SSW [ANOVA]

The Problem

Example #456: Let us test the null hypothesis that the average yield does not depend upon the treatment used. Since there are 5 treatments in this experiment, the hypotheses are

H₀ : μ₁ = μ₂ = μ₃ = μ₄ = μ₅
H_A : At least one mean differs from the others.

To test this hypothesis, we collect data. The data consist of two measurements on each unit: yield value and treatment level. (Note that yield is assumed numeric and treatment is assumed categorical.) It is typical to group the experimental units by treatment level. Thus, our data are

Treatment 1:
13, 4, 11, 6, 10, 11, 4, 6, 8, 13, 8, 5

Treatment 2:
11, 5, 7, 10, 7, 14, 10, 14, 6

Treatment 3:
7, 10, 10, 12, 8, 11

Treatment 4:
9, 11, 7, 11, 6, 14, 5

Treatment 5:
6, 11, 11, 8, 13, 17, 14, 10, 8

With this information, calculate the sum of squares within (SSW) for the data.

Information given:

To summarize the above, the values of import are:

Summary statistics from the problem
$ \bar{x} $	=	9.3488

$ \bar{x}_1 $	=	8.25
$ \bar{x}_2 $	=	9.3333
$ \bar{x}_3 $	=	9.6667
$ \bar{x}_4 $	=	9
$ \bar{x}_5 $	=	10.8889

$ n_1 $	=	12
$ n_2 $	=	9
$ n_3 $	=	6
$ n_4 $	=	7
$ n_5 $	=	9

Your Answer

In the box below, please enter the sum of squares within (SSW) for the data, then click on the “Check your answer!” button. Please round your answer to the ten-thousandths place.

Your guess:

Another Set of Data

Would you like to continue working on this topic? If so, click here for another data set.

