Mean Square F Ratio

Between samples SSB k - 1 MSB

W

B

MSMS

Within samples SSW n - k MSW Total SST n - 1

Working Method

STEP I: Set up Null Hypothesis kH ==== 210 (Population means are equal) STEP II: Set up Alternative Hypothesis kH = 211 (Population means are not equal) STEP III: Take l.o.s =

STEP IV: Find total number of observations n.

STEP V: Calculate T, the Grand Total number of observations.

STEP VI: Calculate the sum of squares SST, SSB, SSW.

STEP VII: Prepare ANOVA Table to calculate F-Ratio.

STEP VIII: Conclusions.

i) If calculated F > F for F, (k -1) + (n k) d.o.f , Reject H0

ii) If calculated F < F for F, (k -1) + (n k) d.o.f , Accept H0

PROBLEM 1

Neuroscience researchers examined the impact of environment on rat development. Rats were

randomly assigned to be raised in one of the four following test conditions: Impoverished (wire mesh

cage - housed alone), standard (cage with other rats), enriched (cage with other rats and toys), super

enriched (cage with rats and toys changes on a periodic basis). After two months, the rats were tested

on a variety of learning measures (including the number of trials to learn a maze to a three perfect trial

criteria), and several neurological measure (overall cortical weight, degree of dendritic branching, etc.).

The data for the maze task is below. Compute the appropriate test for the data provided below.

Impoverished Standard Enriched Super Enriched

22 17 12 8

19 21 14 7


Page 5

15 15 11 10

24 12 9 9

18 19 15 12

Solution:


Between samples SSB= 323.35 3 107.7833 71.12=W

B

MSMS

Within samples SSW = 135.6 16 8.475

Total SST = 458.95 19

Null Hypothesis 43210 ====H Alternative Hypothesis H1: At least two means differ

Test Statistic: Fc = 12.71

Table Value F0.05,(3,16)= 3.49

Conclusion: Fc > F0.05,(3,12) , Reject Null Hypothesis

1. What is your computed answer? F = 12.71 (3,16) p < .01

2. What would be the null hypothesis in this study? Environment will have no impact on learning

ability as operationalized by maze performance in rats.

3. What would be the alternate hypothesis? Environment will have an impact on learning ability

as operationalized by maze performance in rats.

4. What is your Fcrit? Fcrit = 5.29

5. Are there any significant differences between the four testing conditions? Yes - There is no

significant difference between the impoverished group and the standard group (Fcomp = 2.32

and qobs= 2.15, n.s.). There is a significant difference between the impoverished group and both

the enriched and supenriched group (Fcomp = 16.15 and qobs= 5.68, p < .01) and Fcomp = 31.90 and

qobs= 7.98, p < .01), respectively). There is no significant difference between the standard group

and the enriched group (Fcomp = 6.24 and qobs= 3.53, n.s.). There is a significant difference

between the standard group and the supenriched group (Fcomp = 17.03 and qobs= 5.83, p < .05).

There is no significant difference between the enriched group and the superenriched group

(Fcomp = 2.65 and qobs= 2.30, p < .05)).

6. Interpret your answer. Environment may have an impact on ability to learn. Differences were

found between groups when each group is compared to a group at least two levels above the

one under study. Thus for example, there is a difference between the impoverished and the

enriched and superenriched but not between the impoverished and the standard groups.

PROBLEM 2

A research study was conducted to examine the clinical efficacy of a new antidepressant. Depressed

patients were randomly assigned to one of three groups: a placebo group, a group that received a low

dose of the drug, and a group that received a moderate dose of the drug. After four weeks of

treatment, the patients completed the Beck Depression Inventory. The higher the score, the more

depressed the patient. The data are presented below. Compute the appropriate test.


Page 6

Placebo Low Dose Moderate Dose

38 22 14

47 19 26

39 8 11

25 23 18

42 31 5

Solution:


Between samples SSB= 1484.9333 2 742.46666 11.26=W

B

MSMS


Total SST = 2275.73333 14

Null Hypothesis 3210 ===H Alternative Hypothesis H1: At least two means differ


Table Value F0.05,(2,12)= 6.93


1. What is your computed answer? F = 11.26 (2,12) p < .01

2. What would be the null hypothesis in this study? There will be no difference in depression

levels between the three groups. The groups taking the drug will not be different than the

groups taking the placebo.

3. What would be the alternate hypothesis? There will be a difference somewhere in depression

levels between the three levels of drug groups.

