1 One-way ANOVA

In this Workbook we deal with one-way analysis of variance (one-way ANOVA) and two-way analysis of variance (two-way ANOVA). One-way ANOVA enables us to compare several means simultaneously by using the $F$ -test and enables us to draw conclusions about the variance present in the set of samples we wish to compare.

Multiple (greater than two) samples may be investigated using the techniques of two-population hypothesis testing. As an example, it is possible to do a comparison looking for variation in the surface hardness present in (say) three samples of steel which have received different surface hardening treatments by using hypothesis tests of the form

$H_0:{\mu }_1={\mu }_2$

$H_1:{\mu }_1\ne {\mu }_2$

We would have to compare all possible pairs of samples before reaching a conclusion. If we are dealing with three samples we would need to perform a total of

${}_{}^3{C}_2=\frac{3!}{1!2!}=3$

hypothesis tests. From a practical point of view this is not an efficient way of dealing with the problem, especially since the number of tests required rises rapidly with the number of samples involved. For example, an investigation involving ten samples would require

${}_{}^{10}{C}_2=\frac{10!}{8!2!}=45$

separate hypothesis tests.

There is also another crucially important reason why techniques involving such batteries of tests are unacceptable. In the case of 10 samples mentioned above, if the probability of correctly accepting a given null hypothesis is 0.95, then the probability of correctly accepting the null hypothesis

$H_0:{\mu }_1={\mu }_2=\dots ={\mu }_{10}$

is ${\left(0.95\right)}^{45}\approx 0.10$ and we have only a 10% chance of correctly accepting the null hypothesis for all 45 tests. Clearly, such a low success rate is unacceptable. These problems may be avoided by simultaneously testing the significance of the difference between a set of more than two population means by using techniques known as the analysis of variance.

Essentially, we look at the variance between samples and the variance within samples and draw conclusions from the results. Note that the variation between samples is due to assignable (or controlled) causes often referred in general as treatments while the variation within samples is due to chance . In the example above concerning the surface hardness present in three samples of steel which have received different surface hardening treatments, the following diagrams illustrate the differences which may occur when between sample and within sample variation is considered. Case 1

In this case the variation within samples is roughly on a par with that occurring between samples.

Figure 1

Case 2

In this case the variation within samples is considerably less than that occurring between samples.

Figure 2

We argue that the greater the variation present between samples in comparison with the variation present within samples the more likely it is that there are ‘real’ differences between the population means, say ${\mu }_1,{\mu }_2$ and ${\mu }_3$ . If such ‘real’ differences are shown to exist at a sufficiently high level of significance, we may conclude that there is sufficient evidence to enable us to reject the null hypothesis $H_0:{\mu }_1={\mu }_2={\mu }_3$ .

1.1 Example of variance in data

This example looks at variance in data. Four machines are set up to produce alloy spacers for use in the assembly of microlight aircraft. The spaces are supposed to be identical but the four machines give rise to the following varied lengths in mm.

Since the machines are set up to produce identical alloy spacers it is reasonable to ask if the evidence we have suggests that the machine outputs are the same or different in some way. We are really asking whether the sample means, say ${\bar{X}}_A,{\bar{X}}_B,{\bar{X}}_C$ and ${\bar{X}}_D$ , are different because of differences in the respective population means, say ${\mu }_A,{\mu }_B,{\mu }_C$ and ${\mu }_D$ , or whether the differences in ${\bar{X}}_A,{\bar{X}}_B,{\bar{X}}_C$ and ${\bar{X}}_D$ may be attributed to chance variation. Stated in terms of a hypothesis test, we would write

$H_0:{\mu }_A={\mu }_B={\mu }_C={\mu }_D$

$H_1:$ At least one mean is different from the others

In order to decide between the hypotheses, we calculate the mean of each sample and overall mean (the mean of the means) and use these quantities to calculate the variation present between the samples. We then calculate the variation present within samples. The following tables illustrate the calculations.

