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Lecture Note

Experimental design for Forestry (SFM 507)(B.Sc. 3rd/1st Semester)

Y.P. TimilsinaAssociate Professor

Institute of Forestry, Pokhara2069

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Sub: Experimental design for Forestry (SFM 507)Level: B.Sc. 3rd/1st semester

Unit 1: Review of basic statistical concepts1.1 Dot notationDot is a subscript on a variable which sums the variable over the entire range.For eg If the sample data are yi, where,i=1,2,3………..tThen by using dot notation,

tTotal of yi=T.= ∑yi

i=1If sample data are yij, where i=1,2,3,…..t

j=1,2,3,…..rThen, rTotal of yij=Ti. = ∑yij where i=1,2,3,…..t

j=1Similarly, tTotal of yij=T.j = ∑yij where j=1,2,3,….r.

i=1 r tGrand total G=∑∑T..= ∑Ti.= ∑T.j=∑ ∑yij(Note: Total no of observations =n= no. of rows x no. of columns= r x t, for equalreplication and n=∑ri for unequal replication)We can use this dot notation for yijk.

Using dot notation in two way table (For equal replication)

Columns j

Rows i

1, 2, 3, 4,…………r Row total Ti.

Row mean yi.

1

2

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t

y11 y12 y13 ……….y1r

y21 y22 y23……… y2r

yt1 yt2 yt3 …………. ytr

T1.

T2.

Tt.

y1

y2.

yt.Columntotal T.j

T.1 T.2 T.r G=∑Ti.= ∑T.j=∑ ∑ yij

ColumnMean y.j y.1 y.2 y.r

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From above table,For totals For means,

rr y1.=1/r∑y1j

T1.=∑y1j j=1j=1r y2.=1/r∑y2j

T2.=∑y2j. j=1 .

. .

.

. .Tt.=∑ytj yt.=1/r∑ytj

j=1Generalization, Generalization,

rTi.=∑yij, Where,i=1,2,3,….t yi.=1/r∑yij where, i=1,2,3,….t

j=1Similarly,

t Similarly,T.j= ∑yij where j=1,2,3,…..r y.j=1/t∑ yij where j=1,2,3,…..r

i=1 _ _Overall mean, y..=1/t∑yi.=1/r∑y.j=1/tr ∑∑yij (for equal replication)

1 ∑∑yij (for unequal replicationAnd y..= ∑ri

Statistical model for ANOVAIn general, there are three types of model in ANOVA,1. Fixed effect model (Model I)2. Random effect model (Model II)3. Mixed effect model (Model III)

1.Fixed effect modelIn this model the investigator or researcher is concerned to draw inferences about ‘t”treatments involved in the experiment or it is the appropriate model to use if the interestof the researcher, inferencewise, is in the’t ‘treatments only.Let us consider the model for one way ANOVA,yij==µ+ αi +eij ………..(1) i=1,2,3,….t

j=1, 2,3,….rThe model (1) is considered as a fixed effect model such that if αi are considered to be aset of t fixed quantities.In case, the main interest lies only in estimating the effect of ttreatments included in the experiment, the above model is a fixed effect model.For e.g. If we run an experiment with the objective of comparing the yield performanceof three varieties of wheat and wish to make inferences on these three varities only, thenwe will used the fixed effect model.

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1. Random effect model (Model II)This model is suitable for an experiment in which a researcher is interested to drawinferences about population of treatments of which only a random sample of ttreatments is selected.For one way classified data, yij==µ+ αi +eij ………..(1) i=1,2,3,….t

j=1, 2,3,….rHere, αi is a random variable such that

αi ~ i.i.d. N(0, σα2 ) and eij~ i.i.d. N(0, σe2 )For eg, if a breeder is interested in assessing the performance of certain varieties,then these varieties have to be taken as a random sample as they have come froma population of varieties. The random effects can belong to finite or infinitepopulation. In this model the main emphasis is on estimating and testing thevariability among the effects of different treatments. This model is suitable forCRD and can be extended to other design.The random effect model is often appropriate for data arising from survey type

investigations where a number of area or units are sampled from which inference is madeconcerning the entire population of area are units.

3. Mixed effect model (Model III)If both the fixed and random effect model is involved, it will be mixed effect

model. In this model, at least one αi is random variable and at least one βj is otherconstant (fixed). In factorial experimental combination, mixed effect model isused.yij==µ+ αi+βj+ (αβ)ij+ eijWhere (αβ)ij= interaction effectThis is mixed if the factor α is fixed and β is random and vice versa.Note: The over all mean µ and eij donot play any role whether the model ismixed, fixed and random.

In short form:Fixed, Random and Mixed ModelsThe methods describe the situation in which the treatments (that is, factor levels) are theonly treatments of interest. This is called the fixed model case.Now, assume the treatments (or factor levels) are randomly selected from a population oftreatments (or factor level) and we are interested in making statistical inferences aboutthe population of treatments. Then this is a random model case.A mixed model case is one with fixed factor(s) and random factor(s).

REVIEW

Hypothesis testing

The science of statistics is broadly studied under Descriptive statistics Inductive/inference statisticsThe decision problem in inductive/ inference statistics is again divided into two parts Estimation

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i.Point estimationii.Interval estimation

Hypothesis testingi. Parametric testii. Non parametric test

Parametric testsUsually assume certain properties of the parent population(s) from which we drawsample (s) i.e. i. Parent population from which observations are taken is normal(Observations

come from the normal population(s) ii.Sample size large or small iii. Population variances known or unknown iv.Use measurement data scale ( interval and ratio scale)

Commonly used parametric tests are z test, t test and F-test (ANOVANon- parametric tests Tests that do not make assumptions about the population distribution are referred

to as non-parametric- tests. Generally used categorical data scale (ordinal and nominal) with frequencies. All commonly used nonparametric tests rank the outcome variable from low to

high and then analyze the ranks. These tests are chi-square, Wilcoxon, Mann-Whitney , Kruskal-Wallis,Fried-mann tests. These tests are also calleddistribution-free tests.

Hypothesis in StatisticsHypothesis: Some statements or assumptions made about the parent population based onsample observations.

• A test of statistical hypothesis is a two action decision problem after theexperimental sample values have been obtained,the two action being theacceptance or rejection of the hypothesis under consideration.

Examples of hypothesis• The fruiting yield obtained by two methods (seed and branch cutting) in Jatropha

is same.• Participation of male and female in forest management activities is same.• The performance of private school is better than that of Government school.• Rich participation is higher in meeting when operational plan was finalized.

Hypothesis example Dependence on environmental income decreases with increasing absolute total

household income. Growth performance/Species richness of MAPs is associated with crown

cover/altitude/aspect factors. Soil carbon depends on depth of soil layers. Basal area increment varies with respect to applied silvicultural treatments.

Terms used in testing of hypothesisNull Hypothesis (Ho): Hypothesis of no difference

H0 : 1= 2,i.e. There is no significance difference between two population means. Alternative Hypothesis (H1): hypothesis of differences

3 cases arises

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i. 1 2 (two tailed)i.e. There is significance difference between two population means.ii. 1 < 2 (left tailed)

i.e. Population mean of 1st is less than population of second.iii. 1 > 2 (right tailed)

i.e. Population mean of 1st is greater than population of second.Research hypothesis Suppose a forest manager suspects a decline in the productivity of forest

plantations of a particular species in a management unit due to continued croppingwith that species.

Set the statistical hypothesisHo: µ= µoH1: : µ< µo (left tailed)Type I and Type II error

Decision from sample

Reject Ho Accept Ho

Truestate

Ho True Wrong (Type I error) Correct

Ho False (H1true)

Correct Wrong (Type II error)

Type I error = Reject H0 when H0 is true Type II error = Accept H0 when H0 is false. P (Reject H0 / H0 true)= Probability of type I error = =Level of significance. it

is called the size of the rejection (critical) region It is generally 1%, 5% & 10%. Itis producer’s risk.

Probability of II error = = P (accept H0 / H0 false). It is consumer’s risk Rejecting a null-hypothesis when it should not have been rejected creates a type I

error. Failing to reject a null-hypothesis when it should have been rejected creates a

type II error. Level of significance, =5 % means that 5 cases out of 100 cases has the

probabilty of reject Ho when Ho is true or 95% confidence that our decision iscorrect.

Degree of freedom (df)

The number of values of a variable that can be chosen freely and given byd.f.()=Total numbers of values of a variable -no. of constraints (known values) =n-k

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Probability and Hypothesis Testing

Steps in hypothesis testing

1. State the null hypothesis.2. Decide whether to test it as a one- or two-tailed hypothesis. If there is

no research evidence on the issue, select a two-tailed hypothesis. Thiswill allow you to reject the null hypothesis in favor of an alternativehypothesis. If there is research evidence on the issue, select aone-tailed hypothesis. This will allow you to reject the nullhypothesis in favor of a directional hypothesis.

3. Set the probability level (α level). Justify your choice.4. Select the appropriate statistical test(s) for the data.5. Collect the data and apply the statistical test(s).6. Report the test results and interpret them correctly.

Testing of hypothesis for large samples

Z testAssumptions

All the sample(s) are taken from the normal parent population(s). Sample size is large i.e. n >30 . Generally population variance 2 is known if unknown, we use sampling variance

s2

Random sampling is assumed.

Application of Z-test• Test for significance of single sample mean, population variance being known or

unknown (z-test for single mean)• Test for significance of difference between two independent sample means( z-test

for double means)

Testing of hypothesis for small samples

t-test:Assumptions

i. Population(s) from which sample(s) is (are) drawn is (are) normal i.e. the parentpopulation(s) is (are) normally distributed.

ii. Samples are regarded as small if n≤30 . iii. Population variance 2 is unknown and it is estimated and replaced by

sample variance s2 such that sampling variance is unbiased estimate of populationvariance.

Random sampling method is used.

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Note: For the large values of n, t-test follows Z- test.

Application of t-test Test for significance of single sample mean, population variance being unknown

(t-test for single mean) Test for significance of difference between two independent sample means (t-test

for two independent sample means) Test for the significance difference between dependent sample means (Paired t-

test) Test for the significance of an observed sample correlation coefficient (t-test for

correlation coefficient)Test for significance of sample regression coefficient (t-test for regressioncoefficient)

F-test F-test is applied in testing

The equality of population variances The equality of several population means (ANOVA)

UNIT TWO::ANALYSIS OF VARIANCE (ANOVA)Analysis of variance, which is the powerful statistical tool for tests of significance. The t-test is used only for testing the significance of the difference between two sample meansbut ANOVA (F-Test) is used to test the homogeneity (equality) of 3 or more than 3sample means or to test that all the samples have been come from the same normalpopulation.

Variation is inherent in nature. The total variation in any set of numerical data is dueto the no.of causes which may be classified as:i. Assignable causes: It is fixed causes and can be detected, measured and control byhuman endeavour.ii. Chance causes: It is beyond the control of human hand and cannot be measuredseparately.Assumptions of ANOVAFor the validity of F-test in ANOVA, following assumptions are made:i. The observations are independent.ii. Parent population from which observations are taken is normal.iii. Various treatments and environmental effects are additive in nature.

Analysis of variance (ANOVA) by one way classification

In one way classification, the effect of only one factor is taken into consideration. Fore.g., the effect of the use of different types of fertilizers or same type with different

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quantity may be considered on several pieces of land (different experimental plots). Here,the different types of fertilizers (treatments) are represented by rows and the yields ofsample plots under each type of fertilizers are placed in the respective columns.For equal replicationLet us consider a set of n=rt observations yij where i=1,2,3,…..t and j=1,2,3,……r

classified as t treatments (rows) in a random manner and re- arranged as given below:

Rows i(treatments)

Row totalTi. Row mean yi.

1

2

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t

y11 y12 y13 ……….y1r

y21 y22 y23……… y2r

yt1 yt2 yt3 …………. ytr

T1.

T2.

Tt.

y1.

y2.

yt.Classified above table such that all yij’s are different and variability may be considered

as due toi. Variability due to treatments (between treatments) : Variability due to treatments are

assignable (fixed) causes and can be detected and control by human endeavor.ii. Variability due to error (within treatments): The variability within the classes or

inherent variation of the random variable within the observations of a class which ischance causes and beyond the control of human hand.Mathematical modelThe linear model is,

yij=µi.+eijyij=µ+(µi.-µ) +eij

Where αi=(µi.-µ)yij==µ+ αi +eij ………..(1) i=1,2,3,….t

j=1,2,3,….rWhere yij=(i,j)th cell observations=Yield or response from the jth unit receiving the ith

treatmentsµ= General mean = overall population mean(i.e. If there is no treatment differences and no chance causes, the yield or response ofeach plot will be µ.)µi.= Fixed mean due to ith treatmentαi=(µi.-µ)=Effect due to ith treatment (row)(ith treatment increases or decreases the yield by an amount αi)eij= error effect due to chance= random error attached to (i,j)th cell observationsn=rt=Total no. of experimental units (for equal replication)

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=∑ri (for unequal replication)Assumptions in the model

All the observations are independent. Different effects are additive in nature. eij are identically & independently distributed N( 0, σe

2).HypothesisHo: There is no significant different in the average yields due to treatments or treatments

effects are same.H0 : 1. = 2. = 3. = ………………….= t. = i.e. α1 = α 2 = α 3 =………………….= α t = 0H1: There is significant different in the average yields due to treatments.H1 : 1. 2. 3. ………………… t.

i.e. α 1 α 2 …………………. α t

Statistical analysis of the ModelIn the case of one way classified data, we estimate the parameters , αi and eij by usingprinciple of least square such as eij (residual) is minimum from equation (1) of above,Then, estimated values are^= y..^

αi =yi.-y.. ( as αi=i.-) and ( yi.= mean of the ith treatment)

^eij=yij-yi.= Random deviation from the mean of the ith treatmentThen the mathematical model becomes,

^ ^ ^yij=+ αi + eij

yij=y..+yi.-y..+yij-yi.

yij- y..=(yi.-y..)+(yij-yi.)

Squaring on both sides and summing over i and j, we get,

∑∑(yij- y..)2= ∑∑[(yi.-y..)+(yij-yi.)] 2

_==∑∑(yi.-y..) 2+∑∑ (yij-yi.) 2+2∑∑[(yi.-y..) (yij-yi.)]

==∑∑(yi.-y..) 2+∑∑ (yij-yi.) 2+2[∑{(yi.-y..)∑(yij-yi.)}]i j

where, ∑(yij-yi.)=0 due to the first property of A.M. that the sum of deviations of a set ofvalues from their A.M. is always zero so that last term=0

_ _ _∑∑(yij- y..)2 =∑∑ (yi.-y..)2+∑∑ (yij-yi.) 2

Total sum of squares= Sum of squares due to treatments+ Sum of squares due to errorTSS=SST+SSE

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Computational or practical used short cut formulas are as follows:TSS=Raw SS-C.F. =∑∑yij2-C.F. Where, C.F.=G2/n=G2/rt=G2/∑riSST=∑Ti. 2/r-C.F. (For equal replication)

=∑Ti2./ri-C.F. (For unequal replication)SSE=TSS-SST

The above statistical analysis and other steps are presented in the following table knownas ANOVA table.