4. What probability level did you choose and why? p = .01. There is a risk involved with a Type I

error. I do not want to erroneously say the drug works and then later find out that it doesn't.

5. What is your Fcrit? Fcrit = 6.93

6. Is there a significant difference between the groups? Yes - a significant difference exists

somewhere between the three groups.

7. If there is a significant difference, where specifically are the differences? There is a significant

difference between the placebo group and the low dose group (Fcomp = 11.75 and qobs= 4.84, p <

.05). There is a significant difference between the placebo group and the moderate dose group

(Fcomp = 20.77 and qobs= 6.44, p < .01). There is no significant difference between the low dose

and the moderate dose groups (Fcomp = 1.27 and qobs= 1.59, n.s.).

8. Interpret your answer. The drug appears to help alleviate depression. However, as there is no

significant difference between taking a low or moderate dose, a low dose would be

recommended.

PROBLEM 3


Page 7

A manufacturer of television sets is interested in the effect on tube conductivity of four different types

of coating for color picture tubes. The following conductivity data are obtained.

Coating Type Conductivity

1 143 141 150 146

2 152 149 137 143

3 134 136 132 127

4 129 127 132 129

Test the null hypothesis that 43210 ====H , against the alternative that at least two of the means differ. Use = 0.05.

Solution:


Between samples SSB= 844.68750 3 281.56250 14.30=

W

B

MSMS


Total SST = 1080.93750 15

Null Hypothesis 43210 ====H Alternative Hypothesis H1: At least two means differ


Table Value F0.05,(3,12)= 3.49


PROBLEM 4

A manufacturer suspects that the batches of raw material furnished by her supplier differ significantly

in calcium content. There is a large number of batches currently in the warehouse. Five of these are

randomly selected for study. A chemist makes five determinations on each batch and obtains the

following data.

Batch 1 Batch 2 Batch 3 Batch 4 Batch 5

23.46 23.59 23.51 23.28 23.29

23.48 23.46 23.64 23.40 23.46

23.56 23.42 23.46 23.37 23.37

23.39 23.49 23.52 23.46 23.32

23.40 23.50 23.49 23.39 23.38

Is there a significant variation in calcium content from batch to batch? Use

= 0.05.

Solution:



W

B

MSMS



Page 8

Total SST = 0.1845760 24

43210 ====H H1: At least two means differ


Table Value F0.05,(4,20)= 2.84

Conclusion: Fc > F0.05,(4,20) , Reject Null Hypothesis.

PROBLEM 5

Four Laboratories measure the tin coating weight of 12 disks and that the results are as follows.

Lab A 0.25 0.27 0.22 0.30 0.27 0.28 0.32 0.24 0.31 0.26 0.21 0.28

Lab B 0.18 0.28 0.21 0.23 0.25 0.20 0.27 0.19 0.24 0.22 0.29 0.16

Lab C 0.19 0.25 0.27 0.24 0.18 0.26 0.28 0.24 0.25 0.20 0.21 0.19

Lab D 0.23 0.30 0.28 0.28 0.24 0.34 0.20 0.18 0.24 0.28 0.22 0.21

Construct an ANOVA table and test the hypothesis , whether there is any difference among the four

sample means can be attributed to chance at 5%

Solution:



W

B

MSMS


Total SST = 0.0809 47

43210 ====H H1: At least two means differ


Table Value F0.05,(3,44)= 2.82

Conclusion: Fc > F0.05,(3,44) , Reject Null Hypothesis.

PROBLEM 5

A production manager wishes to test the effect of 5 similar milling machines on the surface of finish of

small casting. So he selected 5 such machines and conducted the experiment with four replication

under each machine as per Completely Randomized Design and obtained the following reading

Machines


Page 9

Relication

M1 M2 M3 M4 M5

25 10 40 27 15

30 20 30 20 8

16 33 49 35 45

36 42 22 48 34

Perform the required ANOVA and state the inference at 5% i.o.s.

Solution:



W

B

MSMS


Total SST = 2831.78 19

543210 =====H H1: At least two means differ


Table Value F0.05,(3,44)= 3.06

Conclusion: Fc > F0.05,(3,44) , Accept Null Hypothesis.

There is no significant difference between machines in terms of surface finish of small castings.

6) A study of depression and exercise was conducted. 3 groups were used: those in a designed

exercise program; a group that is sedentary and a group of runners. A depression rating was good one

to the members in each group.