$H_0:{\mu }_A={\mu }_B={\mu }_C={\mu }_D$

$H_1:$ At least one mean is different from the others

The mean of the means is clearly

$\overline{\stackrel{̄}{X}}=\frac{50+56+52+54}{4}=53$

so the variation present between samples may be calculated as

Note that the notation $S_{Tr}^2$ reflects the general use of the word ‘treatment’ to describe assignable causes of variation between samples. This notation is not universal but it is fairly common.

Variation within samples

We now calculate the variation due to chance errors present within the samples and use the results to obtain a pooled estimate of the variance, say $S_E^2$ , present within the samples. After this calculation we will be able to compare the two variances and draw conclusions. The variance present within the samples may be calculated as follows.

Sample A

$\sum {\left(X-{\bar{X}}_A\right)}^2={\left(46-50\right)}^2+{\left(54-50\right)}^2+{\left(48-50\right)}^2+{\left(46-50\right)}^2+{\left(56-50\right)}^2=88$

Sample B

$\sum {\left(X-{\bar{X}}_B\right)}^2={\left(56-56\right)}^2+{\left(55-56\right)}^2+{\left(56-56\right)}^2+{\left(60-56\right)}^2+{\left(53-56\right)}^2=26$

Sample C

$\sum {\left(X-{\bar{X}}_C\right)}^2={\left(55-52\right)}^2+{\left(51-52\right)}^2+{\left(50-52\right)}^2+{\left(51-52\right)}^2+{\left(53-52\right)}^2=16$

Sample D

$\sum {\left(X-{\bar{X}}_D\right)}^2={\left(49-54\right)}^2+{\left(53-54\right)}^2+{\left(57-54\right)}^2+{\left(60-54\right)}^2+{\left(51-54\right)}^2=80$

An obvious extension of the formula for a pooled variance gives

$S_E^2=\frac{\sum {\left(X-{\bar{X}}_A\right)}^2+\sum {\left(X-{\bar{X}}_B\right)}^2+\sum {\left(X-{\bar{X}}_C\right)}^2+\sum {\left(X-{\bar{X}}_D\right)}^2}{\left(n_A-1\right)+\left(n_B-1\right)+\left(n_C-1\right)+\left(n_D-1\right)}$

where $n_A,n_B,n_C$ and $n_D$ represent the number of members (5 in each case here) in each sample. Note that the quantities comprising the denominator $n_A-1,\cdots ,n_D-1$ are the number of degrees of freedom present in each of the four samples. Hence our pooled estimate of the variance present within the samples is given by

$S_E^2=\frac{88+26+16+80}{4+4+4+4}=13.13$

We are now in a position to ask whether the variation between samples $S_{Tr}^2$ is large in comparison with the variation within samples $S_E^2$ . The answer to this question enables us to decide whether the difference in the calculated variations is sufficiently large to conclude that there is a difference in the population means. That is, do we have sufficient evidence to reject $H_0$ ?

1.2 Using the $F$ -test

At first sight it seems reasonable to use the ratio

$F=\frac{S_{Tr}^2}{S_E^2}$

but in fact the ratio

$F=\frac{n{S}_{Tr}^2}{S_E^2}$ ,

where $n$ is the sample size, is used since it can be shown that if $H_0$ is true this ratio will have a value of approximately unity while if $H_0$ is not true the ratio will have a value greater that unity. This is because the variance of a sample mean is ${\sigma }^2∕n$ .

The test procedure (three steps) for the data used here is as follows.

1. Find the value of $F$ ;
2. Find the number of degrees of freedom for both the numerator and denominator of the ratio;
3. Accept or reject depending on the value of $F$ compared with the appropriate tabulated value.

Step 1

The value of $F$ is given by

$F=\frac{n{S}_{Tr}^2}{S_E^2}=\frac{5\times 6.67}{13.13}=2.54$

Step 2

The number of degrees of freedom for $S_{Tr}^2$ (the numerator) is

$\text{Number of samples}-1=3$

The number of degrees of freedom for $S_E^2$ (the denominator) is

$\text{Number of samples}\times \left(\text{sample size}-1\right)=4\times \left(5-1\right)=16$

Step 3

The critical value (5% level of significance) from the $F$ -tables (Table 1 at the end of this Workbook) is $F_{\left(3,16\right)}=3.24$ and since $2.54<3.224$ we see that we cannot reject $H_0$ on the basis of the evidence available and conclude that in this case the variation present is due to chance. Note that the test used is one-tailed.