ONE WAY ANOVA TABLESource ofvariation

d.f. SS MSS=SS/d.f Fcal

Treatments

Error

t-1

t(r-1)=(n-t)

SST

SSE

MST=SST/t-1

MSE=SSE/t(r-1)

Fcal=MST/MSE

Total n-1 TSSTabulated F- value= F tab= Fα (at t-1 and n-t d.f.)Decision rule: If F cal≤ Ftab, Ho is accepted otherwise rejected.Rejection means treatment effects differ significantly.

Note: A non- significant F in the analysis of variance indicates the failure of theexperiment to detect any differences among treatments. It doesn’t, in any way, prove thatall treatments are the same, because the failure to detect treatment differences based onnon- significant F- test could be the result of either a very small or no difference amongthe treatments or due to large experimental error, or both. Thus, whenever the F test isnon- significant, the researcher should examine the size of the experimental error and thenumerical differences among the treatment means. If both values are large, the trial maybe repeated and efforts made to reduce the experimental error so that the differencesamong treatments, if any, can be detected. On the other hand, if both values are small, thedifferences among treatments are probably too small to be of any economic value and,thus, no additional trials are needed.Example of One way ANOVA2. A set of data involving four tropical feed stuffs A, B, C and D tried on 20 chicks is

given below. All the 20 chicks are treated alike in all respects except the feedingtreatments and each feeding treatment is given to 5 chicks. Weight gain of babychicks fed on different feeding materials composed of tropical feed stuffs:

ObservationsTi. Yi.

A 55 49 42 21 52 219 43.8B 61 112 30 89 63 355 71

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C 42 97 81 95 92 407 81.4D 169 137 169 85 154 714 142.8

Grand total G=1,6951. Give linear model for this experiment and explain what each model component

refers to.Also explain assumptions used in this model.2. Analyse the data to test the mean yield of each treatment differ significantly for=.05

Solution: According to question, it is one way ANOVA so that only one treatment(food stuff) is involved.Therefore, required linear model for this experiment is,

yij==µ+ αi +eij ………..(1) i=1,2,3,….tj=1,2,3,….r

Where yij= (i, j) th cell observations=Yield or response from the jth unit receiving theith foodstuffsµ= General mean = overall population mean(i.e. If there is no treatment differences and no chance causes, the yield or response ofeach plot will be µ.)µi.= Fixed mean due to ith foodstuffαi=(µi.-µ)=Effect due to ith foodstuff(ith treatment increases or decreases the yield by an amount αi)eij= error effect due to chance= random error attached to (i,j)th cell observationsn=rt=Total no. of experimental units (for equal replication)

Assumptions in the model: All the observations yij are independent. Different effects are additive in nature. eij are identically & independently distributed N( 0, σe

2).Hypothesis:H0: 1. = 2. = 3. = 4. i.e. the treatment effects are same. In other words, all the

treatments (A, B, C and D) are alike as regards their effects on increase in weight.H1 : 1. 2.3. 4.

Working formula:TSS= Total sum of squares= Yij2 –C.F

Where C.F=correction factor=G2/nSST=Sum of squares due to treatments=Ti .2/r –C.FSSE= Sum of squares due to error=TSS-SST

Working table:Observations(weight gained) yij

Ti.A 55 49 42 21 52 219B 61 112 30 89 63 355

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C 42 97 81 95 92 407D 169 137 169 85 154 714

Grand total G=1,695

Calculation:First of all, we have to compute C.F., TSS, SST and SSE and prepare ANOVA table.TSS= Total sum of squares= Yij2 –C.F.=552+462+…….+1542= 37,793.75where C.F=correction factor=G2/n=1695x1695/20=143651.25SST=Sum of squares due to treatments=Ti .2/r –C.F.=2192/5+3552/5+4072/5+7142-143651.25=26,234.95SSE= Sum of squares due to error=TSS-SST=11,558.80

One way ANOVA TABLESource ofvariation

d.f. SS MSS=SS/d.f Fcal

Treatments

Error

3

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26234.95

11558.80

8744.98

722.42

Fcal=MST/MSE=12.105

Total 19 37793.75F tab=F(3,16 ) at 5% level of significance=3.26

Decision: Fcal>Ftab, Ho is rejected i.e. treatments A, B, C and D differ significantly. Orall the treatments (A, B, C and D) are not alike as regards their effects on increase inweight.

TWO WAY ANOVALet us consider the case when there are two factors which may affect the variate valuesyij e.g. the yield of rice may be affected by difference in fertilizers (treatments or rows)and different types of seed (varieties or columns or blocks or replicates).For this ,let us consider a set of n=rt observations yij where i=1,2,3,……t andj=1,2,3,……r classified into t treatments(rows) each with r varieties(columns) as givenbelow table:

Columns j(varieties)

1, 2, 3, 4,…………r Row total Ti.

Row mean yi.

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Rows i(treatments)

1

2

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t

y11 y12 y13 ……….y1r

y21 y22 y23……… y2r

yt1 yt2 yt3 …………. ytr

T1.

T2.

Tt.

y1.

y2.

yt.Columntotal T.j

T.1 T.2 T.r G=∑ ∑ yij=∑Ti.=∑T.j

ColumnMean y.j y.1 y.2 y.rClassified above table such that all yij’s are different and variability may be considered asdue toi. Variability due to treatments (between treatments or rows)ii.Variability due to varieties (columns)ii. Variability due to error

Mathematical modelIn two way ANOVA, the linear model will be,

yij==µ+ αi +βj+eij ………..(2) i=1,2,3,….tj=1,2,3,….r

Where yij=(i,j)th cell observations=Yield or response of the jth variety(column) by theuse of ith treatment(fertilizer or row)µ= General mean = overall population meanµi.= Fixed mean due to ith treatmentαi=(µi.-µ)=Effect due to ith treatment (row)µ.j= Fixed mean due to jth variety (column)

βj==(µ.j-µ)=Effect due to jth variety (column)eij= Error effect due to chance= random error attached to (i,j)th cell observationsn=rt=Total no. of experimental units

Assumptions in the model: All the observations are independent. Different effects are additive in nature. eij are identically & independently distributed N( 0, σe

2).

Hypothesis:For treatmentsHoT:There is no significant difference in the average yields due to treatments.HoT: 1. = 2. = 3. = ………………….= t. = (say)i.e. α1 = α 2 = α 3 =………………… .= α t = 0

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H1T: There is significant difference in the average yields due to treatments.H1 T : 1. 2. 3. ………………… t.

i.e H1 T: α 1 α 2 …………………. α t

For blocks (varieties)HoB:There is no significant difference in the average yields due to blocks.HoB: .1 = .2= .3 = ………………….= .r = (say)HoB: β1= β2= β3 =………………… = βr =0H1B: There is significant difference in the average yields due to variety (blocks).H1B: .1 ≠ .2≠ .3 ≠………………….≠ .rH1B: β1≠ β2≠ β3 ≠………………… ≠βr

Statistical analysis of the ModelIn the case of two way classified data, we estimate parameters , αi ,βj and eij by usingprinciple of least squares such that eij is minimum from equation (2) of above,Then, estimated values are^= y..^

αi =yi.-y.. (Where αi=i.-) and yi.= sample mean of the ith treatment)^βj =y.j-y.. (Where βj =.j-) and y.j= sample mean of the jth variety)^eij=yij-yi.- y.j+y..Then, the mathematical model becomes,

^ ^ ^ ^yij=+ αi + βj + eij

yij=y..+(yi.-y..)+ (y.j-y..)+ (yij-yi.- y.j+y..)

yij- y..=(yi.-y..)+ (y.j-y..)+ (yij-yi.- y.j+y..)Squaring on both sides and summing over i and j, we get,∑∑(yij- y..)2=∑ ∑ [{(yi.-y..)+ (y.j-y..)+ (yij-yi.- y.j+y..)}]2

∑∑(yij- y..)2=∑ ∑ [{(yi.-y..)2+∑ ∑ (y.j-y..) 2+ ∑ ∑ (yij-yi.- y.j+y..) 2+ALL THEPRODUCT TERMS WILL BE VANISHED DUE TO THE FIRST PROPERTY OFA.M.

Total sum of squares= Sum of squares due to treatments+ Sum of squares due tovarieties(blocks) +Sum of squares due to errorTSS=SST+SSB+SSE

Computational or practical used shortcut formula formulas are as follows:TSS=Raw SS-C.F. =∑∑yij2-C.F. Where, C.F.=G2/n=G2/rtSST=∑Ti. 2/r-C.F.SSB=∑T.j 2/t-C.F.

SSE=TSS-SST-SSBThe above statistical analysis and others steps are presented in the following table known

as ANOVA table.

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TWO WAY ANOVA TABLESourcesofvariation

d.f. SS MSS=SS/d.f Fcal

Treatments

Variety

Error

(t-1)

(r-1)

(r-1) (t-1)

SST

SSB

SSE

MST=SST/t-1

MSB=SSB/r-1

MSE=SSE/(t-1)(r-1)

FT=MST/MSE

FB=MSB/MSE

Total n-1=rt-1 TSSTabulated F- valuesF α(T)= Fα (T)[at t-1 and(r-1) (t-1) d.f.] for treatments

and Fα(B)= Fα (B)[at r-1 and(r-1) (t-1) d.f.] for varietyDecision rule: If F* ≤ Fα(*), Ho* is accepted otherwise rejected. Where*=(T,B)Unit 1 and 2: Practical problems based on one and two way ANOVA

1. (Bio –Physical Analysis)For establishing the rectangular PSP of size (20X25 m) in Banpale Dada of IOF, a

pilot survey was conducted and trees with DBH ≥10 cm were measured on the someplots of different aspects as follows:

DBH (in cm) in Different StrataNorth West South East12 23 12 2312 34 32 3443 23 12 5432 12 34 3223 45 50 6043 65 11 2142 32 12 3431 12 33 3421 23 23 3223 42 28 4534 23 23 1235 54 56 3436 43 43 3237 32 32 4538 21 5339 5540

a. Give linear model for this experiment and explain what each model componentrefers to.Also explain assumptions used in this model.b. Test whether the applied stratification i.e. aspect effects are significant or not at the5 % level of significant.

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2. Average crown cover percentage of different plots measured by crownnometer(crown periscope) with no. of seedlings are given the following table (Vegetationanalysis)

Frequency of seedlings in different crown cover percentage10-20 20-40 40-60 60-80 80-10031 30 12 9 841 40 23 8 950 55 43 10 1031 60 24 23 1120 72 15 12 1210 29 33 23 10a. Give linear model for this experiment and explain what each model component refersto.Also explain assumptions used in this modelb.Test whether the crown cover effects are significant or not (i.e. Mean frequency ofseedling are alike in different crown cover percentages) at 1% and 5 % level ofsignificant. (Note: you can do it for herb, shrub, grass, mushroom etc. at the same time ofobservations).3. Socio –Economic analysis

Following is the CF income (000,Rs/yr) of different households with differentincome groups (household categories) .

CF income(Rs/yr) of the different household categoriesPoor(<RS 50,000 /yr) Middle(Rs.50,000-1,50,000) Rich(>Rs.1,50,000)33 11 103 12 93 10 836 21 2337 22 2134 34 2035 3236a. Give linear model for this experiment and explain what each model component refersto.Also explain assumptions used in this modelb. Test the significance of difference of CF income according to different income groups(which may be other categories caste, education, forest condition etc)(Note: If significant, it cleary indicates the different level of contribution of CF income tothe households of different economic class.)4.. Following table shows the pilot survey records of the distribution of Chiraito (no. ofseedlings/1x1 m.square) in different aspects and attitude in midhill CF Nagi, Myagdi ofNepal by preparing 1x1 m.square Nested plots within 20x25 m .square.Aspects/Altitude(m)

1000-1500 1500-2000 2000-2500 2500-3000

N 12 13 9 51S 17 31 27 48W 19 11 30 37E 29 24 25 40

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23 29 25 41a. Give linear model for this experiment and explain what each model componentrefers to.Also explain assumptions used in this model

b. Test whether the chiraito distribution differs considerably with respect to altitudes andaspects at 5 % and 10% level of significance.

5. In a green house experiment on wheat, 4 fertilizer treatments of the soil and 4 chemicaltreatments of the seed were used. The table gives yield in some suitable unit. Analyse the2 way ANOVA with model and assumptions.

Fertilizer/chemicaltreatment

1 2 3 4

1 21 20 19 172 12 13 13 133 13 14 13 124 12 15 14 12

6.Following table shows the assessment survey records of the distribution ofRubia cordifolia local named Majhitho (no. of seedlings/1x1 m.square) indifferent crown cover and slope in one of the renowned midhill CF Nagi, Myagdidistrict of Nepal by preparing 1x1 m.square Nested plots within 20x25 m.squaresample plot.

Crown cover (%)

Slope(Degree)

10-20 20-30 30-40 40-50

0-20 24 37 40 48

20-40 12 32 41 50

40-60 20 35 46 60

Test whether the average distribution of Rubia cordifolia vary considerably with respectto crown cover as well as slope at 5% level of significance. (1+9)7. . Following table shows laboratory results of quantity of soil carbon stocks(SOC, t/ha) in different soil depths sothat each depth consists of 4 plots inMidhioll CF of Nepal.

Soildepths(cm)

Quantity of Soil carbon stocks (t/hac)

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0-20 51 50 52 4720-40 33 36 37 3540-60 30 35 34 33

(i) Give the linear model, assumptions for this classification.(ii)Test whether mean soil carbon stocks vary noticeably with respect to soildepths at 5 % level of significance.

(iii) Compare the mean effects by using least significance difference test (LSD)procedure. Interpret your result in practical manner in this midhill forest bylinking with soil carbon assessment.

8.A research theme based on distance of forest from users’ households’ as welleconomic status identifies the quantity of fuelwood collection by the Communityforest user groups (CFUGs). Based on this statement, three types of distance usercategory and with four types weath being ranking are used with their averagequantity of fuelwood collected (00, qtl, per hh/yr) from the 12 households aregiven in the following table.

User Category(Distant/wealth)

Quantity of fuelwood collection (00,qtl, per hh/yr)

Ultra poor Poor Middle RichNear user 16 13 11 10

Middle user 15 14 10 9Far user 12 10 9 8

(i) Give the linear model, assumptions for this classification.(ii)Test whether average quantities of fuelwood collection vary considerably withrespect to distant user category as well as wealth being ranking at 10 % level ofsignificance. Draw your conclusion in practical manner and discuss the scenarioof this result with reference to CF of Nepal.