Exercise Group 63 58 61 60 62 59 363

Sedentary Group 71 64 68 65 67 67 402

Runners 49 52 47 51 48 247

Total 183 174 176 176 177 126 1012

Does the data provide sufficient evidence to indicate difference among the population means at 1%

level of significance?

Answer: H: No difference among the population mean.

=1%

We shift the origin to 60 and subtracted the given values with 60.

Depression Ti Ti/n Xij

Exercise

Group 3 -2 1 0 2 -1 3 13.5 19

Sedentary

Group 11 4 8 5 7 7 42 294 324

Runners -11 -8 -13 -9 -12 -53 468.1 579

Tj 3 -6 -4 -4 -3 6 -8 775.6 922

Tj/n 3 12 5.3 5.3 3 12 40.6


Page 10

=

=775.6+0.44

=776.04

=

=922+0.4

=922.04

= =922.04-776.04

=146

Source of variation Sum of squares Degrees of

freedom

Mean square F ratio

Between samples 776.04 2 388.02 =13.2 Within samples 146 5 29.2

Total 922.04 18 51.22

F.=4.33. > Therefore Reject (there is no difference between population means).

7) Three special ovens in a metal working shop are used to heat metal

specimens. All the ovens are supposed to operate at the same

temperature. It is known that the temperature of an ovens. The table

below shows the temperature, in degrees centigrade, of each of the three

ovens on a random sample of heating.

Test for difference between mean oven temperatures at 5% los?

Solution:-

OVEN TEMPERATUREc

1 494 497 481 496 487

2 489 494 479 478

3 489 483 487 472 472 477

OVEN TEMPERATUREc

494 497 481 496 487

2 489 494 479 478

3 489 483 487 472 472 477

OVEN TEMPERATURE c

1 5 8 -8 7 -2 10 16.66 206

2 0 5 -10 -11 -16 42.66 246

3 0 6 -2 -17 -17 -12 -54 486 762

5 7 -20 -21 -19 -12 =

=

=1294 /R 8.33 16.33 133.33 147 120.33 48 /R=473.3

2


Page 11

we shift the origin to 489 and subtract 489 from the given values.

s = -

= 1214 - ! = 1194

s" =

= 545.32

! = ##.

s =

" = 473.32 -

! = %#.

s& = s '" ' = (% ##. %#. = 215.36

ANOVA TABLE:- Source

of

variation

Sum of squares d.0.f Mean

square

F Ratio

Between

rows 525.32 2 262.66 = . (

Between

columns 453.32 5 90.664 = %.

Error 215.36 10 21.536

TOTAL 1194

.#,(2,10) = 4.10 .#,(5,10)=3.33

545.3

= (%


Page 12

> .# = 12.19 > 4.10 Reject > .#= 4.20 > 3.33 Reject

Hence, there is difference between mean oven temperatures.

8 ) A manufacturer of television sets is interested in the effect on tube conductivity of four different

types of coating for color picture tubes. The following conductivity data are obtained.

Coating type conductivity

1 143 141 150 146

2 152 149 137 143

3 134 136 132 127

4 129 127 132 129

Test the null hypothesis that H==== against the alternative that at least 2 of the means differ.

Use =0.05.

Answer:

Step1: Null hypothesis- H==== (population means are equal)

Step2: Alternative hypothesis H:ij (popula_on means are not equal)

=0.05

Coating

type

Conductivity Ti Tin Xij

1 143 141 150 146 580 84100 84146

2 152 149 137 143 581 84390.25 84523

3 134 136 132 127 529 69960.25 70005

4 129 127 132 129 517 66822.25 66835

Ti 558 553 551 545 2207 305272.75 305509

SSb= (Ti)-(Tn)

=305272.75-304428.06

=844.687

SST= (Xij)-(Tn)

=305509-304428.0625

=1080.9375

SSW=SSt-SSb

=1080.93-844.68

=236.25

Source of variation Sum of squares Degrees of

freedom

Mean square F ratio

Between samples 844.687 3 281.56 Fc=14.30

Within samples 236.25 12 19.681

Total 1080.93 15 72.02

F.=3.34. Fc>FTherefore Reject H (population means are not equal).