Assistance

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$$ \begin{align} \text{SSW} &= \sum_{i=1}^g \sum_{j=1}^{n_i}\ (x_{i,j} - \bar{x}_i)^2 \\[3em] &= \sum_{i=1}^{5} \sum_{j=1}^{n_i}\ (x_{i,j} - \bar{x}_i)^2 \\[1em] &= \sum_{j=1}^{12}\ (x_{1,j} - 8.25)^2 + \sum_{j=1}^{9}\ (x_{2,j} - 9.3333)^2 + \sum_{j=1}^{6}\ (x_{3,j} - 9.6667)^2 + \sum_{j=1}^{7}\ (x_{4,j} - 9)^2 + \sum_{j=1}^{9}\ (x_{5,j} - 10.8889)^2 \\[1em] &= (x_{1,1} - 8.25)^2\ + (x_{1,2} - 8.25)^2\ + (x_{1,3} - 8.25)^2\ + (x_{1,4} - 8.25)^2\ + (x_{1,5} - 8.25)^2\ + (x_{1,6} - 8.25)^2\ + (x_{1,7} - 8.25)^2\ + (x_{1,8} - 8.25)^2\ + (x_{1,9} - 8.25)^2\ + (x_{1,10} - 8.25)^2\ + (x_{1,11} - 8.25)^2\ + (x_{1,12} - 8.25)^2\ + \\ & \qquad (x_{2,1} - 9.3333)^2\ + (x_{2,2} - 9.3333)^2\ + (x_{2,3} - 9.3333)^2\ + (x_{2,4} - 9.3333)^2\ + (x_{2,5} - 9.3333)^2\ + (x_{2,6} - 9.3333)^2\ + (x_{2,7} - 9.3333)^2\ + (x_{2,8} - 9.3333)^2\ + (x_{2,9} - 9.3333)^2\ + \\ & \qquad (x_{3,1} - 9.6667)^2\ + (x_{3,2} - 9.6667)^2\ + (x_{3,3} - 9.6667)^2\ + (x_{3,4} - 9.6667)^2\ + (x_{3,5} - 9.6667)^2\ + (x_{3,6} - 9.6667)^2\ + \\ & \qquad (x_{4,1} - 9)^2\ + (x_{4,2} - 9)^2\ + (x_{4,3} - 9)^2\ + (x_{4,4} - 9)^2\ + (x_{4,5} - 9)^2\ + (x_{4,6} - 9)^2\ + (x_{4,7} - 9)^2\ + \\ & \qquad (x_{5,1} - 10.8889)^2\ +\ (x_{5,2} - 10.8889)^2\ +\ (x_{5,3} - 10.8889)^2\ +\ (x_{5,4} - 10.8889)^2\ +\ (x_{5,5} - 10.8889)^2\ +\ (x_{5,6} - 10.8889)^2\ +\ (x_{5,7} - 10.8889)^2\ +\ (x_{5,8} - 10.8889)^2\ +\ (x_{5,9} - 10.8889)^2 \\[1em] &= (13 - 8.25)^2\ + (4 - 8.25)^2\ + (11 - 8.25)^2\ + (6 - 8.25)^2\ + (10 - 8.25)^2\ + (11 - 8.25)^2\ + (4 - 8.25)^2\ + (6 - 8.25)^2\ + (8 - 8.25)^2\ + (13 - 8.25)^2\ + (8 - 8.25)^2\ + (5 - 8.25)^2\ + \\ & \qquad (11 - 9.3333)^2\ + (5 - 9.3333)^2\ + (7 - 9.3333)^2\ + (10 - 9.3333)^2\ + (7 - 9.3333)^2\ + (14 - 9.3333)^2\ + (10 - 9.3333)^2\ + (14 - 9.3333)^2\ + (6 - 9.3333)^2\ + \\ & \qquad (7 - 9.6667)^2\ + (10 - 9.6667)^2\ + (10 - 9.6667)^2\ + (12 - 9.6667)^2\ + (8 - 9.6667)^2\ + (11 - 9.6667)^2\ + \\ & \qquad (9 - 9)^2\ + (11 - 9)^2\ + (7 - 9)^2\ + (11 - 9)^2\ + (6 - 9)^2\ + (14 - 9)^2\ + (5 - 9)^2\ + \\ & \qquad (6 - 10.8889)^2\ +\ (11 - 10.8889)^2\ +\ (11 - 10.8889)^2\ +\ (8 - 10.8889)^2\ +\ (13 - 10.8889)^2\ +\ (17 - 10.8889)^2\ +\ (14 - 10.8889)^2\ +\ (10 - 10.8889)^2\ +\ (8 - 10.8889)^2 \\[1em] &= (4.75)^2\ + (-4.25)^2\ + (2.75)^2\ + (-2.25)^2\ + (1.75)^2\ + (2.75)^2\ + (-4.25)^2\ + (-2.25)^2\ + (-0.25)^2\ + (4.75)^2\ + (-0.25)^2\ + (-3.25)^2\ + \\ & \qquad (1.6667)^2\ + (-4.3333)^2\ + (-2.3333)^2\ + (0.6667)^2\ + (-2.3333)^2\ + (4.6667)^2\ + (0.6667)^2\ + (4.6667)^2\ + (-3.3333)^2\ + \\ & \qquad (-2.6667)^2\ + (0.3333)^2\ + (0.3333)^2\ + (2.3333)^2\ + (-1.6667)^2\ + (1.3333)^2\ + \\ & \qquad (0)^2\ + (2)^2\ + (-2)^2\ + (2)^2\ + (-3)^2\ + (5)^2\ + (-4)^2\ + \\ & \qquad (-4.8889)^2\ +\ (0.1111)^2\ +\ (0.1111)^2\ +\ (-2.8889)^2\ +\ (2.1111)^2\ +\ (6.1111)^2\ +\ (3.1111)^2\ +\ (-0.8889)^2\ +\ (-2.8889)^2 \\[1em] &= (22.5625)\ + (18.0625)\ + (7.5625)\ + (5.0625)\ + (3.0625)\ + (7.5625)\ + (18.0625)\ + (5.0625)\ + (0.0625)\ + (22.5625)\ + (0.0625)\ + (10.5625)\ + \\ & \qquad (2.7778)\ + (18.7778)\ + (5.4444)\ + (0.4444)\ + (5.4444)\ + (21.7778)\ + (0.4444)\ + (21.7778)\ + (11.1111)\ + \\ & \qquad (7.1111)\ + (0.1111)\ + (0.1111)\ + (5.4444)\ + (2.7778)\ + (1.7778)\ + \\ & \qquad (0)\ + (4)\ + (4)\ + (4)\ + (9)\ + (25)\ + (16)\ + \\ & \qquad (23.9012)\ +\ (0.0123)\ +\ (0.0123)\ +\ (8.3457)\ +\ (4.4568)\ +\ (37.3457)\ +\ (9.679)\ +\ (0.7901)\ +\ (8.3457) \\[1em] &= 120.25\ + 88\ + 17.3333\ + 62\ + 84.5432 \\[1em] &= 380.4722 \\[1em] \end{align} $$

From these calculations, the within sum of squares is SSW = 380.4722.