Unit 3: Design of ExperimentsThe subject matter of the design of experiment includes:i. Planning of the experimentii. Obtaining relevant information from it regarding the statistical hypothesis

under studyiii. Making a statistical analysis of the data and draw valid conclusion.Thus, Design of Experiment is defined as the planning an experiment to obtainappropriate data and drawing inference out of the data with respect to any problem

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under investigation. This might range anywhere from the formulation of theobjectives of the experiment in clear terms to the final stage of drafting reportsincorporating the important findings of the enquiry.The design of experiment based on analysis of variance, which is the powerfulstatistical tool for tests of significance. The t- test is used only for testing thesignificance of the difference between two sample means but ANOVA (F-Test) isused to test the homogeneity of 3 or more than 3 sample means or to test that all thesamples have been come from the same normal population.Variation is inherent in nature. The total variation in any set of numerical data is dueto the no.of causes which may be classified as:i. Assignable causes: It is fixed causes and can be detected, measured and control byhuman endeavour.ii. Chance causes: It is beyond the control of human hand and cannot be measuredseparately.Assumptions of ANOVAFor the validity of F-test in ANOVA, following assumptions are made:i. The observations are independentii. Parent population from which observations are taken is normaliii. Various treatments and environmental effects are additive in nature

Terminology in Experimental designs:Experiment: It is a device or means of getting an answer to the problem underconsideration. It is a process of examining the truth of a statistical hypothesis,relating to some research problem, is known as Experiment. For e.g. we can conductan experiment to examine the usefulness of a certain newly developed seed. It isclassified into two categories as follows:

i.. Absolute : Absolute value of some characteristics like average mark, correlationcoefficient etc.ii. Comparative: To compare the effect of two or more objects on some population

characteristics e.g. comparison of different manures or fertilizers, different varietiesof crop etc. It is also called Design of experiment.Treatments: Various objects of comparison in a comparative experiment, are termedas treatments, e.g. in field experimentation different fertilizers or different varietiesof crop or different methods of cultivation are the treatments.Experimental unit (EU): The pre-determined plots or blocks, where differenttreatments are used, or the smallest division of the experimental material to which weapply the treatments and on which we make observations on the variable under studyis called Experimental unit (EU) For e.g. leaf, a tree or a collection of adjacent treesmay be an EU.An experimental unit is also sometimes referred as plot. A collectionof plots is called block. Observations made on experimental units vary considerably.These variations are partly produced by the manipulation of certain variables ofinterest generally called treatments, built-in and manipulated deliberately in theexperiment to study their influences.Experimental error: Besides the variations produced in the observations due to theknown sources, variations are also produced by a large number of unknown sourcessuch as uncontrolled variation in extraneous factors related to the environment,

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genetic variations in the experimental material other than that due to treatments, etc.These variations because of their undesirable influences are called Experimentalerror thereby meaning not an arithmetical error but variations produced by a set ofunknown factors beyond the control of the experimenter.Yield/Response: The measurements of the variable under study on differentexperimental units are termed as yields/responses.Experimental and control group: In an experimental hypothesis testing research,when a group is exposed to usual conditions is known as control group but whenexposed to special conditions is called experimental group.Extraneous variable: It is that independent variable, which is not related to thepurpose of the study but affect the dependent variable and that effect noticed ondependent variable as a result of extraneous variable is technically described asexperimental error.Other terms are level of significance, factor, levels, residual variation etc.

Basic Principles of experimental designReplicationAccording to this principle the experiment should be repeated more than once, thus eachtreatment is applied in many experimental units instead of one. By doing so the statisticalaccuracy of the experiment is increased. In other words when a treatment is applied to anumber of experimental units in the experiment, it is said to be replicated, i.e. repetitionof treatments under investigations. The functions of replications are: To provide an estimate of experimental errors To improve the precision of the experiment by reducing the standard error of the

mean To control the effect of error variance To increase the scope of the inference of the experimentThe no. of replications to be included in any experiment depends upon many factorslike the homogeneity of experimental material, the no. of treatments, the degree ofprecision required etc. As a rough rule, it may be stated that the number ofreplications in a design should provide at least 1o to 15 degrees of freedom forcomputing the experimental error variance.The reciprocal of the variance of the mean is termed as the precision or the amount ofinformation of the design.

Precision=1/var(y)=1 =r/ σ2

σ2/rThis shows the efficiency or precision of the design can be increased by;1. Controlling i.e. decreasing σ2, the error variance per unit. This is done by arrangingthe plots into small homogeneous blocks, and2. Increasing r, the no. of replications

RandomizationIt refers to the manner in which the treatments are assigned to the experimental units. Inother words, this principle indicates that we should design or plan the experiment in such

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a way that the variations caused by extraneous factors can all be combined under thegeneral heading of chance. Randomization in the context of random sampling means,allocation by some chance device of experimental units among treatments such that everytreatment has an equal chance of being assigned to any experimental unit. Randomizationallows one to assume that the experimental error comprises a random sample from somepopulation, which allows us to make valid inferences from them. The main function ofrandomization is to avoid bias in the estimate of experimental error and to ensurecomparisons among treatment means.It is commonly used when experimental units are homogeneous or an experimental areahappens to be homogeneous.Local controlThe process of dividing the relatively heterogeneous experimental area into homogeneousblocks so as to minimize the experimental error is known as local control. It increases theefficiency of the design.Local control includes: Blocking Proper plot technique Data analysisLocal control means the control of all factors except the ones about which we areinvestigating. Local control, like replication is yet another device to reduce or controlthe variation due to extraneous factors and increase the precision of the experiment.If, for instance, an experimental field is heterogeneous with respect to soil fertility,then the field can be divided into smaller blocks such that the plots within each blocktend to be more homogeneous. This kind of homogeneity of plots(experimental units)ensures an unbiased comparison of treatment means, as otherwise it would bedifficult to attribute the mean difference between two treatments solely to differencesbetween treatments when the plot differences also persist. This kind of local controlto achieve homogeneity of experimental units will not only increase the accuracy ofthe experiment, but also help in arriving at valid conclusions.

Formal Experimental designThe type of design to be used depends on the nature of the experimental treatments, sitecharacteristics and blocks sites.

UNIT 4: Completely Randomized Design (CRD)It is the simplest form of formal experimental design. It is also known as non- restrictionand single factor design The essential characteristic of this design is that treatments areassigned completely at random so that each experimental unit has the same chance ofreceiving any one treatment. One way classified data come out from the experimentallayout in a C.R.D. Data obtained from experiments and surveys which are classifiedaccording to only one independent factor are called one way classified data. Here anydifference among the experimental units receiving the same treatment is considered as the“experimental error”. CRD is one in which all the experimental units are taken in a single

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group which is homogeneous as far as possible. For example, the entire field plotsconstituting the group having the same soil fertility, soil depth, soil texture, soil moistureetc. All the cows forming a group are of the same breed, same age, same weight, andsame lactation etc.It is commonly used when experimental units are homogeneous or an experimental areahappens to be homogeneous.It involves only two principles of experimental design, they are replication &randomization.

Randomization & layout Whole experimental material is divided into n number of experimental units.

n = r t, r is number of replication, t is number of treatments for equal replicationn = ri, for unequal replication

Assign the plots from 1 to n Assign the treatment to the experimental plots randomly

Let us take an example of CRD with four treatments A, B, C& D each replicated fivetimes look like this.

A D A BC B C AD A C BC B D CD B D A

Lay out of CRD (with equal replication)

Similarly, four treatments T1, T2, T3& T4 each are replicated 4,3,3,5 times respectivelythen layout of CRD have been 15 plots (units) as shown below:

T2 T2 T2T3 T4 T3T1 T1 T1T4 T4 T1T3 T4 T4

Layout of CRD (Unequal replication)

Mathematical modelIts mathematical model is analogous to the ANOVA of one-way classified data.The linear model is,

yij=µi.+eijyij=µ+(µi.-µ) +eij

Where αi=(µi.-µ)yij==µ+ αi +eij ………..(1) i=1,2,3,….t

j=1,2,3,….r

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Where yij=(i,j)th cell observations=Yield or response from the jth unit receiving the ithtreatmentsµ= general mean = overall population mean(i.e. If there is no treatment differences and no chance causes, the yield or response ofeach plot will be µ.)µi.= Fixed mean due to ith treatmentαi=(µi.-µ)=Effect due to ith treatment (row)(ith treatment increases or decreases the yield by an amount αi)eij= error effect due to chance= random error attached to (i,j)th cell observationsn=rt=Total no. of experimental units (for equal replication)

=∑ri (for unequal replication)

Assumptions All the observations are independent Different effects are additive in nature eij are identically & independently distributed N( 0, σe

2) α i = 0, or αi N (0, σα2)

HypothesisHo:There is no significant different in the average yields due to treatments.H0 : 1. = 2. = 3. = ………………….= t. = i.e. α1 = α 2 = α 3 =………………….= α t = 0H1: There is significant different in the average yields due to treatments.H1 : 1. 2. 3. ………………… t.

i.e. α 1 α 2 …………………. α t

Statistical analysis of the ModelStatistical analysis is analogous to the ANOVA of one-way classified data.

It is the same case of one way classified data, we estimate , αi and eij by using principleof least square such that eij is minimum from equation (1) of above,Then, estimated values are^= y..^

αi =yi.-y.. ( as αi=i.-) and ( yi.= sample mean of the ith treatment)^

eij=yij-yi.

Then the mathematical model becomes,^ ^ ^

yij=+ αi + eij

yij=y..+yi.-y..+yij-yi.

yij- y..= (yi.-y..)+(yij-yi.)

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Squaring on both sides and summing over i and j, we get,

∑∑(yij- y..)2= ∑∑[(yi.-y..)+(yij-yi.)] 2

==∑∑[(yi.-y..) 2+∑∑ (yij-yi.) 2+2∑∑[(yi.-y..) (yij-yi.)]_

==∑∑[(yi.-y..) 2+∑∑ (yij-yi.) 2+2[∑{(yi.-y..)∑(yij-yi.)}]i j

(Where last term=0 due to the first property of A.M. that the sum of deviations of a set ofvalues from their A.M. is always zero.)

_ _∑∑(yij- y..)2 = ∑∑ (yi.-y..)2+∑∑ (yij-yi.) 2

Total sum of squares= Sum of squares due to treatments+ Sum of squares due to errorTSS=SST+SSE

Computational or practical used formulas are as follows:TSS=Raw SS-C.F. =∑∑yij2-C.F. Where, C.F.=G2/n=G2/rt=G2/∑riSST=∑Ti. 2/r-C.F. (For equal replication)

=∑Ti2./ri-C.F. (For unequal replication)SSE=TSS-SST

The above statistical analysis and others steps are presented in the following table knownas ANOVA table.

ANOVA TABLE FOR CRDSource ofvariation

d.f. SS MSS=SS/d.f Fcal

Treatments

Error

t-1

t(r-1)=(n-t)

SST

SSE

MST=SST/t-1

MSE=SSE/t(r-1)

Fcal=MST/MSE

Total n-1 TSSTabulated F- value= F tab= Fα (at t-1 and n-t d.f.)Decision rule: If F cal≤ Ftab, Ho is accepted otherwise rejected.Note: Note that a non- significant F in the analysis of variance indicates the failure of theexperiment to detect any differences among treatments. It doesn’t, in any way, prove thatall treatments are the same, because the failure to detect treatment differences based onnon- significant F- test could be the result of either a very small or no difference amongthe treatments or due to large experimental error, or both. Thus, whenever the F test isnon- significant, the researcher should examine the size of the experimental error and thenumerical differences among the treatment means. If both values are large, the trial maybe repeated and efforts made to reduce the experimental error so that the differencesamong treatments, if any, can be detected. On the other hand, if both values are small, thedifferences among treatments are probably too small to be of any economic value and,thus, no additional trials are needed.

Advantages

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Simple & easy layout Utilization of whole experimental material Complete flexible Simple analysis Missing data create no problem in analysis Specially suitable even some units are destroyed or failed to respond Mostly used in laboratory or green house experiment Gives maximum degrees of freedom for experimental error

Disadvantage Only suitable for small number of treatments Homogeneous experimental units can rarely obtained Less informative for heterogeneous fields Seldom suitable for field experiments All extraneous variations included in the error variation (residual variation)

UNIT 7: Multiple –comparison:When an ANOVA F-test reject Ho shows that there is significance different amongtreatment means but it doesn’t specify which pairs of treatments that differ significantly.To obtain this information, procedure for comparing treatment means like leastsignificant difference (LSD) test and Duncan’s New multiple range test (DNMRT) areneeded for Biological research.Pair comparison is the simplest and most commonly used comparison in research. Thereare two types:.Planned pair comparison, in which the specific pair of treatments to be compared wasidentified before the start of the experiments. A common example is comparison of thecontrol treatment with each of the other treatments..Unplanned pair comparison, in which no specific comparison is chosen in advance.Instead, every possible pair of treatment means is compared to identify pairs oftreatments that are significantly different.

Least significant difference (LSD) test:It is simple and most commonly used procedure for making pair comparisons. Theprocedure provides for a single LSD value, at a prescribed level of significance, whichserves as the boundary between significant and non-significant differences between anypair of treatment means. That is two treatments are declared significantly different at aprescribed level of significance if their difference exceeds the computed LSD value,otherwise not.LSD test is used when i. F-test shows rejection of Ho

ii. Numbers of treatments is not too large i.e. less than 6Formula :(i) For equal replication

LSD=r(2, α ,γ)√MSE/rWhere, r(2, α ,γ)= Special range value of rank of order 2 at α level of significance with

error degree of freedom(d.f) γ ,MSE= Mean sum of squares due to errorr=Replication of each treatment

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ii. For unequal replication ( For CRD)LSD=r(2, α ,γ)√MSE(1/ri+1/ri’)

√2Where ri and ri’ denote the replication of ith and i’th treatments

_ _Decision: If the observed value (Yi-Yi’) is greater than computed LSD, Ho is rejected i.e.related treatment pairs are judged to be significantly different.Duncan’s New Multiple Range Test (DNMRT) :. It is especially true when the total no.of treatments is large i.e >6.Procedure of DNMRT is similar as that of LSD. In LSD, onlysingle LSD value is required for any pair comparison but DNMRT requires a series ofvalues.This test is most widely used when several multiple range tests available. This test usesranges of rank of order 2,3……….t according to the ranking of pairs of means when theyare arranged.

Formula :(i) For equal replicationDNMRT=r(p, α ,γ)√MSE/rWhere, r(p, α ,γ)= Special range value of rank of order p( where p=2.3……t) at α levelof significance with error degree of freedom(d.f) γ ,MSE= Mean sum of squares due to errorr=Replication of each treatmentMSE/r =Syi=Standard error of single treatment mean

ii. For unequal replication ( for CRD)

DNMRT=r(p, α ,γ)√MSE(1/ri+1/ri’)√2

Where ri and rj denote the replication of ith and i’th treatments

Decision: If the observed value (Yi-Yi’) is greater than computed DNMRT, Ho isrejected i.e. related treatment pairs are judged to be significantly different.Example of CRD

3. A set of data involving four tropical feed stuffs A, B, C and D tried on 20 chicksis given below. All the 20 chicks are treated alike in all respects except thefeeding treatments and each feeding treatment is given to 5 chicks. Analyse thedata to test the mean yield of each treatment differ significantly for =.05 usingLSD and DNMRT test.

Weight gain of baby chicks fed on different feeding materials composed of tropical feedstuffs:

B(61) A(52) C(97) A(49) B(89)C(81) C(95) B(63) B(112) B(30)D(169) D(137) C(42) D(169) A(42)

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D(154) A(55) D(85) C(92) A(21)

Solution This is CRD with equal replication. Then, linear Model,yij==µ+ αi +eij ………..(1) i=1,2,3,….t

j=1,2,3,….r

Hypothesis:H0: 1. = 2. = 3. = 4. i.e. the treatment effects are same. In other words, all the

treatments (A, B, C and D) are alike as regards their effects on increase in weight.H1 : 1. 2. 3. 4.Arrangement of data by one way ANOVA

ObservationsTi. Yi.