Page 13

RANDOMIZED BLOCK DESIGN OR TWO-WAY ANALYSIS OF VARIANCE

Two-way (or multi-way) ANOVA is an appropriate analysis method for a study with a quantitative

outcome and two (or more) categorical explanatory variables. This is an extension of the one factor

situation to take account of second factor. As such it is often called a Blocking Factor because it places

subjects or units into homogeneous groups called Blocks. The design itself is called a Randomized

Block Design. The usual assumptions of Normality, equal variance, and independent errors apply. If an

experiment has a quantitative outcome and two categorical explanatory variables that are defined in

such a way that each experimental unit (subject) can be exposed to any combination of one level of

one explanatory variable and one level of the other explanatory variable, then the most common

analysis method is two-way ANOVA. Because there are two different explanatory variables the effects

on the outcome of a change in one variable may either not depend on the level of the other variable

(additive model) or it may depend on the level of the other variable (interaction model).

Assumptions

1. The population at each factor level combination is (approximately Normally Distributed)

2. These normal populations have a common variance, 2.

3. The effect of one factor is the same at all levels of the other factor.

Notations

Number of levels of row factor r

Number of levels of column factor c

Total number of observations r x c

Observation in (ij) th cell of the table xij (ith level of row factor and

j th level of column factor)

i = 1,2,,r

j = 1,2,,c

Sum of c observations in i thi row =j

ijRi xT

Sum of r observations in j th column =i

ijCj xT

Sum of all r x c observations ===i j i j

CjRi TTxijT

Computational Formulae

Total Sum of Squares

=i j

ijTcr

TxSS

22

Between Rows Sum of Squares =

i

RiR

cr

Tc

TSS22

Between Columns Sum of Squares =

i

RiC

cr

Tr

TSS22


Page 14

Error(residual) Sum of Squares SSE = SST SSR SSC

ANOVA TABLE


Between rows SSR r - 1 MSR

E

R

MSMS

E

C

MSMS

Between Columns SSC c - 1 MSC

Error(residual)

SSE

(r 1) x (c 1)

MSE

Total SST r x c -1

H0 : No effect due to row factor H0 : No effect due to column factor

H1: An effect due to row factor H1: An effect due to column factor

Critical region F > F,(r-1,(r-1)(c-1)) Critical region F > F,(c-1,(r-1)(c-1))

Test Statistic E

RR MS

MSF = Test Statistic E

CC MS

MSF =

PROBLEM 1

Three laboratories, A, B, and C, are used by food manufacturing companies for making nutrition

analyses of their products. The following data are the fat contents (in grams) of the same weight of

three similar types of peanut butter.

Laboratory

Peanut

Butter

A B C D

Brand 1 16.6 17.7 16.0 16.3

Brand 2 16.0 15.5 15.6 15.9

Brand 3 16.4 16.3 15.9 16.2

Analyse the data at 5% significance by (a) carrying out a one-way ANOVA to see if there is a difference

between the fat content of the three brands; (b) performing a two-way ANOVA to see if there is any

difference between the Brands using the laboratories as blocks. (c) Do you think there is any evidence

that the results were not reasonably consistent between the four laboratories?

a) One-way ANOVA

Laboratory

Peanut Butter A B C D Mean

Brand 1 16.6 17.7 16.0 16.3 16.65

Brand 2 16.0 15.5 15.6 15.9 15.75

Brand 3 16.4 16.3 15.9 16.2 16.20


Page 15

Mean 16.33 16.50 15.83 16.13 16.20

Sums of squares

Total SS: Inputting all the individual values into the calculator gives the following summary statistics: n

= 12, x = 16.20, sn = 0.546 nsn2 = 3.58

Between Brands SS: The mean scores 1x = 16.65, 2x = 15.75 and 3x

= 16.20

Each of these means came from 4 values so inputting the means with a frequency of 4 gives: n = 12, x

= 16.20, sn = 0.367 nsn2 = 1.62 (n and x for checking)

Error SS: 3.58 1.62 = 1.96

Anova table In this example: (k =3 brands, N =12 values)

Source S.S. d.f. M.S.S. F

Between

brands

1.62 3 - 1 = 2 1.62/2 = 0.81 0.81/0.22 = 3.72

Errors 1.96 11 - 2 = 9 1.96/9 = 0.22

Total 3.58 12 - 1 = 11

Hypothesis test

H0: 1 = 2 = 3 H1: At least two of them are different.

Critical value: F0.05 (2,9) = 4.26 (Deg. of free. from 'between brands' and 'errors'.)

