TSS [ANOVA]

The Problem

Example #472: Let us test the null hypothesis that the average yield does not depend upon the treatment used. Since there are 3 treatments in this experiment, the hypotheses are

H₀ : μ₁ = μ₂ = μ₃
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:
0, -5, 1

Treatment 2:
1, 2, 1, 12, 4, 13, 3, 7

Treatment 3:
14, 5, 3, 0, -1, -2, 11, 10, 8

With this information, calculate the total sum of squares (TSS) for the data.

Information given:

To summarize the above, the values of import are:

Summary statistics from the problem
$ \bar{x} $	=	4.35

$ n_1 $	=	3
$ n_2 $	=	8
$ n_3 $	=	9

Your Answer

In the box below, please enter the total sum of squares (TSS) 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

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$$ \begin{align} \text{SSW} &= \sum_{i=1}^g \sum_{j=1}^{n_i}\ (x_{i,j} - \bar{x})^2 \\[3em] &= \sum_{i=1}^{3} \sum_{j=1}^{n_i}\ (x_{i,j} - \bar{x})^2 \\[1em] &= \sum_{j=1}^{3}\ (x_{1,j} - 4.35)^2 + \sum_{j=1}^{8}\ (x_{2,j} - 4.35)^2 + \sum_{j=1}^{9}\ (x_{3,j} - 4.35)^2 \\[1em] &= (x_{1,1} - 4.35)^2\ + (x_{1,2} - 4.35)^2\ + (x_{1,3} - 4.35)^2\ + \\ & \qquad (x_{2,1} - 4.35)^2\ + (x_{2,2} - 4.35)^2\ + (x_{2,3} - 4.35)^2\ + (x_{2,4} - 4.35)^2\ + (x_{2,5} - 4.35)^2\ + (x_{2,6} - 4.35)^2\ + (x_{2,7} - 4.35)^2\ + (x_{2,8} - 4.35)^2\ + \\ & \qquad (x_{3,1} - 4.35)^2\ +\ (x_{3,2} - 4.35)^2\ +\ (x_{3,3} - 4.35)^2\ +\ (x_{3,4} - 4.35)^2\ +\ (x_{3,5} - 4.35)^2\ +\ (x_{3,6} - 4.35)^2\ +\ (x_{3,7} - 4.35)^2\ +\ (x_{3,8} - 4.35)^2\ +\ (x_{3,9} - 4.35)^2 \\[1em] &= (0 - 4.35)^2\ + (-5 - 4.35)^2\ + (1 - 4.35)^2\ + \\ & \qquad (1 - 4.35)^2\ + (2 - 4.35)^2\ + (1 - 4.35)^2\ + (12 - 4.35)^2\ + (4 - 4.35)^2\ + (13 - 4.35)^2\ + (3 - 4.35)^2\ + (7 - 4.35)^2\ + \\ & \qquad (14 - 4.35)^2\ +\ (5 - 4.35)^2\ +\ (3 - 4.35)^2\ +\ (0 - 4.35)^2\ +\ (-1 - 4.35)^2\ +\ (-2 - 4.35)^2\ +\ (11 - 4.35)^2\ +\ (10 - 4.35)^2\ +\ (8 - 4.35)^2 \\[1em] &= (-4.35)^2\ + (-9.35)^2\ + (-3.35)^2\ + \\ & \qquad (-3.35)^2\ + (-2.35)^2\ + (-3.35)^2\ + (7.65)^2\ + (-0.35)^2\ + (8.65)^2\ + (-1.35)^2\ + (2.65)^2\ + \\ & \qquad (9.65)^2\ +\ (0.65)^2\ +\ (-1.35)^2\ +\ (-4.35)^2\ +\ (-5.35)^2\ +\ (-6.35)^2\ +\ (6.65)^2\ +\ (5.65)^2\ +\ (3.65)^2 \\[1em] &= (18.9225)\ + (87.4225)\ + (11.2225)\ + \\ & \qquad (11.2225)\ + (5.5225)\ + (11.2225)\ + (58.5225)\ + (0.1225)\ + (74.8225)\ + (1.8225)\ + (7.0225)\ + \\ & \qquad (93.1225)\ +\ (0.4225)\ +\ (1.8225)\ +\ (18.9225)\ +\ (28.6225)\ +\ (40.3225)\ +\ (44.2225)\ +\ (31.9225)\ +\ (13.3225) \\[1em] &= 117.5675\ + 170.28\ + 259.38 \\[1em] &= 560.55 \\[1em] \end{align} $$

From these calculations, the total sum of squares is TSS = 560.55. Note that if we had already calculated the between sum of squares (SSB) and the within sum of squares (SSW), then we could have used the relation TSS = SSB + SSW.

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(0, -5, 1)

treatment2 = c(1, 2, 1, 12, 4, 13, 3, 7)

treatment3 = c(14, 5, 3, 0, -1, -2, 11, 10, 8)

## Change to Long Format

mmt = c( treatment1, treatment2, treatment3 )

grp = c( rep("trt1",3), rep("trt2",8), rep("trt3",9) )

## Model the data

mod = aov(mmt~grp)

summary(mod)

In the R output, the value of the total sum of squares is the sum of the within and the between sums of squares. To have R do that calculation for you, run:

modSummary = summary(mod)

modSummary[[1]][1,2] + modSummary[[1]][2,2]

Here, the number outputted is the sum of the between and the within sums of squares. 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 sums the between ([1,2]) and within ([2,2]) sums of squares to get the total sums of squares.

Method 2: Long Format