A 55 49 42 21 52 219 43.8B 61 112 30 89 63 355 71C 42 97 81 95 92 407 81.4D 169 137 169 85 154 714 142.8

Grand total G=1,695

First of all, we have to compute C.F., TSS, SST and SSE and prepare ANOVA table.TSS= Total sums of square= Yij2 –C.F.=37,793.75 where C.F=correction factor=G2/nSST=Sums of square due to treatments=Ti 2/ri –C.F.=26,234.95SSE= Sums of square due to error=TSS-SST=11,558.80

ANOVA TABLE for CRDSource ofvariation

d.f. SS MSS=SS/d.f Fcal

Treatments

Error

3

16

26234.95

11558.80

8744.98

722.42

Fcal=MST/MSE=12.105

Total 19 37793.75F tab=F(3,16 ) at 5% level of significance=3.06Decision: Fcal>Ftab, Ho is rejected i.e. treatments A, B, C and D differ significantly.Or all the treatments (A, B, C and D) are not alike as regards their effects on increasein weight. Since Ho is rejected in this case, we proceed further to find out which of

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the treatments means differ significantly. For this we use LSD and DNMRT asfollows:

The treatments means are arranged in descending order of magnitude as given below:Treatment

Yi. |Yi.-YA| |Yi.-YB| |Yi.-YC|

D 142.8 99.0(p=4) 71.8(p=3) 61.4(p=2)C 81.4 37.6(p=3) 10.4(p=2) -B 71.0 27.2(p=2) - -A 43.8 - - -

Using LSD,For equal replication on CRD

LSD=r(2, α ,γ)√MSE/r =r(2,.05,16) √722.42/5=36.06Where, r(2, α ,γ)= Special range value of rank of order 2 at .05 level of significance with

error degree of freedom(d.f)16 = r(2,.05.16) (see table)=3.00MSE= Mean sum of squares due to errorr=Replication of each treatmentDecision: Thus, by comparing LSD value and above difference of mean table values, thetreatment pairs,(D, A),(C,A),(D,B) and(D,C) are judged to be significantly different.Using DNMRT,

For equal replication on CRDDNMRT test can be used for rank of order (p) of 2,3 and 4.DNMRT=r(p, α ,γ)√MSE/rDNMRT=r(4, .05 ,16)√722.42/5 =3.23 √722.42/5 =38.825DNMRT=r(3, .05 ,16√ 722.42/5 =3.15 √722.42/5 =37.86DNMRT=r(2, .05 ,16)√722.42/5 =3 √722.42/5 =36.06

Decision: Thus, by comparing DNMRT and the above difference of mean values, thetreatment pairs (D,A),(D,B) and (D,C) are significant.Practical problems

1. Mycelial growth in terms of diameter of the colony (mm) of Rizoctonia Solani(RS) on PDA medium after 14 hrs of incubation are given below:

R. Solani isolatesMycelial growth

RS 1 29.0 28.0 29.0RS 2 33.5 31.5 29.0

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RS 3 26.5 30.0 -RS 4 48.5 46.5 49.0RS 5 34.5 31.0 -

Test the significance of % isolates or R.solani w.r.t. mycelial growth on PDAmedium. Also, use LSD and DNMRT test at α = 5% level of significance.(Ans: Fcal=57.38)2. The following table gives the plan and yield of barbley in grams of 5 different

varities. Test the varietal effect using CRD at α=5 % level of significance.

3200(V1) 3400(V4) 3980(V5) 3600(V4) 3500(V3)3720(V2) 4550(V2) 4170(V3) 4200(V1) 3580(V5)4000(V3) 3530(V1) 3340(V5) 3310(V1) 3580(V4)3700(V4) 3400(V5) 3750(V2) 3200(V5) 3250(V3)4300(V5) 3580(V1) 3780(V3) 3950(V4) 3280(V2)

3820(V2) 2750(V3) 3750(V4) 3080(V2) 4000(V1)(For easy calculation you can use change of origin and scale)Which pair of treatment means would be judged significantly difference by LSD andDNMRT test.(use only if Ho is rejected).

3. Complete the following equal replication CRD and test treatments significance at10 % level of significance.

Sources of variation d.f. SS MSTreatments 4 229.25Error - 44356Total 29

4. Four silviculture thinning treatments T1=Control, T2=10% thinning, T3=20% thinningand T4=30% thinning tried on 20 plots of homogeneous forest.. Plot level mean basalareas (cm square) gained after 2 years is measured as follows.

16 ( T3 ) 27 ( T4) 8 ( T1) 9 ( T2) 8(T1)24( T4) 6 ( T1 ) 10 (T2) 20 (T3) 11(T2)4 ( T1 ) 10 ( T2 ) 20 ( T3) 26 (T4) 17(T3)10 (T2 ) 20 ( T3) 25 ( T4 ) 7 (T1) 29(T4)

Test whether there is marked distinct between silvicultural treatments effects at 5 % levelof significance.Also,test LSD for multiple comparison.

5. Three treatments as A=control, B=Compost, C = Chemical fertilizer are applied to seethe growth performance (height increment) of a highly demand Biofuel plant speciesnamed jatropha curcas at initial homogeneous height of 12 branch cutting plants at thenursery of khairenitar Tanahun district. All other conditions are being kept constantexcept applied treatments. After six months, the increased heights ( 0, cm) of that speciesare recorded as below.B(8) C(6) A(3) C(5)

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Test whether there is remarkable difference in the average height increment of thatspecies due to applied treatments at 5% level of significance. Also, compare pair wisedifferent by using LSD test procedure. Interpret your result in statistical manner.

UNIT 5: RANDOMIZED BLOCK DESIGN (RBD) or RCBDThe RBD is one of the most widely used experimental designs in forestry & biologicalresearch. The design is especially suitable for field experiments where the number oftreatments is not large and there exists a remarkable factor based on which homogeneoussets of experimental units can be identified. The primary distinguished feature of theRBD is the presence of block of equal size each of which contains all the treatments.Blocking TechniqueThe purpose of blocking is to reduce the experimental error by eliminating thecontribution of known sources of variation among the experimental units. This is done bygrouping the experimental units into blocks such that variability within a block isminimized & variability among the blocks is maximized. Since, only the variationswithin a block become the part of the experimental error blocking is most effective whenthe experimental area has a predictable pattern of variability.Two important points should be kept in mind while blocking Selection of the source of variability

Heterogeneity i.e. slope.altitude, aspect, soil, forest strata, direction ofinsect migration, crown cover, soil layer, light intensity, land type etc.

Block shape Unidirectional – use long & narrow Two directional – ignore the weaker one Equally strong – square blocks

“When the productivity gradient is expected within the experimental area, block shouldbe laid across the gradient & plots within the blocks be laid parallel to the gradient”.Randomization & layoutThe randomization process for RBD is applied separately & independently to each block.For example let us take a field experiments with 6 treatments A, B, C, D, E & F and 3replications.

Gradient

Layout of RCBDMathematical model & statistical analysisIts mathematical model & statistical analysis is analogous to the ANOVA of two-wayclassified data.The linear model will be,

A(4) B(7) B(6) C(5)A(5) A(3) C(7) B(6)

Block I Block II Block IIIC A FD E DF F CE C AB D BA B E

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yij==µ+ αi +βj+eij ………..(2) i=1,2,3,….tj=1,2,3,….r

Where yij=(i,j)th cell observations=Yield or response of the jth variety(column) by theuse of ith treatment(fertilizer or row)µ= general mean = overall population meanµi.= Fixed mean due to ith treatmentαi=(µi.-µ)=Effect due to ith treatment (row)µ.j= Fixed mean due to jth variety (column)βj==(µ.j-µ)=Effect due to jth variety (column)eij= error effect due to chance= random error attached to (i,j)th cell observationsn=rt=Total no. of experimental units

Assumptions All the observations are independent Different effects are additive in nature eij are identically & independently distributed N( 0, σe

2) αi = 0, j = 0, or αi N (0, σα2) ,j N (0, σ2)

Hypothesis:For treatmentsHoT:There is no significant difference in the average yields due to treatments.HoT: 1. = 2. = 3. = ………………….= t. = (say)i.e. α1 = α 2 = α 3 =………………… .= α t = 0H1T: There is significant difference in the average yields due to treatments.H1 T : 1. 2. 3. ………………… t.

i.e H1 T: α 1 α 2 …………………. α t

For blocks (varieties)HoB:There is no significant difference in the average yields due to blocks.HoB: .1 = .2= .3 = ………………….= .r = (say)HoB: β1= β2= β3 =………………… = βr =0H1B: There is significant difference in the average yields due to variety (blocks).H1B: .1 ≠ .2≠ .3 ≠………………….≠ .rH1B: β1≠ β2≠ β3 ≠………………… ≠βrStatistical analysis of the ModelIt is the same as the case of two way classified data, we estimate parameters , αi ,βj and

eij by using principle of least squares such that eij is minimum from equation (2) ofabove,

Then, estimated values are^= y..^ _

αi =yi.-y.. (Where αi=i.-) and yi.= Sample mean of the ith treatment)^ _

βj =y.j-y.. (Where βj =.j-) and y.j=Sample mean of the jth variety)^

eij=yij-yi.- y.j+y..

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Then, the mathematical model becomes,^ ^ ^ ^

yij=+ αi + βj + eij

yij=y..+(yi.-y..)+ (y.j-y..)+ (yij-yi.- y.j+y..)

yij- y..=(yi.-y..)+ (y.j-y..)+ (yij-yi.- y.j+y..)Squaring on both sides and summing over i and j, we get,∑∑(yij- y..)2=∑ ∑ [{(yi.-y..)+ (y.j-y..)+ (yij-yi.- y.j+y..)}]2

∑∑(yij- y..)2=∑ ∑ [{(yi.-y..)2+∑ ∑ (y.j-y..) 2+ ∑ ∑ (yij-yi.- y.j+y..) 2+ALL THEPRODUCT TERMS WILL BE VANISH DUE TO THE FIRST PROPERTY OF A.M.

Total sum of squares= Sum of squares due to treatments+ Sum of squares due tovarieties(blocks) +Sum of squares due to errorTSS=SST+SSB+SSE

Computational or practical used shortcut formulas are as follows:TSS=Raw SS-C.F. =∑∑yij2-C.F. Where, C.F.=G2/n=G2/rtSST=∑Ti. 2/r-C.F.SSB=∑T.j 2/t-C.F.

SSE=TSS-SST-SSBThe above statistical analysis and others steps are presented in the following table known

as ANOVA table.ANOVA TABLE FOR RCBD

Sourcesofvariation

d.f. SS MSS=SS/d.f Fcal

Treatments(row)

Variety (block)

Error

(t-1)

(r-1)

(r-1) (t-1)

SST

SSB

SSE

MST=SST/t-1

MSB=SSB/r-1

MSE=SSE/(t-1)(r-1)

FT=MST/MSE

FB=MSB/MSE

Total n-1=rt-1 TSS

Tabulated F- valuesF α(T)= Fα (T)[at t-1 and(r-1) (t-1) d.f.] for treatments

and Fα(B)= Fα (B)[at r-1 and(r-1) (t-1) d.f.] for variety (block)Decision rule: If F* ≤ Fα(*), Ho* is accepted otherwise rejected. Where*=(T,B)Relative Efficiency of RCBD over CRD (Blocking efficiency)Blocking maximizes the difference among blocks, leaving the difference among plots ofthe same block as small as possible. Thus, the result of every RBD experiment should beexamined to see how this objective has been achieved. For this

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1. Determine different blocks differ significantly or not2. Then, determine the magnitude of the reduction in experimental error due to

blocking by computing Relative efficiency (R.E.) given by,R.E.= (r-1) MSB+r(t-1) MSE

(rt-1) MSEWhere, t=no. of treatments

r= no. of blocksMSB= Mean sum of squares due to blocksMSE= Mean sum of squares due to error

If R.E.=1.60, this result indicate that the use of RBD instead of CRD design increasedexperimental precision by 60%.In other words, this result indicate that, if the CRD had been used, an estimated .60 timesmore replications (repeatations) would have been required to detect the treatmentdifference of the same magnitude as that detected with RBD design.Gain (loss) in precision= [(R.E.-1) x100] %=+ve value (in above case, gain in precision=60%)

If R.E.=.60Loss in precision=(.60-1)100%= -40%, this shows RCBD is 40% insufficient than CRD.

Blocking is not to be applied and RCBD is not suitable design.

If error degree of freedom is < 20, The R.E. value should be multiplied by adjustmentfactor k given asK= [(r-1)(t-1)+1] [t(r-1)+3]

[(r-1)(t-1)+3] [t(r-1)+1]Adjusted R.E.=k x R.E.Then, Gain (loss) in precision=[Adj.R.E.-1) x100]%The principle advantage of RCBD Blocking reduce error variance and provides more accurate result. Any number of treatments & any number of replication may be included. This is

the most popular design in view of simplicity, flexibility & validity. No otherdesign has been used as frequently as RBD

Control treatments can easily be included without causing any complication in theanalysis of the data.

Disadvantages If the blocks are not homogeneous the error term will be large It cannot accommodate large number of treatment since in this situation the

homogeneity of blocks or groups is always in danger or hazard.In many situations the criteria for blocking or grouping is not easily selectable.

Example of RCBD( Analysis procedures is same as two way ANOVA)2. Linthrust conducted a green house experiment on the growth of spartira alterni flora,an economically important salt march plant species, to evaluate the effect of salinity. Thevariable reported is biomass, the dried weight of all arial plant material.

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Blocks Treatments1(15 ppt) 2(30 ppt) 3(45ppt)

I 11.8 8.8 2.2II 8.1 8.1 3.3III 22.6 2.1 11.1IV 4.1 10.0 2.7

Display the ANOVA and test the effect of salinity at 5 % level of significance.Solution: For treatments

H0 : 1. = 2. = 3. i.e salinity effects are same.H1 : 1. 2. 3.

For blocksH0 : .1 = .2 = .3 = .4

H1 : .1 .2 .3 .4

TABLE

Blocks/TreatmentsI II III IV Ti. Ti.2

1(15 ppt) 11.8 8.1 22.6 4.1 46.6 2171.562(30 ppt) 8.8 8.1 2.1 10.0 29.0 841.03(45 ppt) 2.2 3.3 11.1 2.7 19.3 372.49T.j 22.8 19.5 35.8 16.8 G=94.9 3385.o5T.j2 519.84 380.25 1281.64 282.24 2463.97

First of all, we have to compute C.F., TSS, SST, SSB and SSE and prepare ANOVAtable.

Correction Factor (C.F.)=G2/rt=94.92/12=750.50TSS= Total sums of square=Yij2–C.F.=11.82 +8.12 +…………+16.82 -750.50=375.61

SST=Sums of square due to treatments=Ti.2/r –C.F.=3385.05/4-750.50=95.76

SSB=Sums of square due to blocks=Tj.2/t –C.F.=2463.97/3-750.50=70.82

SSE= Sums of square due to error=TSS-SST-SSB=209.03

ANOVA TABLESource ofvariation

d.f. SS MSS=SS/d.f Fcal Ftab at α =.o5

Treatments 2 95.76 47.88 Ft=MST/MSE F.05(t) at(2,6)at 2,6=5.1 =5.1433

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BlocksError

36

70.82209.03

23.6034.83

=2.028Fb=MSB/MSE F.05 (b) at 3,6=0.677 =4.7571

Total 11 375.61Decision: F tab (t)>Fcal (t), Then, Ho is accepted i.e. there is no significancedifferences of mean biomass values due to different treatments.Similarly, also, F tab (b)>Fcal (b), Then, Ho is accepted i.e. there is no significancedifferences of mean biomass values due to different blocks.