Test Statistic: 3.72

Conclusion: T.S. < C.V. so H0 not rejected. There is no difference between the fat content of the

brands.

b) Two-way ANOVA

Sums of squares

From (a): Total SS: nsn2 = 3.58 Between Brands SS: nsn

2 = 1.62

Between Labs Sum of Squares: Mean scores Ax = 16.33, Bx = 16.50, Cx

= 15.83, Dx = 16.13

Each of these means came from 3 values so inputting the means with a frequency of 3 gives: n = 12, x

= 16.20, sn = 0.249 nsn2 = 0.75 (n and x for checking)

Error SS: 3.58 (1.62 + 0.75) = 1.21

Anova table In this example: (k =3 brands, N =12 values)

Source S.S. d.f. M.S.S. F


Page 16

Between

brands

1.62 3 - 1 = 2 1.62/2 = 0.81 0.81/0.20 = 4.05

Between

labs

0.75 4 1 = 3 0.75/3 = 0.25 0.25/0.20 = 1.25

Errors 1.21 11 - 5 = 6 1.21/6 = 0.20

Total 3.58 12 - 1 = 11

Hypothesis test for Brands

H0: 1 = 2 = 3 H1: At least two of them are different.



Conclusion: T.S. < C.V. so H0 not rejected. There is no difference between the fat content of the

brands. Blocking has not changed to conclusion even though the test statistic has increased.

c) Hypothesis test for Laboratories

H0: A = 2 = C = D H1: At least two of them are different.



Conclusion: T.S. < C.V. so H0 not rejected. The results between the different laboratories are

consistent.

PROBLEM 2

The following data represent the number of units of production per day turned out by 5 different workers using 4 different types of machines

MACHINE TYPE W O R K E R S

A B C D 1 44 38 47 36

2 46 40 52 43

3 34 36 44 32 4 43 38 46 33 5 38 42 49 39

a) Test whether the five men differ with respect to mean productivity. b) Test whether the mea productivity is same for four different machine types. Take = 5%


Page 17

Solution We shift the origin to 40 and subtract 40 from the given values and work out with new values of xij.

MACHINE TYPE Ti r

Ti2

j

ijx2

W O R K E R S

A B C D

1 4 -2 7 -4 5 6.25 85 2 6 0 12 3 21 110.5 189 3 -6 -4 4 -8 -14 49.0 132 4 3 -2 6 -7 0 0 98 5 -2 2 9 -1 16 16 90

Ti 5 -6 38 -17 T = 20

r

Ti2

=181.1 594

c

Ti2

5 7.2 288.8 139

c

Ti2

=358.8

i

ijx2

101 28 326 139 594

=i

RiR

cr

Tc

TSS22

181.5 20 = 161. 5

=i

RiC

cr

Tr

TSS22

358.5 20 = 338. 8

SSE = SST SSR SSC 574 (161.5 + 338.8)= 73.7

Source of

Variation

S.S. d.o.f. M.S.S F

Between row

Workers

161.5 c- 1 = 4 40.375 40.374/6.142 = 6.57

Between Columns

(Machines)

338.8 r 1 = 3 11.933 112.933/6.142 = 18.39

Errors 73.7 12 6.142 -

Total 574 19 - -

F0.05,(4, 12) = 3.26 F0.05,(3, 12) = 3.49

F > F0.05,(4, 12) with respect to rows, hence 5 workers differ significantly.

F > F0.05,(3, 12) with respect to columns, hence 4 machine types also differ significantly in mean productivity.


Page 18

An experiment was designed to study the performance of four

different detergents for cleaning injectors. The following

cleanliness readings were obtained with specially designed

equipment for 12 tanks of gas distributed three different models

of engines.

Obtain appropriate ANOVA table and test at 1% LOS whether

there are differences in the detergents on the engines.

Solution :

We choose 45 as origin.

Now,

Engine 1 Engine 2 Engine 3

Detergent A 45 43 51

Detergent B 47 46 52

Detergent C 48 50 55

Detergent D 42 37 49

Engine

1

Engine

2

Engine

3

Ti *+, - .+/

Detergent A 0 -2 6 4 5.33 40

Detergent B 2 1 7 10 33.33 54

Detergent C 3 5 10 18 108 134

Detergent D -3 -8 4 -7 16.33 89

Tj 2 -4 27 T=25 162.99 317 */0

1 2 182.25 185.25

- .+/ 22 94 201 317


Page 19

SST = .+//+ xij2

12 = 317 52.08

= 264.92

SSR = 32+ 12

= 162.99 52.08

= 110.91

SSC = 2/12/ 12

= 185.25 52.08

= 138.17

SSE = SST- SSR - SSC

= 264.92 110.91 133.17

= 20.84

ANOVA Table :

Source of

Variance

Sum of

sequence

d.o.f. Mean of

square

F ratio

Between

rows

110.91 3 36.97 FR=456457

= 10.65

FC = 458457

= 19.19

Between

Columns

133.17 2 66.58

Error 208.84 6 3.47

Total 264.92

9%(?,A) = 9.78 9%(,A) = 10.92 Now,


Page 20

Let H0 : No difference in the detergent on the engines

Let H1 : There is difference in the detergent on the engine

But, FR>9%(?,A) H0 is rejected at 1% LOS.

