# Analysis of Variance One-Way:

The total variation present in a set of observable quantities may under certain circumstances be partitioned into a number of disjoint components associated with the nature of classification of the data. This systematic methodology by which one can partition the causes of variation into several components is called analysis of variance.

Let us consider an example of yield of paddy. Suppose the yield is carried out using three kinds of seeds. So, the yield variation occurs due to variation of seed and also due to some random error (the position of seed was suitable for germination). This is a classic example of one-way layout of ANOVA.

Assumptions:

1. The observations recorded were independent
2. Parent population from which observations were taken have normal distribution.
3. Homogeneity of variances in the different treatment groups i.e. the variance of all the treatment groups are equal.

One Way ANOVA:

The main theory behind CRD is One-way ANOVA.

Let us consider the following Example.

A spots analyst wanted to know if the physical weight of players differs due to different training strategy of 5 different clubs. For this purpose, he gathered 5 groups from each of the 5 clubs.

Theory:

Here Weight of a player is influenced by a single treatments/Factor, Club, A and the factor has 5 levels.

A (Factor): Clubs from which a professional football player plays for.

Level 1 (Cowboys): players from the Dallas Cowboys

Level 2 (Packers): players from the Green Bay Packers

Level 3 (Broncos): players from the Denver Broncos

Level 4 (Dolphins): players from the Miami Dolphins

Level 5 (Niners): players from the San Francisco Forty Niners

So, there are 5 treatments and ith treatment (where i=1, 2, 3, 4, 5) is replicated ri =17 times. i.e., it can look upon as a similar setup like ANOVA one may fixed effect model, where a single factor has 5-levels and each level consists of ri =17
observations.

For ith level, let there be ni observations,

We represent the observations in the following array data.

One-way fixed effect model is given by:

yij = response corresponding to jth observation of ith level of A

µi = effect due to ith level of A.

αi = additional effect due to i-th level of A

µ= general effect

eij = error in model

Re-parameterization essentially leads to separation of error and exact effect due the factor A

Generally, the two hypotheses are considered as:

H0: α1 = α2 = α3= α4= α5 v/s H1: at least one inequality

The following two hypotheses can be described as:

H0: there is no difference between mean weights of players from different clubs

Vs.

H1: at least one of the mean weights is different from another

One -way ANOVA using R:

Code:

[Note:

Do not choose Excel formal for this case as it doesn’t read the levels of a factor

Consider the following code and output for illustration:

+ col_types = c(“numeric”, “text”))

> View(player_weight)

> names(player_weight)

[1] “Weight” “Club”

> levels(player_weight\$Club)

NULL

Note that R cannot read its levels. (Giving output as NULL)]

``````#Choosing dataset (if the dataset is in .csv Format)
``````my_data <- read.csv(file.choose())
``````View(my_data)
``````

Preview of the data:

 Weight Club 250 Cowboys 255 Cowboys 255 Cowboys 264 Cowboys 250 Cowboys 265 Cowboys

71 more columns

``````# Show the levels
``````names(my_data)
``````levels(my_data\$Club)
``````
```Output:
[1] "Weight" "Club"
[1] "Broncos"  "Cowboys"  "Dolphins" "Niners"
[5] "Packers"
```
``````#Estimation of Model
``````model1 <- aov(Weight ~ Club, data = my_data)
``````summary(model1)
``````
 Df Sum sq. Mean Sq. F value Pr(>F) Club 4 1714 428.4 1.575 0.189 Residuals 80 21761 272.0
`Signif. codes:  0     '***' 0.001     '**' 0.01     '*' 0.05     '.' 0.1      ' ' 1`

The output includes the columns F value and Pr(>F) corresponding to the p-value of the test. From the p-value we cannot reject the Null Hypothesis at 5% level of Significance.

So, we can conclude that at 5% level of Significance the physical weight of plyers does not differ significantly due to different training strategy of 5 different clubs.

[Note:

But if the Null hypothesis was rejected then we can perform a paired comparison with the help of Tukey test and can find out which club’s training strategy differs significantly. The R code for this is:

``````TukeyHSD(model1, conf.level = 0.95)
``````

We will consider an example in the Next portion of 2-way ANOVA so that we can perform the Tukey test.]

``````##Checking ANOVA assumptions
``````# 1. Homogenity of variances
``````install.packages("car")
``````library(car)
``````leveneTest(Weight ~ Club, data = my_data)
``````
```Output:
Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)group  4  0.0956 0.9836      80
```

From the output above we can see that the p-value is not less than the significance level of 0.05. This means that there is no evidence to suggest that the variance across groups is statistically significantly different. Therefore, we can assume the homogeneity of variances in the different treatment groups.

``````# 2. Normality
``````plot(model1, 2)
``````# Extract the residuals
``````aov_residuals <- resid(model1 )
``````# Run Shapiro-Wilk test
``````shapiro.test(x = aov_residuals )
``````

Output:

```Shapiro-Wilk normality test
data:  aov_residuals
W = 0.94462, p-value = 0.001161
```

From the p-value and the plot we can see that the data does not violate the Normality assumption.

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