(Due to accepting of Ho for treatments, we cannot use LSD and DNMRT tests fortreatments)

Practical problems related to RCBD

1. Linthrust conducted a green house experiment on the growth of spartira alterniflora, an economically important salt march plant species, to evaluate the effect ofsalinity. The variable reported is biomass, the dried weight of all arial plantmaterial.

i. Give the model for this experimentii. Write down the basic assumptions regarding the problemiii. Display the ANOVA and test the effect of salinity at 5 % level of significance.iv. Compute the relative efficiency of the design due to blocking.v. Use LSD and DNMRT to make pairwise comparison.2. Consider the result given in the following table for an experiment involving six

treatments in four randomized blocks. The treatments are indicated by numberswithin parentheses.

Yield for a randomized block experimentBlocks Treatment and yield

1 ( 1) ( 3) (2 ) (4 ) (5 ) ( 6)24.7 27.7 20.6 16.2 16.2 24.9

2 ( 3) (2 ) ( 1) (4 ) ( 6) ( 5)22.7 28.8. 27.3 15.0 22.5 17.0

3 ( 6) (4 ) (1 ) (3 ) ( 2) ( 5)26.3 19.6 38.5 36.8 39.5 15.4

4 ( 5) ( 2) ( 1) (4 ) (3 ) ( 6)17.7 31.0 28.5 14.1 34.9 22.6

i. Give the model for this experiment

BlocksTreatments

1 (15 ppt) 2 (30ppt) 3(45 ppt)I 12.8 9.8 3.2II 9.1 9.1 4.3III 23.6 3.1 12.1IV 5.1 11.0 3.7

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ii. Write down the basic assumptions regarding the problemiii. Display the ANOVA and test whether the treatments differ significantly at 5 %level of significance.iv. Compute the relative efficiency of the design due to blocking.vi. Use LSD and DNMRT to make pairwise comparison.3. Fill in the blanks in the following analysis of variance table of RBD.

Sources ofvariation

d.f. SS MS

Block 4 26.8 -Treatment 3 - -Error - - 2.5Total - 85.3 -Test the hypothesis that treatment effects are equal to zero at 5% level of significance.4. Four silviculture thinning treatments T1=Control, T2=10% thinning, T3=20%thinning and T4=30% thinning tried on 20 plots of Forest. From the known sources ofvariation, it is known that basal area increment due to applied treatments is alsoaffected by altitudinal variations. Therefore, 5 different altitudes are taken for theblocking. Plot level mean basal areas (cm square) gained after 2 years is measured asfollows.

Basal area (cm square) in different altitudinal blocks1200-1400 1400-1600 1600-1800 1800-2000 2000-220016 ( T3 ) 27 ( T4) 8 ( T1) 9 ( T2) 8(T1)24( T4) 6 ( T1 ) 10 (T2) 20 (T3) 11(T2)4 ( T1 ) 10 ( T2 ) 20 ( T3) 26 (T4) 17(T3)10 (T2 ) 20 ( T3) 25 ( T4 ) 7 (T1) 29(T4)

(a) Test whether there is marked distinct between silvicultural treatments effects at 5 %level of significance.also use LSD test procedure for multiple comparison.(b) Does this design gain the efficiency compared to Completely Randomized Design?

UNIT 6 : LATIN SQUARE DESIGN (LSD)A Latin square design is a balanced two-way classification scheme. This design has twosuperimposed blocking systems called rows & columns. The number of rows and thenumber of columns must both be equal to number of treatments and each treatmentoccurs once in each row & once in each column. The principle use of LSD in forestryresearch is in nursery & glass house experiments in which rows & columns are easilyidentified with the natural layout. This design is specially used when the variations are intwo direction & perpendicular to each other. There are many situations in which twodirectional variations has to be control to have valid estimates or comparisons oftreatments, such type of control is well maintained through LSD. For example, if a forestresearcher wants to estimate the effect of 4 different fertilizers (say) A, B, C & D in thegrowth of nursery seedlings of 4 different species (row) & 4 different age groups(column), he has to use the design LSD.Randomization & layout

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The process of randomization & layout for LSD is shown below for the experiment withfour different treatments A, B, C & D cited above. In this experiment the researcher hasto divide the total experimental land in 44 = 16 experimental units.Step 1: The whole experimental area is divided into 42 = 16 experimental units arrangedin a square so that each row as well as each column contains 4 units.Step 2: The 4 treatments are then allocated to these rows & columns in such a way thatevery treatment comes once & only once in each column. The randomization can beshown as follows:

GradientColumns (age of seedlings)

1(3month) 2(4month) 3(5month) 4(6month)

Rows(species)

1 A B C D2 B C D A3 C D A B4 D A B C

Layout of LSD

Mathematical ModelThe linear model will be,

yijk==µ+ αi +βj+γk+eijk ………..(3) i=1,2,3,….tj=1,2,3,….tk=1,2,3,…t

Where yijk=(i,j,k)th cell observations=Yield or response of the jth column and ith rowby the use of kth treatmentµ= general mean = overall population meanµi..= Fixed mean due to ith rowαi=(µi..-µ)=Effect due to ith rowµ.j.= Fixed mean due to jth columnβj=(µ.j.-µ)=Effect due to jth columnµ..k=Fixed mean due to kth treatmentγk=(µ..k-µ)=effect due to kth treatmentseijk= error effect due to chance= random error attached to (i,j,k)th cell observationsn=t2=Total no. of experimental unitsAssumptions All the observations are independent Different effects are additive in nature eijk are identically & independently distributed N( 0, σe

2) i = 0, j = 0, k = 0,or i N (0, σ2) ,j N (0, σ2) , k N (0, σ2)HypothesisH0R: There is no significant difference in the mean yields due to rows.

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H0R : 1..= 2.. = 3.. = ………………….= t.. = i.e. α1 = α 2 = α 3 =………………….= α t = 0H1R: 1.. 2.. 3.. …………………. t..

i.e. α 1 α 2 …………………. t

H0C: There is no significant difference in the mean yields due to columns.H0C : .1. = .2. = .3. = ………………….= .t. =

i.e. 1 = 2 = 3 =………………….= t = 0H1C : .1. .2. .3. ………………… .t.

i.e. 1 2 3 …………………. tH0T= There is no significant difference in the mean yields due to treatments.H0T : ..1 = ..2 = …3 = ………………….= ..t =

i.e. 1 = 2 = 3 =………………….= t = 0H1T : ..1 ..2 ...3 ………………… ..t

i.e. 1 2 3 …………………. t

Statistical analysis of the ModelIn this LSD,we estimate parameters , αi ,βj ,γk and eijk by using principle of least

squares such that eijk is minimum from equation (3) of above,Then, estimated values are^= y…^

αi =yi..-y… (Where αi=i..-) and yi..= mean of the ith row)^βj =y.j.-y… (Where βj =.j.-) and y.j.= mean of the jth column )^γk =y..k-y… (Where γk =..k-) and y..k= mean of the kth treatment)

^eijk=yijk-yi..- y.j.- y..k.+2y…Then, the mathematical model becomes,

^ ^ ^ ^ ^yijk=+ αi + βj + γk + eijk

yijk=y…+(yi..-y...)+ (y.j.-y…)+ (y..k-y…)+ (yijk-yi..- y.j.- y..k.+2 y…)

yijk- y…=(yi..-y...)+ (y.j.-y...) + (y..k-y…) + (yijk-yi..- y.j.- y..k.+2 y…)Squaring on both sides and summing over i ,j and k, we get,

_∑∑∑ (yijk- y…)2=∑ ∑∑ [{(yi..-y…)+ (y.j.-y…)+ (y..k-y…) + (yij-yi..-

y.j.+y..k+2y…)}]2

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=∑∑∑ [{(yi..-y…)2+∑∑(y.j.-y…) 2 +(y..k-y…) 2+ ∑ ∑ (yijk-yi..-y.j.+y..k+2 y…) 2+ALLTHE PRODUCT TERMS WILL BE VANISHED DUE TO THE FIRST PROPERTYOF A.M.

Total sum of squares= Sum of squares due to rows+ Sum of squares due to columns+sum of squares due to treatments +Sum of squares due to errorTSS=SSR+SSC+SST+SSE

Computational or practical used shortcut formulas are as follows:TSS=Raw SS-C.F. =∑∑∑ yijk2-C.F. Where, C.F.=G2/n=G2/t 2

SSR=∑Ti.. 2/t-C.F.SSC=∑T.j. 2/t-C.F.SST= =∑T..k 2/t-C.F

SSE=TSS-SSR-SSC-SSTThe above statistical analysis and others steps are presented in the following table known

as ANOVA table.ANOVA TABLE FOR LSD

Sourcesofvariation

d.f. SS MSS=SS/d.f Fcal

RowColumnTreatmentError

(t-1)(t-1)(t-1)(t-1) (t-2)

SSRSSCSSTSSE

MSR=SSR/t-1MSB=SSC/t-1MST=SST/t-1MSE=SSE/(t-1)(t-2)

FR=MSR/MSEFB=MSC/MSEFT=MST/MSE

Total n-1=t2-1 TSS

Tabulated F- valuesF α(R)= Fα (R)[at t-1 and(t-1) (t-2) d.f.] for rows

and Fα(C)= Fα (C)[at t-1 and(t-1) (t-2) d.f.] for columnsand Fα(T)= Fα (T)[at t-1 and(t-1) (t-2) d.f.] for treatmentsDecision rule: If F* ≤ Fα(*), Ho* is accepted otherwise rejected. Where *=(R,C,T)

Relative efficiency of LSD over RCBD and CRDCase IR.E. of LSD (CRD) =MSR+MSC+(t-1)MSE

(t+1)MSESince the error d.f. is less than 20, R.E. has to be adjusted by multiplying the R.E.with adjustment factor k which is given as follows:K= [{(t-1)(t-2)+1}{(t-1)2+3}]

[{(t-1)(t-2)+3}{(t-1)2+1]so, Adjusted R.E of LSD(CRD)=Adj.R.E.=R.E. x kGain (loss) in precision (efficiency)=[(Adj.R.E.-1) x100]%

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Case II

R.E. of LSD (RCBD when rows are taken as blocks) =MSR+ (t-1)MSEt x MSE

R.E. of LSD (RCBD when column are taken as blocks) =MSC+ (t-1)MSEt x MSE

Since the error d.f. is less than 20, R.E. has to be adjusted by multiplying the R.E.with adjustment factor k which is given as follows:K= [{(t-1)(t-2)+1}{(t-1)2+3}]

[{(t-1)(t-2)+3}{(t-1)2+1]So, Adjusted R.E of LSD (RBD)=Adj.R.E.=R.E. x kGain (loss) in precision (efficiency)=[(Adj.R.E.-1) x100]%Advantages Because of the two way blocking or stratification LSD controls more of the

variations than CRD & RBD. Greater sensitivity – row & column variation is removed from error Easy analysis Several LSD of the same size may be combined and it is suitable for 5-9 no. of

treatmentsDisadvantages To obtain equal number of row, column & treatment is often difficult.

When number of treatment is large, design become impracticable because of the largenumber of replication required and when number of treatment is small, design gives fewerror degrees of freedom.

Practical for LSD1. Complete the following table for the analysis of variance of affixed effects LSD.

Sources ofvariation

d.f. SS MS

Columns 5 - -Rows - 4.20 -Treatments - - 2.43Error - - 0.65Total - 39.65

The columns as representing schools, the row as classes, the treatments as methods ofteaching spelling, and the observations as grades based on 100 points. Test thehypothesis that the treatment effects are equal to zero, showing all steps in generalprocedure.

42

2. The data below gives the field layout and the yield in tones of sugar beet producedunder 5 different irrigation treatments designated as A,B,C,D and E.Yields (tons) of sugar beet produced under 5 different irrigation treatments

Rows columns1 1 2 3 4 52 18.52( E ) 19.46( D) 20.66( A) 22.68( B) 18.65( C )3 21.68( C ) 14.29( E ) 18.82(D ) 2.02(A ) 20.58(B )4 26.04( A ) 17.49( C ) 21.06( B ) 18.91(E ) 20.03( D)5 22.51(D ) 22.98( B ) 17.15( E ) 17.14(C ) 20.62(A )6 24.44( B ) 20.25( A ) 18.92( C) 19.73(D ) 14.07( E )

i. Give the model for this experiment. Explain what each term in the modelrefers to.ii. Display the ANOVA table.iii. What are the basic assumptions regarding the problem?iv. Are there considerable variations in yield due to different types ofirrigation treatments? Also test the multiple comparisons by LSD andDNMRT tests.v. Does this design gained the efficiency compared to CRD and RCBD?

3. Four silviculture thinning treatments A=Control, B=15%Thinning, C=25%thinning and D=35% thinning tried on 16 plots of Forest. From the known sourceof variations, it is known that basal area increment due to applied treatment is alsoaffected by altitude and aspect. Therefore,4 different altitudes and 4 aspects aretaken. Plot level mean Basal areas (cm square) gained after 2 years is measured asfollows.

Aspects

Altitudes (m)

North North east South South East

1200-1400 18 ( C ) 29 ( D) 8 ( A) 10 ( B)1400-1600 26 ( D ) 8 ( A ) 11 (B) 21 (C)1600-1800 6 ( A ) 12 ( B ) 21 ( C) 28 (D)1800-2000 12 (B ) 22 ( C ) 27 ( D ) 7 (A)

a.Test whether the silvicultural treatments effects differ remarkably at 5 % level ofsignificance.b.Compare the treatment mean effects by using least significance test (LSD) procedurec.Does this design gained the efficiency compared to CRD and RCBD?

4. The data below gives the field layout and the yield of seed (kg) from thehighly demand bio-fuel plant species named jatropha curcas under 4 differenttreatments designated as control (A), compost (B), cow dung (C) and urea(D) under two superimposed known sources of variations blockings asspacing and age at planting.

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Age atplanting(month)

Spacing (m2)2x2 3x3 4x4 5x5

1.5 10 ( C ) 11 (D) 6(A) 14(B)2 13 ( D ) 7 (A ) 10(B) 13(C)

2.5 9 ( A ) 12 ( B ) 13 (C) 11(D)3 14 (B ) 14 (C ) 9(D ) 9 (A)

i. Are there remarkable distinct in the average yield due todifferent types of applied treatments at 1 % level of significance?

ii. Test the multiple comparisons by Least Significance Differencetest (LSD).

iii. Does this design gained the efficiency compared to CRDUnit 8:Factorial experiment (Complex experiment)IntroductionThe term factorial refers to experiments where the treatments have a special structure.It

refers to the situations where the treatment consists of all combination of levels of two ormore factors.It doesn’t only study the main effects due to factors but also the interactioneffects of factors.

In general, if the levels of various factors are equal then, sn factorial experiment meansan experiment with n factors each at s levels where n is any positive integer greater than orequal to 2, e.g. 22 Factorial design means 2 factors each at 2 levels,23 Factorial designmeans 3 factors each at 2 levels and 2x3x4 factorial means 3 factors so that first factor has2 levels, 2nd has 3 levels and 3rd has 4 levels etc.

