One-Way ANOVA

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Calculating the Sum of Squares Within (SSW)

The within-sample sum of squares (SSW) is a measure of the remaining variablility in the data after applying the model. It is also a non-standardized measure of how well the model fits the data. Larger values of SSW indicate the model fits the data worse, all things being equal. The SSW is sometime called the sum of square due to error (SSE) by some sources.

As usual, these calculations refer only to the one-way completely randomized design, the most basic of sample designs.

The Problem

Example #484: Let us test the null hypothesis that the average yield does not depend upon the treatment used. Since there are 4 treatments in this experiment, the hypotheses are

H₀ : μ₁ = μ₂ = μ₃ = μ₄
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:
20, 17, 10, 15, 14, 21, 12, 1, 14, 20, 15, 1

Treatment 2:
4, 5, 21, 5, 15, 5, 6, 15, 23

Treatment 3:
24, 16, 29, 19, 17, 18, 9, 0

Treatment 4:
16, 13, 15, 7, 38, 5, 20, 16

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
$\bar{x}$	=	14.0811

$\bar{x}_1$	=	13.3333
$\bar{x}_2$	=	11
$\bar{x}_3$	=	16.5
$\bar{x}_4$	=	16.25

$n_1$	=	12
$n_2$	=	9
$n_3$	=	8
$n_4$	=	8

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:

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Assistance

Show Formula

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Here is the formula you can use to calculate the within sum of squares.

$\text{SSW} = \sum_{i=1}^g \sum_{j=1}^{n_i}\ (x_{i,j} - \bar{x}_i)^2$

In this formula, there are a few symbols to know:

$g$

the number of groups; here, that is 4

$n_i$

the sample size for group $i$

$x_{i,j}$

the value of data point $j$ from group $i$

$\bar{x}_i$

the sample mean of group $i$

$\sum$

summing everything in the expression to the right

Show Solution

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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}^{4} \sum_{j=1}^{n_i}\ (x_{i,j} - \bar{x}_i)^2 \\[1em] &= \sum_{j=1}^{12}\ (x_{1,j} - 13.3333)^2 + \sum_{j=1}^{9}\ (x_{2,j} - 11)^2 + \sum_{j=1}^{8}\ (x_{3,j} - 16.5)^2 + \sum_{j=1}^{8}\ (x_{4,j} - 16.25)^2 \\[1em] &= (x_{1,1} - 13.3333)^2\ + (x_{1,2} - 13.3333)^2\ + (x_{1,3} - 13.3333)^2\ + (x_{1,4} - 13.3333)^2\ + (x_{1,5} - 13.3333)^2\ + (x_{1,6} - 13.3333)^2\ + (x_{1,7} - 13.3333)^2\ + (x_{1,8} - 13.3333)^2\ + (x_{1,9} - 13.3333)^2\ + (x_{1,10} - 13.3333)^2\ + (x_{1,11} - 13.3333)^2\ + (x_{1,12} - 13.3333)^2\ + \\ & \qquad (x_{2,1} - 11)^2\ + (x_{2,2} - 11)^2\ + (x_{2,3} - 11)^2\ + (x_{2,4} - 11)^2\ + (x_{2,5} - 11)^2\ + (x_{2,6} - 11)^2\ + (x_{2,7} - 11)^2\ + (x_{2,8} - 11)^2\ + (x_{2,9} - 11)^2\ + \\ & \qquad (x_{3,1} - 16.5)^2\ + (x_{3,2} - 16.5)^2\ + (x_{3,3} - 16.5)^2\ + (x_{3,4} - 16.5)^2\ + (x_{3,5} - 16.5)^2\ + (x_{3,6} - 16.5)^2\ + (x_{3,7} - 16.5)^2\ + (x_{3,8} - 16.5)^2\ + \\ & \qquad (x_{4,1} - 16.25)^2\ +\ (x_{4,2} - 16.25)^2\ +\ (x_{4,3} - 16.25)^2\ +\ (x_{4,4} - 16.25)^2\ +\ (x_{4,5} - 16.25)^2\ +\ (x_{4,6} - 16.25)^2\ +\ (x_{4,7} - 16.25)^2\ +\ (x_{4,8} - 16.25)^2 \\[1em] &= (20 - 13.3333)^2\ + (17 - 13.3333)^2\ + (10 - 13.3333)^2\ + (15 - 13.3333)^2\ + (14 - 13.3333)^2\ + (21 - 13.3333)^2\ + (12 - 13.3333)^2\ + (1 - 13.3333)^2\ + (14 - 13.3333)^2\ + (20 - 13.3333)^2\ + (15 - 13.3333)^2\ + (1 - 13.3333)^2\ + \\ & \qquad (4 - 11)^2\ + (5 - 11)^2\ + (21 - 11)^2\ + (5 - 11)^2\ + (15 - 11)^2\ + (5 - 11)^2\ + (6 - 11)^2\ + (15 - 11)^2\ + (23 - 11)^2\ + \\ & \qquad (24 - 16.5)^2\ + (16 - 16.5)^2\ + (29 - 16.5)^2\ + (19 - 16.5)^2\ + (17 - 16.5)^2\ + (18 - 16.5)^2\ + (9 - 16.5)^2\ + (0 - 16.5)^2\ + \\ & \qquad (16 - 16.25)^2\ +\ (13 - 16.25)^2\ +\ (15 - 16.25)^2\ +\ (7 - 16.25)^2\ +\ (38 - 16.25)^2\ +\ (5 - 16.25)^2\ +\ (20 - 16.25)^2\ +\ (16 - 16.25)^2 \\[1em] &= (6.6667)^2\ + (3.6667)^2\ + (-3.3333)^2\ + (1.6667)^2\ + (0.6667)^2\ + (7.6667)^2\ + (-1.3333)^2\ + (-12.3333)^2\ + (0.6667)^2\ + (6.6667)^2\ + (1.6667)^2\ + (-12.3333)^2\ + \\ & \qquad (-7)^2\ + (-6)^2\ + (10)^2\ + (-6)^2\ + (4)^2\ + (-6)^2\ + (-5)^2\ + (4)^2\ + (12)^2\ + \\ & \qquad (7.5)^2\ + (-0.5)^2\ + (12.5)^2\ + (2.5)^2\ + (0.5)^2\ + (1.5)^2\ + (-7.5)^2\ + (-16.5)^2\ + \\ & \qquad (-0.25)^2\ +\ (-3.25)^2\ +\ (-1.25)^2\ +\ (-9.25)^2\ +\ (21.75)^2\ +\ (-11.25)^2\ +\ (3.75)^2\ +\ (-0.25)^2 \\[1em] &= (44.4444)\ + (13.4444)\ + (11.1111)\ + (2.7778)\ + (0.4444)\ + (58.7778)\ + (1.7778)\ + (152.1111)\ + (0.4444)\ + (44.4444)\ + (2.7778)\ + (152.1111)\ + \\ & \qquad (49)\ + (36)\ + (100)\ + (36)\ + (16)\ + (36)\ + (25)\ + (16)\ + (144)\ + \\ & \qquad (56.25)\ + (0.25)\ + (156.25)\ + (6.25)\ + (0.25)\ + (2.25)\ + (56.25)\ + (272.25)\ + \\ & \qquad (0.0625)\ +\ (10.5625)\ +\ (1.5625)\ +\ (85.5625)\ +\ (473.0625)\ +\ (126.5625)\ +\ (14.0625)\ +\ (0.0625) \\[1em] &= 484.6667\ + 458\ + 550\ + 711.4375 \\[1em] &= 2204.1667 \\[1em] \end{align}$

From these calculations, the within sum of squares is SSW = 2204.1667.

Show the R Code

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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(20, 17, 10, 15, 14, 21, 12, 1, 14, 20, 15, 1)

treatment2 = c(4, 5, 21, 5, 15, 5, 6, 15, 23)

treatment3 = c(24, 16, 29, 19, 17, 18, 9, 0)

treatment4 = c(16, 13, 15, 7, 38, 5, 20, 16)

## Change to Long Format

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

grp = c( rep("trt1",12), rep("trt2",9), rep("trt3",8), rep("trt4",8) )

## 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(20, 17, 10, 15, 14, 21, 12, 1, 14, 20, 15, 1, 4, 5, 21, 5, 15, 5, 6, 15, 23, 24, 16, 29, 19, 17, 18, 9, 0, 16, 13, 15, 7, 38, 5, 20, 16)

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', 'trt3', 'trt3', 'trt4', 'trt4', 'trt4', 'trt4', 'trt4', 'trt4', 'trt4', 'trt4')

## 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]

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.

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