Show the R Code

Hide the R Code

There are two ways of performing these calculations in R. The method you select will depend on how your data are stored.

Method 1: Wide Format

Copy and paste the following code into your R script window, then run it from there.

## Import data

treatment1 = c(13, 4, 11, 6, 10, 11, 4, 6, 8, 13, 8, 5)

treatment2 = c(11, 5, 7, 10, 7, 14, 10, 14, 6)

treatment3 = c(7, 10, 10, 12, 8, 11)

treatment4 = c(9, 11, 7, 11, 6, 14, 5)

treatment5 = c(6, 11, 11, 8, 13, 17, 14, 10, 8)

## Change to Long Format

mmt = c( treatment1, treatment2, treatment3, treatment4, treatment5 )

grp = c( rep("trt1",12), rep("trt2",9), rep("trt3",6), rep("trt4",7), rep("trt5",9) )

## Model the data

mod = aov(mmt~grp)

summary(mod)

In the R output, the value of the sum of squares within is the number in the table under Sum Sq and to the right of Residuals. If you would like better precision for that value, or if you would like to have only that value, run the following code in addition to that above:

modSummary = summary(mod)

modSummary[[1]][2,2]

Here, the number outputted is the sum of squares between. How did you get the number? The summary table (also known as an ANOVA table) is just a table. Thus, the first line saves the table as the variable modSummary the last line looks inside that variable, selects the ANOVA table ([[1]]), and then selects the row 2, column 2 value.

Method 2: Long Format

Copy and paste the following code into your R script window, then run it from there.

## Import data

yields = c(13, 4, 11, 6, 10, 11, 4, 6, 8, 13, 8, 5, 11, 5, 7, 10, 7, 14, 10, 14, 6, 7, 10, 10, 12, 8, 11, 9, 11, 7, 11, 6, 14, 5, 6, 11, 11, 8, 13, 17, 14, 10, 8)

grp = c('trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt1', 'trt2', 'trt2', 'trt2', 'trt2', 'trt2', 'trt2', 'trt2', 'trt2', 'trt2', 'trt3', 'trt3', 'trt3', 'trt3', 'trt3', 'trt3', 'trt4', 'trt4', 'trt4', 'trt4', 'trt4', 'trt4', 'trt4', 'trt5', 'trt5', 'trt5', 'trt5', 'trt5', 'trt5', 'trt5', 'trt5', 'trt5')

## Model the data

mod = aov(yields~grp)

summary(mod)

As discussed above, in the R output, the value of the sum of squares within is the number in the table under Sum Sq and to the right of Residuals. If you would like better precision for that value, or if you would like to have only that value, run the following code in addition to that above:

modSummary = summary(mod)

modSummary[[1]][2,2]

Here, the number outputted is the sum of squares between. How did you get the number? The summary table (also known as an ANOVA table) is just a table. Thus, the first line saves the table as the variable modSummary the last line looks inside that variable, selects the ANOVA table ([[1]]), and then selects the row 2, column 2 value.

Note: The difference between wide and long formats is this: In wide formatted data, each group has its own variable. In long formatted data, the group number is a variable.

\( \bar{x} \)	=	9.3488

\( \bar{x}_1 \)	=	8.25
\( \bar{x}_2 \)	=	9.3333
\( \bar{x}_3 \)	=	9.6667
\( \bar{x}_4 \)	=	9
\( \bar{x}_5 \)	=	10.8889

\( n_1 \)	=	12
\( n_2 \)	=	9
\( n_3 \)	=	6
\( n_4 \)	=	7
\( n_5 \)	=	9

© Ole J. Forsberg, Ph.D. 2025. All rights reserved.		.

Calculating the Sum of Squares Within (SSW)

The Problem

Information given:

Your Answer

Another Set of Data

Assistance

Method 1: Wide Format

Method 2: Long Format