They are especially important in several economic, biophysical and social phenomenawhere usually a large numbers of factors affect a particular problem. A fully factorialexperiment is a highly efficient way of obtaining information on each of the treatmentfactor & on the extent to which they interact with each other.For example, in 22 Factorial design, suppose we want to find out the effect of twodifferent fertilizers (factors) nitrogen (N) & potash (K) on the production of certain cropby using the two different amount of each fertilizer as 20 kg & 25 kg.Here, we can usefactorial experiment.Let each level of each factor is denoted as N0, N1, K0, K1 then we obtain four-treatmentcombinations as shown below;

Fac

tor

K

LevelsFactor N

20kg( N0) 25kg(N1)20kg(K0) N0K0=1 N1K0=n25kg (K1) N0K1=k N1 K1=nk

These 4 treatment combinations can be compared by laying out the experiment in(i) CRD (ii) R.B.D., with r replicates (blocks), each replicate containing 4

units, or (iii) 4x4 L.S.D., and ANOVA can be carried out accordingly. In theabove cases, there are 3 d.f. associated with the treatment effects. In factorial

44

experiment our main objective is to carry out separate tests for the maineffects N, K and interaction NK.

Replication I Replication II Replication IIIN0K0=1 N1K0=n N0K0=1N0K1=k N1 K1=nk N1 K1=nkN1K0=n N0K1=k N1K0=nN1 K1=nk N0K0=1 N0K1=k

A sample layout of 2x2 factorial experiments in a RCBD with 3 replicationsSimilarly, ANOVA of a two- factor experiment on bamboo involving two levels ofspacing (Factor A) and three levels of age at planting (Factor B) can be laid out in RCBDwith required replications where treatments combination are given below:

Age at planting (month)(FactorB)

Levels

Spacing (m)(Factor A)10 mx10 m(a1)

12m x 12m(a2)

6 (b1)12 (b2)24 (b3)

a1b1a1b2a1b3

a2b1a2b2a2b3

The 2x3 factorial treatment combinations of two levels of spacing and three levels ofageAdvantages Flexibility- any number of factors as well as any number of levels can be used

subject to the available resources. Factorial treatments may be used in any experimental design. Interaction of the treatments can be investigation. In the absence of interaction number of replication increases.

Disadvantages More complex if any observation is missing. If interaction is present, the results are more difficult to interpret. If the number of factors & the levels are large, the size of the experiment is large.

22 Factorial designIn this design, we consider 2 factors say A and B each at two levels (0,1) say

(ao, a1), and (bo, b1) respectively so that there are 22=4 treatment combinations in all.

Factor A

Factor BLevels b0 b1

a0 a0b0=1 a0b1=ba1 a1b0=a a1b1=ab

4 treatment combinations are:i. aobo=1, Factors A and B both at first levelii. a1bo=a, Factor A at 2nd level and Factor B at 1st level

45

iii. aob1=b, Factor A at 1st level and Factor B at 2nd leveliv. a1b1=ab, Factors A and B both at 2nd level22 factorial experiments can be performed as a CRD with 4 treatments or RBD with r

replicates or blocks (say), each replicate containing 4 treatments or 4x4 LSD and data canbe analyzed accordingly. In 2x2 design, we split up the treatment SS with 3 d.f. into 3orthogonal components corresponding to the two main effects A and B,one 1st orderinteractions AB each carrying 1 d.f. In the case of 22 design, the letters A, B and AB whenthey refer to numbers will represent the corresponding factorial effects.

Simple, main and interaction effectsSuppose, the factorial experiment with 2x2=4 treatments is conducted in r blocks or

replicates and it is factorial experiment with RCBD.Let, [1], [a], [b] and [ab] denote total yields of r units receiving the treatments 1, a, b, and

ab respectively and let the corresponding mean values obtained on diving these totals by rbe denoted by (1), (a),(b) and (ab) respectively.

The simple effect of one factor can be represented by the difference between mean yieldsobtained at each levels of other factor.

Simple effect of factor A at bo=(a1bo)-(aobo) =(a)-(1)…….. (i)Simple effect of factor A at b1 =(a1b1)-(aob1)=(ab)-(b)…… (ii)Simple effect of factor B at ao= (aob1)-(aobo)=(b)-(1)…….. (iii)

Simple effect of factor B at a1= (a1b1)-(a1bo)=(ab)-(a) ……. (iv)Main effects: Main effects of each factor as the average of the simple effect of that

factor over all levels of other factor.The main effects of factors A and B are defined as the average of these 2 simple effects.Thus, main effect of A=1/2[(ab)-(b) + (a)-(1)]…………… (v)Similarly, main effect of B=1/2[(ab) - (a)+(b)-(1)]………….(vi)Interaction effectsIf the factors A and B are dependent, equation (i) and (ii) will not be the same and the

average of the difference between these 2 numbers gives the 2 factor interaction or 1st orderinteraction between A and B or B and A.Interaction doesn’t depend on the order of thefactors i.e. means interaction effects AB and BA are same.

Then, the interaction effect of AB=1/2[(ab)-(b)-(a)+(1)] ……..(vii)Again, the interaction effect of BA=1/2[(ab)-(a)-(b)+(1)] ……..(viii)

(vii) and (viii) are same.If M for the mean yield of all the 4 treatment combinations, then

M=1/4[(ab) + (a) + (b) + (1)]

Table of signs and divisors giving M and factorial effects in terms of treatment means for

2x2 design;

Treatment meansFactorial effects (1) (a) (b) (ab) Divisor

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M + + + + 4A - + - + 2B - - + + 2AB + - - + 2

Model of 22 designIf Yijk is the response observed at the ith level of A,jth level of B in the kth

replicates(blocks) , then the linear model for a 2x2 design becomes,Yijk= +i+j+()ij+k +eijk, for all i , j =0,….1 , k=1,2,3, ……rWhere, =general mean

i=Effect of the ith level of Aj=Effect of the jth level of B()ij=Interaction effect of the ith level of A with jth level of Bk=Effect due to the kth replicate (Block)

eijk=Error effect due to the chanceHypothesis:

Ho*:There is no significant difference in the mean response due to blocking(replicate)and treatments i.e. main effects A and B and interaction effect AB.

Where *=R, T, A, B, AB

Statistical analysis of 22 designThe factorial effect totals for different treatment combinations are computed as of Yate’s

method,Treatmentcombination(1)

Treatmenttotal (2)

(3) (4) Effect totals

‘1’ [1] [1]+[a]=u1 u1+u2=v1 Ga [a] [b]+[ab]=u2 u3+u4=v2 [A]b [b] [a]-[1]=u3 u2-u1=v3 [B]ab [ab] [ab]-[b]=u4 u4-u3=v4 [AB]

Now, the sum of squares due to any of the factorial effect is given by multiplying thesquare of the effect total by the factor 1/4r each with 1 d.f.

SS due to main effect A=SSA=SA2=[A]2/4r

SS due to main effect B=SSB=SB2=[B]2/4r

SS due to interaction effect AB=SSAB=SAB2=[ AB]2/4r

Also, the sums of square due to total, replication (block), treatment and error arecomputed as usual manner.

TSS= Total sum of squares =Yijk2- C.F. Where C.F.=G2/4r

SST= Sum of squares due to treatment= 1/rTij.2-C.F.

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Where, also, SST=SSA+SSB+SSAB

SSR=Sum of squares due to replication (block) =1/4T..k2-C.F.Where T..k= total of kth replicate (block)

SSE= TSS-SSR-SST

ANOVA TABLE for 22experiment in RBD with r replicates (blocks)S.V. d.f. SS MS Fcal Ftab1.Blocks(replicate) r-1 SSR MSR FR=MSR/MSE F (R)at (r-1) and

3(r-1) d.f.2.Treatments 3 SST MST FT=MST/MSE F (T)at 3 and

3(r-1) d.f.i .Main effect A 1 SSA MSA FA=MSA/MSE F (A)at 1 and

3(r-1) d.f.ii. Main effect B 1 SSB MSB FB= MSB/MSE F (B) at 1 and

3(r-1) d.f.iii.1st order interaction effect AB 1 SSAB MSAB FAB

=MSAB/MSEF (AB)at 1 and3(r-1) d.f.

3.Error 3(r-1)

SSE MSE

Total 4r-1 TSS

Decision: If F* <F (*) for every sources of variations, then Ho* is accepted otherwiserejected. Where *=R, T, A, B and AB

Note:Contrast: Linear combination of treatment means such that the sum of the coefficients is

zero.Orthogonal contrast: Contrasts are orthogonal if the sum of the product of the

coefficients of corresponding treatment means is zero.

23 Factorial designIn this design, we consider 3 factors say A, B and C each at two levels (0,1) say

(ao,a1), (bo,b1) and (co,c1) respectively so that there are 23=8 treatment combinations inall.

Level of A Level of B Level of Cao bo co

a1 b1 c18 treatment combinations are: aoboco=1, a1boco=a, aob1co=b, a1b1co=ab,

aoboc1=c, a1boc1=ac, aob1c1=bc, a1b1c1=abc23 factorial experiments can be performed as a CRD with 8 treatments or RBD with r

replicates (blocks), each replicate containing 8 treatments or 8x8 LSD and data can beanalyzed accordingly. In 23 design, we split up the treatment SS with 7 d.f. into 7orthogonal components corresponding to the three main effects A,B and C,three 1st orderinteractions AB,AC and BC and one 2nd order interaction ABC, each carrying 1 d.f. In thecase of 23 design, the letters A, B, C, AB, AC, BC, and ABC, when they refer to numberswill represent the corresponding factorial effects.

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Simple, main and interaction effectsThe simple effect of A (say) is given by the differences in the main yields of A as a result

of increasing the factor A from the level ao to a1, at other levels of the factors B and C.Level of B Level of C Simple effect of Abo co (a1boco)-(aoboco) =(a)-(1)b1 co (a1b1co)-(aob1co)=(ab)-(b)bo c1 (a1boc1)-(aoboc1)=(ac)-(c)b1 c1 (a1b1c1)-(aob1c1) = (abc)-(bc)The main effect of A is defined as the average of these 4 simple effects.Thus, main effect of A=1/4[(abc)-(bc) + (ac)-(c) + (ab)-(b) + (a)-(1)]Similarly, main effect of B=1/4[(abc) + (bc)-(ac) - (c) + (ab) + (b)-(a)-(1)]And main effect of C=1/4[(abc) + (bc) + (ac) +(c)-(ab)-(b)-(a)-(1)]For interaction effectsThe average effect of A (one level of C) at the level bo of B=1/2[(ac)-(c) + (a)-(1)]The average effect of A (one level of C) at the level b1 of B=1/2[(abc)-(bc)+(ab)-(b)]If the factors A and B are not independent, then a measure of their interaction AB is

given by the ½ of the difference between the average effect of A at the 2nd and 1st level ofB.

Then the interaction effect of AB=1/4[{abc)-(bc)+(ab)-(b)}-{(ac)-(c)+(a)-(1)}]=1/4[{(abc)+(ab)+(c)+(1)}-{(bc)+(b)+(ac)+(a)}]

Interaction effect of BC=1/4[{(abc)+(bc)+(a)+(1)-{(ac)+ (c)+(ab)+(b)}]Interaction effect of CA=1/4[{abc)+(ac)+( b) +(1)}-{(bc)+(c)+(ab)+(a)}]From above 4 simple effects of A, we obtain the expressions for the interaction of AB at

the level co and c1 of the factor C as follows:Interaction of AB at level co of C=1/2[(ab)-(b)-(a)+(1)]Interaction of AB at level c1 of C=1/2[(abc) -(bc)-(ac)+(c)]

Thus, the interaction effect of AB with C i.e. interaction ABC is given by the half thedifference of the first expression from the second expression. Thus, symbolically,

Interaction effect of ABC= ¼[(abc)-(bc)-(ac)+(c)-(ab)+(b)+(a)-(1)]

If M for the mean yield of all the 8 treatment combinations, then

M=1/8[(abc)+(ab)+(ac)+(bc)+(a)+(b)+(c)+(1)]

Table of signs and divisors giving M and factorial effects in terms of treatment means for

23 design;

Factor effectsTreatment means

(1) (a) (b) (ab) (c) (ac) (bc) (abc) Divisor

M + + + + + + + + 8A - + - + - + - + 4B - - + + - - + + 4C - - - - + + + + 4AB + - - + + - - + 4

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AC + - + - - + - + 4BC + + - - - - + + 4ABC - + + - + - - + 4

Model of 23 designIf Yijkl is the response observed at the ith level of A,jth level of B, kth level of C in the

lth replicate (blocks), then the linear model for a 23 design becomes,Yijkl= +i+j+k+()ij+()ik+()jk+()ijk+l+eijkl, i , j ,k =0,….1

l=1,2,3, ……rWhere the symbols have their usual meanings.

HypothesisHo*:There is no significant difference in the mean response due to blocking(replicate)

and treatments i.e. main effects A, B,C and interaction effects AB,BC,AC and ABC.Where *=R, T, A, B, C, AB, BC, AC and ABC.

Statistical analysis of 23 designThe factorial effect totals for different treatment combinations are computed as in the

case of 22 design as follows of Yate’s method,Treatmentcombination(1)

Treatmenttotal(2)

(3) (4) (5) Effect totals

‘1’ [1] [1]+[a]=u1 u1+u2=v1 v1+v2=w1 Ga [a] [b]+[ab]=u2 u3+u4=v2 v3+v4=w2 [A]b [b] [c]+[ac]=u3 u5+u6=v3 v5+v6=w3 [B]ab [ab] [bc]+[abc]=u4 u7+u8=v4 v7+v8=w4 [AB]c [c] [a]-[1]=u5 u2-u1=v5 v2-v1=w5 [C]ac [ac] [ab]-[b]=u6 u4-u3=v6 v4-v3=w6 [AC]bc [bc] [ac]-[c]=u7 u6-u5=v7 v6-v5=w7 [BC]abc [abc] [abc]-[bc]=u8 u8-u7=v8 v8-v7=w8 [ABC]

Now the sums of square due to any of the factorial effect is given by =[ ]2/8r each with 1d.f.

e.g. SS due to main effect A=[A]2/8rSS due to interaction effect AB=[ AB ]2/8r, each with 1 d.f.

SS due to interaction effect ABC=[ ABC ]2/8r, each with 1 d.f.SSR=Sums of squares due to replication (SR2) =1/8T…l2-C.F.

Where T…l= total of lth replicate(block)TSS= Yijkl2- C.F. Where C.F.=G2/8rSST=1/rTijk.2-C.F.SSE= TSS-SSR-SSTANOVA TABLE for 23 experiment in`r’ randomized blocksS.V. d.f. SS MS Fcal Ftab1. Blocks(replicate) r-1 SSR MSR FR=MSR/MSE F(R) at (r-1) and

7(r-1) d.f.2.Treatments 7 SST MST FT=MST/MSE F(T) at 7 and

7(r-1) d.f.i .Main effect A 1 SSA MSA FA=MSA/MSE F(**) at 1 and

,, B 1 SSB MSB FB= MSB/MSE

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,, C 1 SSC MSC FC= MSC/MSE 7(r-1) d.f.**= A, B, C, AB,

BC, AC and ABC.ii.1st orderinteraction effectAB

1 SSAB MSAB FAB=MSAB/MSE

,, BC 1 SSBC MSBC FBC=MSBC/MSE

,, AC 1 SSAC MSAC FAC=MSAC/MSE

iii. 2nd orderinteraction effectABC

1 SSABC MSABC FABC=MSABC/MSE

3.Error 7(r-1) SSE MSETotal 8r-1 TSS

Result: If Fcal <Ftab, for every sources of variations, then Ho is accepted otherwiserejected.