4) An industrial engineer is conducting an experiment on eye focus time. He is interested in the effect of the distance of the object from the eye focus time. Four different distances are of interest. He has five subjects available for the experiment. Because there may be differences among individuals, he decides to conduct the experiment in a randomized block design. The data obtained follow

subjects Distances(ft) 1 2 3 4 5 4 10 6 6 6 6 6 7 6 6 1 6 8 5 3 3 2 5 10 6 4 4 2 3 Can we say distance affects the eye focus time@ 5% L.o.s.

SOLUTION: subjects Distances(ft) 1 2 3 4 5 Ti

-

4 10 6 6 6 6 34 231.2 244 6 7 6 6 1 6 26 135.2 158 8 6 3 3 2 5 18 64.8 72 10 5 4 4 2 3 19 72.2 81 Tj 28 19 19 11 20 =97 503.4

196 90.25 90.25 30.25 100 506.75

- 210 97 97 45 106 555

SST=

=555470.45 SST=84.55

SSR= "

=503.4470.45


Page 21

SSR=32.95

SSC=

=506.75470.45 SSC=36.3

SSE=SSTSSRSSC

=84.5532.9536.3 =15.3 ANOVA TABLE: Source of variance

Sum of squares

Degrees of freedom

Mean square F ratio

Between rows 32.95 3 10.98 Fr=8.612 Between columns

36.3 4 9.075 FC=7.118 Error 15.3 12 1.275

F0.05,(3,12)=8.74 F0.05,(4,12)=5.91 FrF0.05,(4,12) Reject H0.

5) Prior to submitting a quotation for a construction project, companies prepare a detailed analysis of

the estimated labour and materials costs required to complete the project. A company which employs

three projects cost assessors, wished to compare the mean values of these assessors cost estimate.

This was done by requiring each assessor to estimate independently the costs of the same four

construction projects. These costs, in 0000s, are shown in the next column.

Assessors

A B C

Project 1 46 49 44

Project 2 62 63 59

Project 3 50 54 54

Project 4 66 68 63

solution :

Ho

i)There is no siginificant difference between the assessors mean cost estimates.

II) There is no siginificant difference between the project

h = 4 ; k=3.

set orgin as 50

N=hk=12


Page 22

Project A B C Ti* Ti*2

1 -4 -1 -6 -11 121 (i=1) 53

2 12 13 9 34 1156 (i=2) 394

3 0 4 4 8 64 (1=3)23

4 16 18 13 47 2209 (1=4)749

T*j 24 34 20 T=78

T*J2 576 1156 400

2132

2

; = 721 ; =18.667

=676.33

=26.

ANOVA TABLE SV SS def MS F ratio

Between rows

h-1=2

Between columns

k-1=3

Errors

(h-1)(k-1)=6

Total V =721 hk-1=11


Page 23

At 5% loss for degree of factor (2,6) is (2,6) = 5.14 At 5% loss for degree of factor (3,6) is (3,6) = 4.76 i) calculated > table value 108.7>5.14 so H0 is rejected . so there is significant difference between the project. ii) calculate < table 2.7845 < 4.76 so H0 is accepted. so There is no significant difference between the assessors mean cost estimates.

LATIN SQUARE DESIGN

A n x n LATIN Square is a square array of n distinct letters, with each appearing once and only once in each row and in column Example: A B C D B C D A C D A B D A B C

NOTATIONS:

Number of levels of row factor n

Number of levels of column factor n


Page 24

Number of levels of treatment factor k

Sum of c observations in 24t h row =j

ijRi xT

Sum of r observations in j th column =i

ijCj xT

Sum of k observations in k th teatment =k

ijK xT

Sum of all r x c observations ===i j i j

CjRi TTxijT

Computational Formulae

Total Sum of Squares

=i j

ijTn

TxSS 2

22

Between Rows Sum of Squares =

i

RiR

n

Tn

TSS 222

Between Columns Sum of Squares =

i

RiC

n

Tn

TSS 222

Between treatment sum of squares

=i

KTk

n

Tn

TSS 222

Error(residual) Sum of Squares SSE = SST SSR SSC - SSTk

ANOVA TABLE


Between rows SSR n - 1 MSR = SSR/(n-1)

