ANOVA is a statistical technique used to determine whether differences exist among three or more population means.
In one-way ANOVA the effect of one factor on the mean is tested. It is based on independent random samples drawn from k – different levels of a factor, also called treatments.

**ANALYSIS OF VARIANCE (ANOVA)**

In chapter 2, testing equality means of two normal populations
based on independent small samples was discussed. When the number of
populations is more than 2, those methods cannot be applied.

ANOVA is used when we want to test the equality of means of more
than two populations. For example, through ANOVA, one may compare the average
yield of several varieties of a crop or average mileages of different brands of
cars.

ANOVA cannot be used in all situations and for all types of
variables. It is based on certain assumptions, and they are listed below:

1. The observations follow normal distribution.

2. The samples are independent.

3. The population variances are equal and
unknown.

According to R.A. Fisher ANOVA is the “Separation of variance,
ascribable to one group of causes from the variance ascribable to other
groups”.

The data may be classified with respect to different levels of a
single factor/or different levels of two factors.

The former is called one-way classified data and the latter is
called two-way classified data.

Applications of ANOVA technique to these kinds of data are
discussed in the following sections.

**One-way ANOVA**

ANOVA is a statistical technique used to determine whether
differences exist among three or more population means.

In one-way ANOVA the effect of one factor on the mean is tested.
It is based on independent random samples drawn from *k* – different
levels of a factor, also called treatments.

The following notations are used in one-way ANOVA. The data can be
represented in the following tabular structure.

*x _{ij} *- the

*k *- number of treatments compared.

*x _{i.} *- the sample total of

*n _{i} *- the number of observations in the

The total variation in the observations *x _{ij}* can
be split into the following two components

(i) variation between the levels or the
variation due to different bases of classification, commonly known as treatments.

(ii) The variation within the treatments *i.e*.
inherent variation among the observations within levels.

Causes involved in the first type of variation are known as
assignable causes. The causes leading to the second type of variation are known
as chance or random causes.

The first type of variation that is due to assignable causes, can
be detected and controlled by human endeavor and the second type of variation
that is due to chance causes, is beyond the human control.

Let the observations *x _{ij}*,

**Step 1 : ****Framing Hypotheses**

**Null Hypothesis ***H _{0}*

That is, there is no significant difference among the population
means of *k* treatments.

**Alternative Hypothesis**

*H _{1}*:

That is, at least one pair of means differ significantly.

**Step 2 : ****Data**

Data is presented in the tabular form as described in the previous
section

**Step 3 : ****Level of significance :**** ***α*

**Step 4 : ****Test Statistic**

*F *=* MST/MSE *which follows* F*_{(k-1,}* _{n}*

To evaluate the test statistic we compute the following:

(iv) Sum of Squares due to Error: *SSE = TSS – SST*

**Degrees of Freedom (d.f)**

**Mean Sum of Squares**

Mean Sum of Squares due to treatment: *MST* = *SST / k *−1

Mean Sum of Squares due to Error: *MSE *=* SSE / n *–* k*

**Step 5 : ****Calculation of Test statistic**

**ANOVA Table (one-way)**

**Step 6 : ****Critical value**

*f _{e} *=

**Step 7 : ****Decision**

If *F*_{0} < *f*_{(k-1,} _{n}_{-k),α}
then reject *H*_{0}.

**Merits**

·
Layout is very simple and easy to understand.

·
Gives maximum degrees of freedom for error.

**Demerits**

·
Population variances of experimental units for different
treatments need to be equal.

·
Verification of normality assumption may be difficult.

**Example 3.4**

Three different techniques namely medication, exercises and
special diet are randomly assigned to (individuals diagnosed with high blood
pressure) lower the blood pressure. After four weeks the reduction in each
person’s blood pressure is recorded. Test at 5% level, whether there is
significant difference in mean reduction of blood pressure among the three
techniques.

*Solution:*

**Step 1 : ****Hypotheses**

**Null Hypothesis: ***H*_{0}: *µ*_{1}* *=* µ*_{2 }= *µ*_{3}

That is, there is no significant difference among the three groups
on the average reduction in blood pressure.

**Alternative Hypothesis: ***H*_{1}:** ***μ _{i}*

That is, there is significant difference in the average reduction
in blood pressure in atleast one pair of treatments.

**Step 2 : ****Data**

**Step 3 : ****Level of significance**** ***α*** **= 0.05

**Step 4 : ****Test statistic**

*F _{0} = MST / MSE
*

**Step 5 : ****Calculation of Test statistic**

**Step 6 : ****Critical value**

*f*_{(2, 12),0.05} = 3.8853.

**Step 7 : ****Decision**

As *F*_{0} = 9.17 > *f*_{(2, 12),0.05} = 3.8853, the null hypothesis is
rejected. Hence, we conclude that there exists significant difference in the
reduction of the average blood pressure in atleast one pair
of techniques.

**Example 3.5**

Three composition instructors recorded the number of spelling
errors which their students made on a research paper. At 1% level of
significance test whether there is significant difference in the average number
of errors in the three classes of students.

*Solution:*

**Step 1 : ****Hypotheses**

Null Hypothesis: *H*_{0} : *µ*_{1} = *µ*_{2} = *µ*_{3}

That is there is no significant difference among the mean number
of errors in the three classes of students.

**Alternative Hypothesis**

*H*_{1}* *:* μ _{i} ≠ μ_{j} *for at one pair (

That is, atleast one pair of groups differ significantly on the
mean number of errors.

**Step 2 : ****Data**

**Step 3 : ****Level of significance**** ***α*** **= 5%

**Step 4 : ****Test Statistic**

*F _{0} = MST /
MSE *

**Step 5 : ****Calculation of Test statistic**

**Individual squares**

**ANOVA Table**

**Step 6 : ****Critical value**

The critical value = *f*_{(15, 2),0.05} = 3.6823.

**Step 7 : ****Decision**

As *F*_{0} = 0.710 < *f*_{(15, 2),0.05} = 3.6823, null
hypothesis is not rejected. There is no enough evidence to reject the null
hypothesis and hence we conclude that the mean number of errors
made by these three classes of students are not equal.

Tags : Test Procedure, Merits and Demerits, Example Solved Problems | Analysis of Variance | Statistics , 12th Statistics : Chapter 3 : Tests Based on Sampling Distributions II

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12th Statistics : Chapter 3 : Tests Based on Sampling Distributions II : One-way ANOVA | Test Procedure, Merits and Demerits, Example Solved Problems | Analysis of Variance | Statistics

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