Practical Problems based on Factorial experiments1. An experiment was planned to study the effect of sulphate of potash and super phosphateon the yield of potatoes. All the combinations of 2 levels of super phosphate and 2 levels of

sulphate of potash were studied in a randomized block design with 4 replications for each ofthe following yields were obtained:

Block I 1 k p kp20 22 19 35

Block II p 1 k kp37 23 27 17

Block III 1 k kp p26 17 27 17

Block IV kp k p 131 28 21 25

Analyze the data and give your conclusion.( Use =5% level of significance)

2. Below is given the plan and yields of 22 – factorial experiment involving 2 factors N and Seach at 2 levels 0 and 1.

Block IN0S0 N0S1 N1S1 N1S0117 106 125 124Block II

N1S1 N0S0 N0S1 N1S0124 120 117 124

Block IIIN0S0 N1S0 N0S1 N1S1

111 127 114 126Block IVN1S1 N1S0 N0S1 N0S0

125 131 112 108Block VN1S1 N0S1 N0S0 N1S0

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95 97 73 138Block VIN1S0 N0S0 N1S1 N0S1158 81 125 117Analyze the design and give your conclusion. (Use =5% level of significance)3. The following table gives the 23 factorial experimental design with 4 blocks. Thepurpose of the experiment is to determine the effect of different kinds of fertilizersNitrogen N, Potash K and Phosphate P on potato crop yield.Block 1nk kp p np 1 k n nkp291 391 312 373 101 265 106 450Block 2kp p k nk n nkp np 1407 324 272 306 89 449 338 106Block 3P 1 np kp nk k n nkp323 87 324 423 334 279 128 471Block 4np nk n p k 1 nkp kp361 272 103 324 302 131 437 435

Analyze the design and give your conclusion (Use =5 %)4. An experiment was planned to study the effect of Nitrogen (N) and Potash (K) onthe yield of tomatoes. All the combinations of 2 levels of N and K were studied in aRBD with 4 replications for each. The following yield (kg) was obtained:

Block IN1k1 N1K2 N2K1 N2K218 20 17 33Block IIN1K2 N1K1 N2K1 N2K235 21 31 33Block IIIN1K1 N1K2 N2K2 N2K124 15 25 15Block IVN2K2 N2K1 N1K2 N1K129 26 19 23

Test whether treatments i.e. main effects N and K and interaction effect NK aswell as block effects differ significantly at 5 % level of significance

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5. Below is given the Plan and yields of 2x2 factorial experiment involving 2factors water management (W) and fertilizer (F) each at two levels 0 as low leveland 1 as high level.

Block I1 w f wf

20 9 32 27Block II

wf 1 w f27 23 20 27

Block III1 w f wf

14 30 17 29Block IV

wf w f 128 34 14 13

Test whether treatments i.e. main effects W and F and interaction effect WF aswell as block effects differ significantly at 5 % level of significance.Unit 9: ANALYSIS OF COVARIANCE (ANOCOVA)IntroductionThe object of experimental design in general happens to be to ensure that the resultsobserved may be attributed to the treatment variable and to no other casualcircumstances. For instances, the researcher studying one independent variable X, maywish to control the influences of some uncontrolled (sometimes called the covariate orthe concomitant variable) Z, which is known to be correlated with the dependent variableY, then he should use the technique of analysis of covariance for a valid evaluation of theoutcome of the experiment. In psychology and education, primary interest in the analysisof covariance rests in its use as a procedure for the statistical control of an uncontrolledvariable.In many experiments, the researcher can exercise control over factors having an impacton the response of interest. Some of this control is accomplished through the use ofblocks where uninteresting effects are balanced out of treatment contrasts. On thepremise that the various biophysical features of an experimental plot do not behaveindependently but are often functionally related to each other, the analysis of covariancesimultaneously examines the variances and covariance among selected variables such thatthe treatment effect on the character of primary interest is more accurately characterizedthan by use of analysis of variance only.

ANOCOVA TECHNIQUEThe ANOCOVA is a procedure that permits us to make adjustments for the effects of anuncontrolled variable. It is the combination of two widely used experimental techniques:( 1) Regression (2) Analysis of variance

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While applying the ANOCOVA technique, the influence of uncontrolled variable isusually removed by simple linear regression method and the residual sums of squares areused to provide variance estimates which in turn are used to make tests of significance.ANOCOVA would be considered for those experiments where treatments effects are tobe analyzed–this is the analysis of variance portion of the procedure. And in addition tothe treatment effects, the response variable is influenced by one or more auxiliary(uncontrolled) variables- this functional relationship represents the regression portion ofthe procedure.The degree of freedom (d.f) associated with adjusted sum of square will be as under;

Between t-1Within N-t-1Total N-2

Assumptions in ANOCOVAThe ANOCOVA technique requires one to assume that there is some sort of relationshipbetween the dependent variable and the uncontrolled variable. We also assume that thisform of relationship is the same in the various treatment groups. Other assumptions are:i. Various treatment groups are selected at random from the populationii. The groups are homogeneous in variability.iii.The regression is linear and is same from group to group.

Example1: The following are paired observations for 3 experimental groups concerningan experiment involving 3 methods of teaching performed on a single class.

Method A to Group I Method B to Group II Method C to Group IIIX Y X Y X Y33 20 35 31 15 1540 32 50 45 10 2040 22 10 5 5 1032 24 50 33 35 15X represents initial measurement of achievement in a subject and Y be the finalmeasurement after subject has been taught. 12 pupils were assigned at random from 3groups of 4 pupils each, one group from one method as shown in the table.Applying the technique of ANOCOVA for analyzing the experimental results and thenstate whether the teaching methods differ significantly at 5% level. Also test H0:β=0against H1: β≠0Ans. F ratio is not significance and hence there is no significance difference inachievement due to teaching methods.

Example 2: Study the three treatment levels: Type I fertilizing procedure and Type IIfertilizing procedure and a control, on seed yield of plants, with the height of plant as thecovariate for adjusting the preexisting difference. Ten replications of each treatmentmethod were observed and the yields and heights are recorded in the following table.Goal: To investigate the difference between release methods.Besides the normality, independent and homogeneity of errors, the additional conditionsthat needto be checked for using a parametric ANOCOVA test are:

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1. The relation between the response and the covariate is linear.2. The regression coefficient for the covariate is the same for all treatments.3. The treatments do not affect the covariate. (The covariate is measured prior to theassignment of treatment.)Control Type I Type IIYield Height Yield Height Yield HeightY X Y X Y X

12.20 45.00 16.60 63.00 9.50 52.0012.40 52.00 15.80 50.00 9.50 54.0011.90 42.00 16.50 63.00 9.60 58.0011.30 35.00 15.00 33.00 8.80 45.0011.80 40.00 15.40 38.00 9.50 57.0012.10 48.00 15.60 45.00 9.80 62.0013.10 60.00 15.80 50.00 9.10 52.0012.70 61.00 15.80 48.00 10.30 67.0012.40 50.00 16.00 50.00 9.50 55.0011.40 33.00 15.80 49.00 8.50 40.00

ANACOVA: Statistical analysis involving a primary variate Y and one or more auxiliaryvariables (Xi’s) which makes the combined effects of ANOVA and the regressionanalysis is called ANACOVA.In ANACOVA, The effect of different treatments is analyzed. Formal relationship between primary variable and the auxiliary variable is

established.

Linear model of ANOCOVA for CRD_

Yij = + i + i (xij-x..)+ eij i=1,2, …………..tj=1,2,……………r

Where Yij =Response for the value of primary variable from (i,j)th cellsxij =Corresponding covariate (uncontrolled) value of ith cell on jth observation =Overall mean of Yiji = Adjusted effect due to ith treatmenti=Slopes of the line Y on X=Regression coefficients of Y on X

_x..=Grand mean of xeij= Residual or error effect due to chance

Hypothesisi. To test the significance of the regression coefficients of Yon X,

Ho1: i=0 Vs Ha: i0(Ho1: There are equal slopes for treatment groups based on x).

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ii.To test the adjusted treatment effectsHo2: i=0 Vs Ha: i0(Ho2: Adjusted y- treatment effects are same).

AssumptionsBesides the normality, independent and homogeneity of errors, the additional conditionsthat needto be checked for using a parametric ANOCOVA test are:1. The relation between the response and the covariate is linear.2. The regression coefficient for the covariate is the same for all treatments.3. The treatments do not affect the covariate i.e. the covariate is measured prior to theassignment of treatments.

Procedure for calculationi. Grand totalFor y –series, Gy=yij, Nx=Ny= N

For x- series, Gx=xijii. Correction factors

a.For y- series ,(c.f.)y =[ yi]2 =[ Gy]2

Ny Nb.For x-series, (c.f)x =[ xij]2 =[Gx]2

Nx Nc. For xy series, ( c.f)xy = (yij xxij) =Gx.Gy

N Niii.Total sum of squaresa. For y-series , Tyy = yij2-(c.f.)yb.For x-series , Txx = xij2 -(c.f)xc.For xy-series , Txy = [yij x xij] -(c.f)xy

iv.Sum of squares due to treatmentsa.For y-series, tyy=yi.2 -(c.f.)y where, yi.=Total of yij values of each treatment

rb.For x-series, txx=xi.2 - (c.f)x where, xi.=Total of xij values of each treatment

rc.For xy-series, txy = ((yi. x xi.) -( c.f)xy

rv. Sum of squares due to error (residuals)

a. For y-series Eyy =Tyy-tyyb.For x-series Exx = Txx -txxc. For xy- series Exy =Txy-txyvi. Error adjustment of y on xa. Error adjusted sums of square = (Exy)2/ Exxb.Adjusted effect of error=E’=Eyy- (Exy) 2/ Exx

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c.Adjusted effect of total =T=Tyy-(Txy)2/ Txxd.Adjusted effect of treatment=T’=T-E’

ANOCOVA TABLE FOR CRDY adjusted on X

S.V. d.f. Sums of productyy xx xy

d.f. SS MS

Treatment t-1 tyy txx txyError N-t Eyy Exx Exy N-t-

1E’ E’/N-t-1= (MSE)`

Total N-1 Tyy Txx Txy N-2 TTreatment adjusted t-1 T’=T-

E’T’/t-1=(MST)’

From ANOCOVA table, we can proceed to test the following kinds of hypotheses byusing given formula as below:i. To test the significance of the regression coefficient of Yon X,

Ho1: i=0 Vs Ha: i0( Ho1: There are equal slopes for treatment groups based on x.Fcal= Error adjusted SS = Error adjusted SS = (Exy)2/Exx

Adjusted error MSE (MSE)’ (MSE)’Ftab at 1 and N-t-1 d.f. =?Result: Since Fcal≤ Ftab, Ho1 is accepted i.e. i=0 means that there are equal slopes forall treatment groups based on x.

ii., Ho2: i=0 Vs Ha: i0(Ho2: Adjusted y- treatment means are same.

Fcal= (MST)’/ (MSE)’Ftab at t-1 and N-t-1 d.f=?Result: Since Fcal≤ Ftab, Ho2 is accepted i.e. i =0 i.e. adjusted y- treatment means aresame.

TU 2063: Practical problem based on ANOCOVASuppose that three different insecticides for the treatment of certain rice disease isapplied. The researcher used the CRD scheme with 21 experimental plots in equalnumber of replication per plot. Before applying the insecticide, the experimenter assessedthe stand of a crop by using a score system, ranging from 1 to 5. The following tableshows the yields in ton/hac and the score X.

ObservationsTreatment1

y

x

6 4 5 3 4 3 63 1 3 1 2 1 4

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2y

x

8 9 7 9 8 5 74 5 5 4 3 1 2

3y

x

6 7 7 7 8 5 73 2 2 3 4 1 4

With these data, Construct the ANOCOVA table andi.Test H0:βi=0 against H1: βi≠0 (There are equal slopes for treatment groups based on x)ii. Ho: i=0 Vs Ha: i0(Adjusted y- treatment means are same)

.

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Unit 10: (Introduction Only)

CONFOUNDING• Large no. of treatment combinations in factorial experiments become

incompatible to be accommodated in randomized blocks as there homogeneity isin risky. Hence, there arises the need of a device which can preserve the feature offactorial arrangement and reduce the block size so as to maintain theirhomogeneity. To fulfill these objective, there is a device known as“Confounding.’’

• In confounding, each replication is subdivided into two or more blocks of suitablesize known as incomplete blocks by giving up the information about a factorialeffect usually higher order interactions.

• some factors, which are not confounded, are estimated and tested with highprecision and some factors which are confounded lose their identity as they aremixed with block differences.

• We know, 2nd and higher order interactions are of little or no importance fromtheir physical interpretation point of view, hence they are primarily chosen forconfounding. Confounding is reducing block size and applicable only in factorialexperiments.

• Techniques of reducing the size of a replication over a number of blocks at thecost of losing some information on some effect which is not of much practicalimportance.Confounding in 23 experimentIn a 2n factorial, the block size in a randomly block design can be reduced to halfby confounding one effect which is of no interest or least interest. In practice, onechooses the highest order interaction. The procedure is very simple .Write thecontrast for the effect which is to be confounded. All these treatmentcombinations with + ve sign are randomly assigned in one half replicate (block)and those with –ve coefficients in the other half. The same entries of the blocksare repeated in other replications with a fresh randomization within blocks. Wegive below the layout of a 23 factorial factors A, Band C each at 2 levels 0 and 1.The effect ABC is confounded to obtain the blocks of size 4 units.

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…..Confounding in 23 experiment

Interaction effect of ABC is given by

ABC= ¼[(a-1) (b-1) (c-1)] =1/4[(abc)+(a)+(b)+(c)- (ab)-(ac)-(bc)-(1)]

ABC confounded with blocks

Rep 1 Rep2 Rep3Block I BlockII Block I Block II Block I BlockIIa (1) c ab b bcabc ab b ac c (1)b ac abc bc a ac

c bc a (1) abc ab

Confounding contd….

ANOVA table will be as follows,S.V. d.f. SS MS Fcal FtabBlock 2r-1A 1B 1C 1

AB 1AC 1BC 1Error 6(r-1)

Total 8r-1

Error SS is obtained as usual bysubtractions,

SE2= TSS-SSR-SA2-SB2-SC2-SAB2-SAC2 -SBC2

Advantages of ConfoundingThe only and the greatest advantage of confounding scheme lies in the fact that it reducesthe experimental error considerably by stratifying the experimental material intohomogeneous subsets or subgroups. The removal of the variation among in completeblocks (freed from treatments) within replicates often results in smaller error mean square

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as compared with a RBD design, thus making the comparisons among some treatmentsmore precise.

SPLIT - PLOT DESIGN

• A split- plot design is a special form of factorial design when there are twotreatment factors and either the effects of one factor is likely to be substantiallylarger than the other or one factor involves operations such as mechanicalcultivation that cannot be easily applied to a small plot.A split plot design can bearranged in randomized blocks, in Latin squares or in any other suitable design.

