Consider the following results of *k *independent random samples each of size *n, *from *k *different populations :

Population 1 : *x _{1, } *x

_{12}, ...

*x*Treatment 1

_{1n}:Population 2 : x_{21}, *x _{22}, *..

*x*Treatment 2

_{2n : }--------------------------------------------------------------------------------------------------

Population *k : x _{k1}, x_{k2}, *...

*x*Treatment

_{kn : }*k*

Here *x _{ij}* refers to the j-th value of the i-th population and the corresponding random variables

*X*which are all independent normally distributed with the respective means Jli and the common variance σ

_{ij}^{2}.

Consider the model,

*x _{ij} *= µ + α

_{i}+

*e*= 1, 2,

_{ij}, i*... ,k, j*= 1

**,**2, ...... ,

*n*

Here, µ is referred to as the Grand mean,

* *

α_{i} is referred to as treatment effects such that ∑ α = 0

and *e _{ij} *random errors are identically distributed as N (0, σ

^{2}).

** **

**Hypothesis **:

H_{0} : Population means are all equal

* *

*i.e.. *µ_{1}= µ_{2}= ··· = *µ _{k}*

* *

*i.e. , *samples were obtained from *k *populations with equal means.

Equivalently α_{i} = 0, *i *= 1, 2, ... , *k *i.e., there is no special effect due to any population.

H_{1} : α_{i} ≠ 0 for at least one value of *i.*

Then we construct the following ANOVA TABLE :

Where SST = Total sum of squares

SS(Tr) = Treatment sum of squares

SSE = Error sum of squares,

SST = SS *(Tr) *+ SSE,

Ti = Total of the values obtained for the *i-th *treatment

T**..** = Grand total of all *nk *observations

MS *(Tr) = *SS *(Tr) / (k-1) *(Treatment mean square)

MSE = SSE / *k *( *n *_ 1) (Error mean square)

F = F_{cal }= MS (Tr) / MSE which follows F distribution.

Let L = level of significance, then

F_{tab} = F_{α,α- 1, k (n-1)}

** **

**Conclusion:**

** **

Reject H_{0} if F_{cal} > F_{tab}

Accept H_{0} if F_{cal} < F_{tab}

** **

**Example 1. ***A test was given to five students taken at random from the Xth class of three schools of a town. The individual scores are*

*Carry out the analysis of variance and state your conclusions.*

** **

**Solution. 1. **H_{0 }: α_{1} = α_{2} = α_{3 }= 0 (No difference between the schools)

H1 : a ≠ 0 for at least one value of *i.*

2. Let a= 0.01, Here *k *= 3, *n *= 5

F_{tab }= F0.01, 2. 12 = 6.93

3. Computations:

T1• = 77 + 81 + 71 + 76 + 80 = 385

T2• = 72 + 58 + 74 + 66 + 70 = 340

T3• = 76 + 85 + 82 + 80 + 77 = 400

T•• = T 1• + T2 • …. T3• = 1125

∑∑ x_{ij}^{2}= 85041

SST = 85041 – 1 / 1.5 (1125)^{2} = 666

SS(Tr) = 1/ 5 [ (385?)^{2}+ (340)^{2} + ( 400)^{2}] – 1/1.5(1125)^{2}

= 390

SSE = SST - SS(Tr) = 666 - 390 = 276

**ANOVATABLE**

4. Conclusion :

Since F _{cal} > F_{tab }=> H_{0 }is rejected.

=> Students of different schools of class X are not same.

**Example 2. ***The following are the number of typing mistakes made in five successive weeks by four typists working for a publishing company.*

*Test at the 0.05 level of significance whether the differences among the four sample means can be attributed to chance.*

** **

**Solution. **1. H_{0 }: µ_{1} = µ_{2} = µ* _{3} *= µ

_{4}

H1 : At least two of them are not equal.

2. Here *k *= 4, *n *= 5, α = 0.05

F_{0.05,3.16} = 3-24

3. Computations :

T1• = 13 + 16 + 12 + 14 + 15 = 70

T 2• = 14 + 16 + 11 + 19 + 15 = 7 5

T 3• = 13 + 18 + 16 + 14 + 18 = 79

T 4• = 18 + 10 + 14 + 15 + 12 = 69

T••= T1• + T2• + T3• + T4• = 293

∑∑ x_{ij}^{2 } = 4407

SST = 4407 -1/ 20 (293)^{2} = 114.55+

SSE = SST - SS(Tr) = 101.6

** **

**ANOVATABLE**

4. Conclusions :

Since Feat < T_{tab}, H_{0 }is accepted and we conclude that the typing mistakes can be attributed to chance.

** **

**Note. **1. Since variance is independent of change of origin, then the origin can be shifted to an arbitrary data. Final variance ratio is independent of change of scale. Hence change of origin and scale, if necessary, can be employed which help to perform simpler arithmetic.

2. Each observations *x _{ij} *can be decomposed as

x_{ij = + (i - ) + ( }*x _{ij} - *

_{ )}

Grand mean Deviation due to Error

Treatment