E

R

MSMS

E

C

MSMS

Between Columns SSC n - 1 MSC = SSC/(n-1)

Between

Treatments

SSE

n - 1

MSE = SSTk/(n-1)

Error(residual)

SSE

(n 1) x (n 2)

MSE = SSE/(n-1) E

Tk

MSMS

Total SST n2 -1

FR, FC, FTk follows (n-1), (n-1)(n-2) d.o.f


Page 25

1) The following data resulted from an experiment to compare three burners B1,B2, and B3,.A Latin square design was resulted was used as the tests were made on three engines and were spreadover three days. Engine 1 Engine 2 Engine 3

DAY 1 B1 16 B2 17 B3 20

DAY 2 B2 16 B3 21 B1 15

DAY 3 B 3 15 B 1 12 B 2 13

Test the hypothesis that there is no difference between the burners at 5% LOS yields.

HO: There is no difference between the burners B1=B2=B3

H1: There is difference between the burners B1B2B3

Engine 1 Engine 2 Engine 3 Ti +

.CD DAY 1 B1 16 B2 17 B3 20 53 936.3 945

DAY 2 B2 16 B3 21 B1 15 52 901.3 922

DAY 3 B 3 15 B 1 12 B 2 13 40 533.3 538

Tj 47 50 48 Tij=145 + =2370

/

736.3 833.3 768

/ =2337

.CD 737 874 794 .CD/+ =2405

SST= .CD /+ -

=2405-2336.1 =68.88

SSR= + -

=2370.9 - 2336.1

=34.8

SSC=/

-

=1.5

Rearrange the data according to treatment

Engine 1 Engine 2 Engine 3 Ti +

.CD DAY 1 B1 16 B1 15 B1 12 43 616.3 625

DAY 2 B2 17 B2 16 B2 13 46 705.3 714

DAY 3 B 3 20 B 3 21 B 3 15 56 1045.3 1066


Page 26

Tj 53 52 40 Tij=145 + =2366

/

936.3 901.3 533.3

/ =2370

.CD 945 922 538 .CD/+ =2405

*E= E

-

=2366.9-2336.1 =30.8 SSE=SST-SSR-SSC-*E =68.88-34.8-1.5-30.8 =1.78 ANOVA TABLE; SOURCE OF VARIATION

SOURCE OF VARIATION

D.O.F MEAN SQUARE F RATIO

BETWEEN ROWS

SSR=34.8 (n-1)=2 MSR= 55F(G)=17.4 9H=45F45I=19.5

BETWEEN COLOUMNS

SSC=1.5 (n-1)=2 MSC= 55J(G)=0.75 9J=45J45I=0.84

BETWEEN TREATMENT

*E=30.8 (n-1)=2 K*E= 55L(G)=15.4 9E=45L45I =17.3

ERROR SSE=1.78 (n-1)(n-2)=2 MSE= 55I(G)(G)=0.89

9(M.MN)(,) =19.0 9E =17.3 9E > 9(M.MN)(,) ACCEPT H0 RESULT; There is no different between the burners at 5%LOS.

2)An oil company tested four different blends of gas online for fuel efficiency according to a latin square design in order to control for the variability of four different drivers and four different models of cars. Fuel efficiency was measured in miles per gallon (mpg) after driving cars over a standard course. Fuel efficiencies (mpg) for 4 blends of gas online (latin square design: blends indicated by letters A-D)

Car models

Drivers I II III IV 1 D 15.5 B 33.9 C 13.2 A 29.1 2 B 16.3 C 26.6 A 19.4 D 22.8 3 C 10.3 A 31.1 D 17.1 B 30.3 4 A 14.7 D 34.0 B 19.7 C 21.6

Analsyse the data and draw you conclusion. Solution:


Page 27

XPQ = .+/-20 n= 4 N=n=16

I II III IV Ti Ti Xij 1 -4.5 13.9 -6.8 9.1 11.7 136.89 342.5 2 -3.7 6.6 -0.6 2.8 5.1 26.01 65.45 3 -9.2 11.1 -2.9 10.3 9.3 86.49 322.3 4 -5.3 14.0 -0.3 1.6 10 100 226.7 Tj -22.7 45.6 -10.6 23.8 T=36.1 349.39 957.0 Tj 515.29 2079.36 112.36 566.44 3273.45

We will rearrange the measurements according to letters Total variation,

V = - - .+/+Q

TN

= 957.05 (?A.)A V= 875.6 VF= T+

V

= W (349.39) (?A.)