Choice of main and sub-plot design Degree of precisionFor a greater degree of precision for factor B than A, assign factor B to the subplot

and Factor A to the main plot. Relative size of the Main effects

If the main effect of one factor (factor B) is expected to be much larger and easier todetect than that of the other factor (factor A),Factor B can be assigned to the main plotand factor A to the sub plot.For eg,in a fertilizer x variety experiment, the researcher mayassign variety to the subplot and fertilizer to the main plot because he expects thefertilizer effect to be much larger than the varietal effect. Management practicesThe cultural practices required by a factor may dictate the use of large plots. For

practical feasibility, such a factor may be assigned to the main plot. For e.g. watermanagement and variety, it may be desirable to assign water management to the mainplot and varieties as sub plot.The chief practical advantage of this design is that it enables to include a factor whichrequire a relatively large amount of materials and another factor which requires only asmall amount of materials to be combined in the same experiment.

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Replication I Replication II Replication IIISample layout of a split plot design involving 4 nitrogen levels (n1,n2,n3

and n4) as main plot treatments and 3 varieties(v1,v2 and v3) as subplottreatments in 3 replications (Blocks)

•v1 v2 v1 v3

v3 v3 v2 v1

v2 v1 v3 v2

v3 v1 v1 v2

v1 v2 v3 v3

v2 v3 v2 v1

V1 v2 v3 v1

v2 v3 v1 v3

v3 v1 v2 v2

Split plot design ……..

n1 n2 n3 n4 n3 n1 n4 n2 n4 n2 n1 n3

• In a split plot design, one of the factors is assigned to the main plot. The assignedfactor is called main plot factor. The main plot is divided into subplot to which thesecond factor, called the subplots factor, is assigned. Thus, each main plotbecomes a block for the subplot treatments (i.e. the levels of the subplot factor).

• If the sample layout of a split plot lay out design involving 3 rice varieties(V1,V2,V3) as sub plot treatments and 4 Nitrogen levels (N1,N2,N3,N4) a mainplot treatments in 3 replications (blocks), ANOVA table will be as follows:

• .

Split plot design ANOVA table with RBD contd….

S.V d.f. SS ms Fcal Ftab• Replication r-1=2

• Main plot factor (A) a-1=3

• Error (a) (r-1) (a-1) = 2x3=6

• Subplot factor (B) b-1=2

• AxB (a-1) (b-1) = 3x2=6

• Error (b) a(r-1) (b-1)=4x2x2=16• Total rab-1=3x4x3-1=35

Decision: If Fcal ≤ Ftab, Ho accepted otherwise rejected

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NESTED DESIGNThe general idea of nesting refers to situations where subdivisions (levels) of a givenfactor change for the various subdivisions of the other factors. A nested designgenerally refers to experiments having several factors whose levels are nested withinthe levels of the other factors. For e.g., Suppose drinking water supply organization isinterested in studying the chlorine content of water in 3 major districts of Nepal. 8villages will be selected at random within each district and 50 household will beselected at random within each selected village for measuring the chlorine content.Here, there are 3 factors; districts, villages and households. The major factor isdistricts and probably is fixed; the other 2 factors are nested: villages within districtsand households within villages. The nested factors would be assumed to exhibitrandom effects, most likely.

We also can think of the nested designs as experiments having various stages ofsampling. In above example, we sample districts to get villages, then sample eachselected villages to get households to measure.Geneally, the ``major factor or first stage’’is assumed to be fixed while the nested factors are assumed to be random.

….

A1B11 B12 B13 B14Y111 Y121 Y131 Y141Y112 Y122 Y132 Y142Y113 Y123 Y133 Y143

A2B21 B22 B23 B24

Y211 Y221 Y231 Y241Y212 Y222 Y232 Y242Y213 Y223 Y233 Y243

The layout of nested design with a=2, b=4 and n=3”

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….

MODELFirst we know the model for an experiment having two factors:

Factor A at “a” levels and factor B at “b’’ levels nested within A.The model for fixed A and random B effects is,

Yijk= +Ai+Bij+eijk, i= 1, 2 ……… a j= 1, 2, ……… bk=1, 2, ...……… n

Where, =Over all meanAi= effect of ith level of A

Bij=random component at the jth level of B withinthe ith level of A

eijk=Residual component of the (i, j, k) thobservation

….

1 m x 1 m herbs plots are nested within 25 m x 20m PSP

Unit 11: Hypothesis Testing: Chi-square,

Chi-square test (2- test)

The tests of significance such as Z, t, F etc. tests are based on the assumptionthat the samples are drawn from a normal population i.e. we make assumptions about thepopulation parameters. Such tests are called parametric tests. However, in manysituations it is not possible to make dependable assumption about the parent populationfrom which samples are drawn. To study these problems some tests called non-parametric tests which do not require any assumptions about the parameters are derived.

Therefore, 2- test is one of the most important non –parametric test that does notrequire any assumptions about the population parameters. Thus it is most commonly used

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to test the hypothesis concerning the frequency distribution of one or more classes. It isbased on attribute data concerned with a finite numbers of discrete classes. The mostcommon types of attribute data are those having two classes, which consist of thepresence or absence of an attribute such as male or female, success or failure, effective orineffective and dead or alive and there may be more than two classifications in manycases.Chi-square tests for differences between categorical variables (i.e. nominal or ordinal).

There are both ‘‘one-way” and ‘’two way” chi-square procedures.

Conditions for applying 2 –test Observations recorded and used are collected on a random basis. The overall number of items must also be reasonably large. It should normally be

at least 50 i.e. N≥50. Each of the observations of the sample must be independent of each other. The expected frequency of any item or cell should not be less than 5. If it is less

than 5, the frequencies of adjacent items or cells should be pooled together inorder to make it 5 or more than 5.In this case, degree of freedom decreases.

Constraints on the cell frequencies, if any, should be linear, e.g.0i=Ei

Application of 2 –test

1. 2 –test as a test of goodness of fitIf we have a set of observed (actual) frequencies under some experiment and we want totest whether a particular distribution like uniform or other standard theoreticaldistribution like binomial, Poisson, normal distribution support the hypothesis (fits thedata), then 2 –test is used to test the goodness of fit of that distribution.. In 2 –test asa test of goodness of fit, we used one way classification tables of observed frequencies ina single row or column.Under this, Ho: The …….. distribution fits the data i.e. there is no significancedifference between observed (actual or experiment) and expected (theoretical)frequencies.

H1: The ………..distribution does not fit the data i.e. there is significancedifference between observed (actual or experiment) and expected (theoretical)frequencies.Karl Pearson proved that the statistic,

2 = (Oi-Ei)2/Ei 2 with d.f = n-k (1 d.f .is lost due to Oi=Ei)gives the magnitude of the discrepancy between theory and experiment and closer thevalues of Oi and Ei, smaller the value of 2 will be and the hypothesized distribution bestfits the data.Result: If 2 cal< 2 tab, Ho is accepted otherwise rejected.Example: Table shows the no. of insects species collected from an undisturbed Bardia

National park in different months. To test whether there is any significance differentbetween the no. of insect species found in different month.Month: Jan Feb March April May June July Aug Sep Octo Nov DecNo. of spec: 67 115 118 72 67 77 75 63 42 24 32 52

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Solution: Ho: The diversity in terms of no. of insect species is the same in all months inthe Bardia National Park. Then, the statistic

2 = (Oi-Ei)2/Ei with n-1 d.f. andUnder uniform distribution, expected frequencies are calculated by, Ei=Oi/n=804/12=67Table:

Month No. ofspecies(Oi)

Ei (Oi- Ei)2/Ei

Jan 67 67 0Feb 115 67 34.38March 118 67 38.82April 72 67 0.37May 67 67 0

June 77 67 1.49July 75 67 0.95Aug 63 67 0.23Sep 42 67 9.3

Octo 24 67 27.59Nov 32 67 18.28Dec 52 67 3.35

=Oi=804 67 (Oi-Ei) 2/Ei= 2=113.84

From above table, Calculated 2 =134.84Tabulated 2 at 0.05 with.12-1=11 d.f=19.7

Decision: Calculated 2> Tabulated 2, H0 is rejected and we conclude that theoccurrence of the no. of insect species in different months is not same in Bardia NP.

2. 2 –test as a test of independent of attributes (Contingency table/Cross tabmethod)In this case, when the observed frequencies occupy r rows and c columns, a two wayclassification table called rxc contingency table. A two-way chi- square is used if twocategorical variables are to be compared. Under this test,

Ho: There is no association between attributes or two attributes are independentlydistributed.

H1: There is association between attributes or two attributes are dependent.Then, test statistic

2=(Oij-Eij)2/EijEij values are computed as follows

Let, 2x3 contingency table,Attribute AAttribute B

A1 A2 A3 Row total (RT)

B1 a b c a+b+cB2 d e f d+e+fColumn total(CT)

a+d b+e c+f N=a+b+c+d+e+f

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Then, expected values of Eij can be calculated by the following procedure,Expected value of cell A1B1 or E (a)=Row total x Column total /N =RTxCT/N=(a+b+c) x (a+d)/NExpected value of cell A2B1 or E (b) =Row total x Column total /N =RTxCT/N=(a+b+c) x (b+e)/NSimilarly others can be calculated and chi –square value can be obtained by using aboveformula.Tabulated 2 at(r-1) (c-1) d.f. =2d.f. =?Decision: If cal 2 > tab2,Ho is rejected i.e. two attributes A and B are dependent(associated).Example: Based on information on 1000 randomly selected fields about the tenancystatus of the cultivators of these fields and use of fertilizers, collected in an agro- ecosurvey. The following classification was noted.

Owned Rented Row total(RT)Usingfertilizer

416 =a 184=b 600=a+b

Not usingfertilizer

64=c 336=d 400=c+d

Columntotal(CT)

480=a+c 520=b+d 1000=a+b+c+d=N

Is there any association between tenancy status and the use of fertilizers at 5% level ofsignificance? Draw your conclusion.Solution: Ho: Ownership of fields and the use of fertilizers are not associated i.e. they areindependent attributes.H1: Ownership of fields and the use of fertilizers are associated i.e. They are dependentattributes.

Under this test,2 =(Oij-Eij)2/Eij

To calculate Eij values, we have to use above table as,Expected value of a=E (a) = RT x CT/N=600x480/1000=288,E(b)=312, E(c)=192 and E(d)=208Now applying 2 table

Oij Eij (Oij-Eij)2/Eij

416 288 56.89

64 192 85.33184 312 52.51

336 208 78.77

2=273.504From above table, calculated 2 =273.504Tabulated 2 at .05 with [(r-1) (c-1) =1] 1 d.f. =3.84

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Result: Calculated 2 is much more than tabulated 2, Ho is rejected and there is highlevel of association between tenancy status and the use of fertilizers. Furthermore, it canbe concluded that the owner cultivators are more inclined towards the use of fertilizer.

3. 2 test as a test of homogeneityThe 2 test as a test of homogeneity is an extension of the 2 test of independence. Tests

of homogeneity are designed to determine whether two or more independent randomsamples are drawn from the same population or from different populations. Instead of onesample as we use with independent problem we shall now have 2 or more samples.For e.g., we may be interested in finding out whether or not university students of variouslevels i.e. I.SC., B.SC,M.SC feel the same in regard to the amount of work required bytheir Professors i.e. too much work, right amount of work , too little work required bytheir professors.We shall take the hypothesis that there exists no difference in opinion among the threeclasses of pupil on the issue.

Examples:1. Hypothesis Ho: there is no significance difference between male and female farmers

in their discussion with males.Discussion patterns of male respondents:Discussion by Discussion with males

Yes No Row totalMale 112 23 135Female 143 24 167Column total 255 47 302Chi-square value=0.4078 NS, df=1χ2- test for discussion patterns showed that there is no significance difference betweenmale and female farmers in their discussion with males.

2. Hypothesis Ho: there is no significance difference between male and female farmersin their discussion with females.Discussion patterns of female respondents:Discussion by Discussion with females

Yes No Row totalMale 79 88 167Female 82 53 135Column total 161 141 302Chi-square value=5.3279 significant, df=1, p<0.05χ2- test for discussion patterns showed that significance difference between male andfemale farmers in their discussion with females.

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PRACTICAL PROBLEMS1..A case study was conducted and following table gives the frequencies of the nearbyvillage people of different wealth being rankings whether they encroach the land or notin the Buffer zone area of Chitwan, Nepal.Wealth being ranking Encroachment of land

Yes NoPoor 40 20

Middle 23 45rich 42 17

Is there any evidence of an association between encroachment of land and wealth beingranking from the above data? How do you conclude your result in the above case?

2. Research based on objective to differentiate vulnerability to climate hazards only basedon landslides across different Ecology sector is performed by the research team of ForestAction, Nepal. The following rating frequencies by the individuals are obtained from thefield survey.Ecology sector Rating frequencies of Climate hazards based on

landslideNot

vulnerableLow vulnerability Medium

VulnerabilityHighvulnerability

Tropical 12 17 43 21Subtropical 13 23 22 11Temperate 11 16 12 9Alpine 9 22 16 13

Is there any association between ecology sector and rating of climatic hazards basedon landslide in the study of climate change vulnerability assessment from the abovedata at 10% level of significance?

3. Following table shows the frequency of participation levels of different income groupin benefit sharing from the questionnaire interview survey of MSc student 2065-67) inthree CFUGs of Parbat district.Income Class Participation Level

Low Medium HighPoor 14 6 6Middle 12 12 22Rich 7 6 12Test at 10% level of significance whether there exists any relationship between incomeclass and participation level in benefit sharing. Interpret your result in conclusive mannerin CF case.

4. Population of 5 cities random sample giving 2x5 contingency table of married andunmarried men as given below was taken. Can it be said that there is a significancedifference among the cities in the tendency of men to marry?

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City A B C D EMarried 133 169 155 106 153Unmarried 36 57 49 40 60

5. A batch of 1500 wooden stakes was divided into three groups and each group receiveda different termite repellent treatment. The treated stakes was driven into the ground, withthe treatment at any particular stake location being selected at random. Two years laterthe stakes were examined for termites. The no. of stakes in each classification is shown inthe following 2x3 table.

Treatment I II III

Attacked 193 148 110Not attacked 307 352 390

Test whether the attack classification is independent of the treatment classification.

6. Research based on objective to measure the influence of climatic changeindicator (variable) temperature based on human health with respect to thewealth being ranking of respondents is performed by the research team of IOF,Nepal. The following response frequencies perceived by the individuals areobtained from the field survey.Wealth beingranking

Human disease influenceLess Medium High

Poor 14 25 28Middle 12 27 15Rich 16 14 8Test whether there is any association between these Wealth being ranking andresponse of human disease influence in the above data table at 10% level ofsignificance? Interpret your result statistically by connecting it to the responses ofpeople of these categories.

7. Research based on the social capital in Community Forestry(CF), following tableillustrates the rating response frequency of the respondents of two CF of middlemountain categorized on distant to market based on statement” Community members’trustworthiness in related to borrowing and lending money.”

CF CategoryRating response frequency

Low Average HighNear from market 25 15 12Far from market 20 28 35

Test whether there exist any significance association between these prescribed CF category andrating response category of that statement at 5% level of significance? Interpret your result inconclusive manner in terms of value of social capital in the above data set.