Let measurements of n treatments are distributed to m blocks where,

yij be the observation of i-th treatment for j-th block

For this, the following layout shall be used,

Blocks:

 B1   B2——-Bj———-Bm Means Treatment 1Treatment 2 y11   y12——-y1j———-y1mY21   y22——-y2j———-y2m .1 Treatment i Yi1   yi2——-yij———-yim .2 Treatment n Yn1yn2——-ynj———-ynm .j Means .1.2——-.j———-.m .m

Consider this model as,

yij = µ + αi + βj + eij  where, i = 1, 2, …….. n and j = 1, 2, …….. m

Here,

µ = Grand Mean

αi = i-thTreatment Effects ()

βj = j-thTreatment Effects ()

eij = Random Errors, these are identically distributed {N (0, σ2)

Hypothesis:

To get the statistic hypothesis, it can be written as-

H01 = α2 = ………. = αn = 0

Or,

H01 = β2 = ………. = βn = 0

H1 :At least one value of these effects should not be zero

Constructing the ANOVA Table:

 Source of Variation Degree of Freedom Sum of Squares Mean Square F Treatments n – 1 SS (Tr) MS (Tr)= FTr = Blocks m – 1 SS (Bl) MS (Bl)= FBl = Error (n – 1) (m – 1) SSE MSE= Total nm – 1 SST

Where,

SSE       = Error Sum of Squares

SST       = SS (Tr) + SSE

It can be calculated as,

SST       = Total Sum of Square

=

SS(Tr)   =

SS(Bl)   =

c          =

Where,

Ti.         = Total of m Observations for i-th Treatment

T.j         = Total of n Observations for j-thBlock

T..         = Grand Total of all ‘nm’ Observations

So,

SSE       = SST – SS (Tr) – SS (Bl)

Set up the value of α

It’s Conclusion:

Reject H0:        αi = 0               i           if FTr> Fα.n – 1. (n – 1) (m – 1)

RejectH0:         βj= 0                j           if FBL> Fα.m – 1. (n – 1) (m – 1)

Example No. 3

When four students appeared for three different IQ test, the following table has been obtained. The scores are tabulated as-

 Students 1 2 3 4 Test-ITest-IITest-III 758386 737261 595653 686970

By using the method of two-way analysis of variance, state the reasons whether these tests are equivalent or not. Assume level of significance = 0.01

Solution:

1. As there are three equivalent tests,

H01 = α2 = α3 = 0

H1i≠ 0 for at least one value of i

Since students’ IQs are equivalent

H01 = β2 = β3 = β4 = 1340 = 0

H1j≠ 0 for at least one value of j

1. Given:α = 0.01, n = 3 and m = 4

Ftab(Tr) = F0.01, 2, 6= 10.92

Ftab(Bl) = F0.01, 3, 6 = 9.78

• Computations:

T1. = 75 + 73 + 59 + 68= 275

T2. = 83 + 72 + 56 + 69 = 280

T3. = 86 + 61 + 53 + 70 = 270

T.1 = 75 + 83 +86= 244

T.2 = 73 + 72 + 61 = 206

T.3 = 59 + 56 + 53 = 168

T.4 = 68 + 69 + 70 = 207

Thus,

T.. = 825

Now,

c = (825)2/ 12 = 56718.75

So,

SST = 57855 – 56718.75 = 1136.25

SS (Tr) = 226925/ 4 – 56718.75

= 12.5

SS (Bl)  = 173045/ 3 – 56718.75

= 962.92

Hence,

SSE = SST – SS (Tr) – SS (Bl)

= 160.83

ANOVA Table:

 Sources ofVariation Degrees of Freedom Sum of Squares Mean Square F Treatment 2 12.5 12.5/ 2 = 6.25 6.25/ 26.81 = 0.23 Blocks 3 962.92 962.92/ 3 = 320.97 320.97/ 26.81 = 11.97 Error 6 160.83 160.83/ 6 = 26.81 Total 11 1136.25

1. Conclusion:

Here,

Fcal (Tr)<Ftab(Tr)

Hence,

H01 = α2 = α3 = 0 is accepted for all the three tests as equivalent

And,

Fcal (Bl) >Ftab (Bl)

Hence,

H01 = β 2 = β 3 = β4 = 0 is rejected for all the three tests as students’ IQs are different.

Note:

The R.B.D is greatly accepted above the C.R.D. because of its greater flexibility in solving different types of experimental works. This is because there are no restrictions on numbers of treatments along with numbers of replicates. But the fact is that- for the large number of treatments or for the wide variable blocks, the R.B.D. is not suitable.

Problems:

1. Three samples are tabulated below. To obtain the hypothesis, consider the population means are equal and level of significance as 5%:
 A B C 81071411 751099 129131214

1. From the group of three main brands of certain powder, 120 sales are examined. Allocate these brands of powder considering the table below:
 Brands Groups A B C D IIIIII 0518 4819 81311 15613

• Fours batches of electric bulbs with 1000 hrs. of lives are tabulated below:
 Batches ABCD 1.61.581.461.51 1.611.641.551.52 1.651.641.61.53 1.681.71.621.57 1.71.751.641.6 1.72 1.661.68 1.8 1.74 1.82

Show that the significance test is accepted by using the analysis of variance method.

1. After conducting five tests in Computer Science paper, the following table has been tabulated with their test scores.
 Batch-I Batch-II Batch-III 8677818469 7978828368 8883868982

The test was conducted in three batches. Take the level of significance as 5% for the analysis of variance.

1. Construct a two-way ANOVA table using the following data-
 Prices of Field Treatment A B C D PQR 454339 404139 384541 373841

If required, you can shift the origin to 40.

1. In a factory, four different workers are hired for working on five different machines. After taking proper observations, some data has been made. You have to calculate in two different ways whether all men are similar in production or not along with the efficiency if each machines. The following has been given-
2. Sum of Squares for Variance between machines = 35.2
3. Sum of Square for Variance between workmen = 53.8
4. Sum of Squares for Total Variance = 174.2

1. Reject H0 : µ1 = µ2 = µ3, Fcal = 4

1. No such difference in the brands

1. Reject H0 as the means are different

1. ANOVA Table:
 Sources ofVariation Degrees of Freedom Sum of Squares Mean Square F BetweenTreatment 3 22.917 7.639 1.0 BetweenFields 2 8.167 4.083 0.534 Error 6 45.833 7.639 Total 11 76.917

1. Workers’ productivity is similar to mean productivity and it is same for all the f machines.

Here,

F (Workers) = 2.53

F (Machines) = 1.24  