A VF =5.89 VJ > TQ

V

= W (3273.45) (?A.)

A VJ =736.9 VX> TY

V

= W (761.73) (?A.)

A VX=108.98 VI=Z ZX ZJ ZF =875.6 108.98736.915.89 = 23.82 ANOVA TABLE:

Tk Tk A 9.1 -0.6 11.1 -5.3 14.3 204.47 B 13.9 -3.7 10.3 -0.3 20.2 408.04 C -6.8 6.6 -9.2 1.6 -7.8 60.84 D -4.5 2.8 -2.9 14 9.4 88.36

T=36.1 T=761.73


Page 28

SV SS DOF MS F

Bet`s rows

ZF>5.89 n-1=3 [6G=1.96 9F>\6

]^=\_

(]^=)(]^) =1.836(reciprocal)

Bet`s columns

ZJ>736.9 n-1=3 [`G=245.63 9F>\8

]^=\_

(]^=)(]^) =68.23

Bet`s letters

ZX>108.9 n-1=3 [aG=36..32 9F>\a

]^=\_

(]^=)(]^) =10.08

Residual error

ZI>23.82 (n-1) (n-2)=6

[_(G)(G)=3.60

total Z>875.6 n-1=15

At 5% loss for D.O.F (Z,Z) (i,e) (3,6) is Fc=4.76, Fc % (3,6) = 4.76 Fd=1.836 HM is accepted. (i,e) There is no significant difference in fuel efficiency between rows. Fg=68.23 Fc=4.76 HM is rejected. (i,e) There is some significant difference in fuel efficiency between columns. Fh=10.08 Fc=4.76 HM is rejected. (i,e) There is some significant difference in fuel efficiency between blends of gas online.

3) The numbers of wireworms counted in the plots of Latin square following soil fumigation (L, M,NO,P)in the previous year were COLUMNS

ROWS

P(4) O(2) N(5) L(1) M(3)

M(5) L(1) O(6) N(5) P(3)

O(4) M(8) L(1) P(5) N(4)

N(12) P(7) M(7) O(10) L(5)

L(5) N(4) P(3) M(6) O(9)

Xij=Xij-1 n=5 N=25

COLUMNS

1 2 3 4 5 Ti Ti


Page 29

ROWS

1 3 1 4 0 2 10 100 30

2 4 0 5 4 2 15 225 61

3 3 7 0 4 3 17 289 83

4 11 6 6 9 4 36 1296 290

5 4 3 2 5 8 22 484 118

Tj 25 17 17 22 19 T=100 = (%

- - xijij

= #!

Tj 625 289 289 484 361 - = !

-

171 95 81 138 97 xij=582

We will rearrange the measurements accurate to letters i i ij L 0 0 0 4 4 8 64 32

M 2 4 7 6 5 24 576 130

N 4 4 3 11 3 25 625 171

O 1 5 3 9 8 26 676 180

P 3 2 4 6 2 17 289 69

Txij 10 15 17 36 22 T = 100 ij=2230 582

- 100 225 289 1296 484 2394

Total varience V= .CD+/ - j =582-

# =182

k" = l j =

# 2394-

# = 78.8

k= l -j =

# 2048 -

# =9.6

km> l n -j =

# 2230 -

# =46

k& = 182-78.8-9.6-46 =47.6

ANOVA TABLE H0 is accepted. Source if

variation

Sum of square Degree of

freedom

Mean square F ratio


Page 30

Between rows

k"=78.8 n-1=4 k"lG =19.7 "=k" lG

k& (lG )(lG) =4.97

Between columns

k=29.6 n-1=4 klG =7.4 =k lG

k& (lG )(lG) =1.86

Between letters

km=46 n-1=4 kmlG =11.5 m=km lG

k& (lG )(lG) =2.90

Residual error k&=47.6 (n-1)(n-2)=12 k&(lG )(lG)=3.96

Total V=182 n=24

At 5% los for dof, F=5.91 "=4.97 H0 is accepted .there is significance difference in soil fumigations between rows. F=5.91 =1.86 H0 is accepted .there is significance difference in soil fumigations between columns. F=5.91 m>2.90 There is some significant different between soil fumigations letters


Page 31

Date post:	04-Mar-2016
Category:	Documents
Upload:	vignanaraj
View:	46 times
Download:	4 times

LECTURE Notes on Design of Experiments

Documents