1.3 ANOVA tables

It is usual to summarize the calculations we have seen so far in the form of an ANOVA table. Essentially, the table gives us a method of recording the calculations leading to both the numerator and the denominator of the expression

$F=\frac{n{S}_{Tr}^2}{S_E^2}$

In addition, and importantly, ANOVA tables provide us with a useful means of checking the accuracy of our calculations. A general ANOVA table is presented below with explanatory notes.

Define $a=$ number of treatments, $n=$ number of observations per sample.

In order to demonstrate this table for the example above we need to calculate

$S{S}_T={\sum }_{i=1}^a{\sum }_{j=1}^n{\left(X_{ij}-\overline{\stackrel{̄}{X}}\right)}^2$

a measure of the total variation present in the data. Such calculations are easily done using a computer (Microsoft Excel was used here), the result being

$S{S}_T={\sum }_{i=1}^a{\sum }_{j=1}^n{\left(X_{ij}-\overline{\stackrel{̄}{X}}\right)}^2=310$

The ANOVA table becomes

It is possible to show theoretically that

$S{S}_T=S{S}_{Tr}+S{S}_E$

that is

${\sum }_{i=1}^a{\sum }_{j=1}^n{\left(X_{ij}-\overline{\stackrel{̄}{X}}\right)}^2=n{\sum }_{i=1}^a{\left({\bar{X}}_i-\overline{\stackrel{̄}{X}}\right)}^2+{\sum }_{i=1}^a{\sum }_{j=1}^n{\left(X_{ij}-{\bar{X}}_j\right)}^2$

As you can see from the table, $S{S}_{Tr}$ and $S{S}_E$ do indeed sum to give $S{S}_T$ even though we can calculate them separately. The same is true of the degrees of freedom.

Note that calculating these quantities separately does offer a check on the arithmetic but that using the relationship can speed up the calculations by obviating the need to calculate (say) $S{S}_T$ . As you might expect, it is recommended that you check your calculations! However, you should note that it is usual to calculate $S{S}_T$ and $S{S}_{Tr}$ and then find $S{S}_E$ by subtraction. This saves a lot of unnecessary calculation but does not offer a check on the arithmetic. This shorter method will be used throughout much of this Workbook.

1.4 Unequal sample sizes

So far we have assumed that the number of observations in each sample is the same. This is not a necessary condition for the one-way ANOVA.

Task!

Three fuel injection systems are tested for efficiency and the following coded data are obtained.

Do the data support the hypothesis that the systems offer equivalent levels of efficiency?

Answer

Exercises

1. The yield of a chemical process, expressed in percentage of the theoretical maximum, is measured with each of two catalysts, A, B, and with no catalyst (Control: C). Five observations are made under each condition. Making the usual assumptions for an analysis of variance, test the hypothesis that there is no difference in mean yield between the three conditions. Use the 5% level of significance.

   Catalyst A	Catalyst B	Control C
   79.2	81.5	74.8
   80.1	80.7	76.5
   77.4	80.5	74.7
   77.6	81.7	74.8
   77.8	80.6	74.9

2. Four large trucks, A, B, C, D, are used to move stone in a quarry. On a number of days, the amount of fuel, in litres, used per tonne of stone moved is calculated for each truck. On some days a particular truck might not be used. The data are as follows. Making the usual assumptions for an analysis of variance, test the hypothesis that the mean amount of fuel used per tonne of stone moved is the same for each truck. Use the 5% level of significance.

   Truck	Obs.
   A	0.21	0.21	0.21	0.21	0.20	0.19	0.18	0.21	0.22	0.21
   B	0.22	0.22	0.25	0.21	0.21	0.22	0.20	0.23
   C	0.21	0.18	0.18	0.19	0.20	0.18	0.19	0.19	0.20	0.20	0.20
   D	0.20	0.20	0.21	0.21	0.21	0.19	0.20	0.20	0.21

Answer