You will remember that calculating the standard deviation is a useful way of measuring the variation within a sample of data and for then making comparisons between other data sets. Variance is always the square of the standard deviation.

There are a number of definitions used to describe ANOVA's but in essence they form a suite of tests used when we want to assess the differences between the means of several samples and to assess whether they have come from a single population. The logic used is similar to that applied to independent samples t-tests

For all ANOVA tests; data should be at least on the Interval scale* and normally distributed but there are a number of other limitations and assumptions we must first consider below.... (See also Levene's test for homogeneity below)

*Note however, that we can use the Kruskal - Wallis one-way analysis of variance for ranked (ordinal data).

ANOVA'S come in at least 5 main designs depending upon:

a) the type of survey or research design being used

b) the comparisons that are being made.

We must define whether the groups under investigation are:

The F-Ratio

The idea of the F-ratio is crucial to understanding how ANOVA works. The rationale is to check whether there is more variation ( i.e. measured variance) between the samples than within them. We will be calculating that ratio (known as the F-ratio) by using the mean squares (MS) of the 'between-groups' variance as the numerator and the 'within-groups' variance as the denominator.

We began to explore the idea of variation 'between' and 'within' back in Focus 4....

Imagine that the Britmax factories in Bournemouth and Reading were to monitor the output per employee per week for a number of months. Firstly we could examine the factories individually and we would see that the outputs vary from week to week (blue arrows). Then we could compare the variation shown between the two factories (green arrows).

[Cautionary note: the 'Boxplot' is not an ideal method of illustrating the idea of 'within' and 'between' because they are really illustrating 'range', interquartile range' and medians. Nevertheless it should convey to you the important idea that 'variation' may be thought of as existing in two forms].

Remember:

Let us imagine two cases: those where the 'between-groups' mean square figure is 12 and then when it is 6.
In both cases let the 'within-groups' figure remain at 3....

If the 'between-groups' figure is a lot larger than the 'within-group' figure, the F-ratio is going to be a large figure e.g. 12 ÷ 3 = 4 but if the variation 'between-groups' is small e.g. 6 ÷ 3 = 2.... then the F-ratio
will be smaller also.

Once again, you need to be sure that your data is Normally distributed to begin with.

We can now use tables to check up on the critical values for F that will allow us to determine the level of statistical significance that the F-ratio value obtained. Remember that a prerequisite of any significance test is that a null hypothesis and an alternative hypothesis must always be constructed.

As usual, firstly the 5% level and then the 1% level is checked. Remember also that SPSS allows
you to take the probability of F directly from the output so you are not restricted to only quoting
a 5% or 1% value.

If using tables; remember that the F-value(calc) must exceed the critical value(tab) in order to achieve significance at that level.

Because there are a number of prescribed steps before the F-ratio and the P value can be calculated, there is a strict ANOVA summary table format which you should learn to recognise and always use.

If we are comparing means why can't we simply use the t-test?

We can !! The problem is that if we had three groups to compare, A, B, and C we would have to do an AB, AC and a BC t-test. This is perhaps not unreasonable but if we had say 5 categories we would have to do 10!

The other problem is that the more t-tests we do in a set (especially once we reach 7 categories = 21 tests) , the more likely we are to obtain a statistically significant result by chance alone. If we have set our critical probability at 0.05 we are likely to reject the null hypothesis at least once for every
20 t-tests we carry out. This would be a type 1 error.

Fortunately, if we do just one ANOVA, we can simultaneously compare all our means and the overall probability of incurring a Type 1 error stays the same.

You should also read the section on Tukey's test below...

Regression ANOVA's

Before dealing directly with one-factor ANOVA's, it will be useful to go back and consider our regression graphs once more where just two variables are involved (Focus 11). Regression ANOVA's look at the strength of the relationship between two variables i.e. just one factor is 'in play'.

Regression ANOVA's are essentially dealing with 2 factor variables and a single dependent variable. We have stated above that a very useful way to compare any two (or more) groups is to look at the variance ratios and to produce a value for the F-ratio.

Study the chart below and make sure that you can see why there are three elements to the variation displayed. There are three particular values involved:

• the point value itself (the blue diamond) at (4, 1)
• the point on the 'line of best fit' (on the dark blue line), this would be the predicted value for 'y' when 'x' is 4 and has a value of 1.4
• the actual value of y at this point is 1

The mean value of y is 3.4 and the pale blue line indicates this position. We are interested to know how that variation (of 2.4) is made up and the chart below explains this.

Naturally this ratio will not apply to the other points on the graph as they are all at different distances from the 'line of best fit' and the horizontal mean of Y. So we need to explore the variation that exists between the points as well as the variation within or unique to each individual point.

We must always start with a null hypothesis:

"The variation in 'y' is not accounted for by the variation in 'x' "

and the alternative hypothesis:

"The variation in 'y' is accounted for by the variation in 'x' "

The calculation:

The first task is to square all true variation values (for each point) in order to make them positive.

Manual method:

Once again, the formulae involved are lengthy rather than complex and so it is wise to begin with a
full tabulation of all that is needed before attempting any calculations because it is easy to muddle the figures.

Although not essential to the calculation, remember that a plot of the data (using Excel in this case) may help to clarify things....

Study the various distances that each point is away from the regression line (only with regards to the 'Y' axis). Note also that, as with any 'line of best fit', explained variations (within each point), when all added together will equal zero.

Check....

The question is: How much of the variation (fourth column above) shown in 'y' can be accounted for by the variation in 'x'

First find: SS (total)..... and then SS (regression), take one from the other to give SS (residual).

You also need to record: n = 5, mean of 'x' = 5.4 and the mean of 'y' = 3.4

So this is all the information needed to undertake a one factor ANOVA other than the formulae themselves:

So: SS(residual) = 33.2 - 26.13 = 7.07.

The final stage is to fill out the ANOVA table which always follows the convention described earlier:

In our case (where n = 5) this becomes:

So our final result gives n F-ratio(calc) of 11.088

* We now look up in the F values table for Regression df: 1 and Residual df: 3 and we find a
figure 10.1 for P = 0.05 and another figure of 34.1 for P =0.01

Our test result (11.088) is higher than 10.1 and so we cannot accept the null hypothesis at the P(0.05) level but the situation is different at the P (0.01) level as we have not exceeded 34.1.

So we can say that a statistically significant amount (P(0.05)) of the variation in 'y' is accounted for by the variation in 'x'.

Now using SPSS....

Note that the P value is given to an exact value (Sig : .045) showing that we have exceeded the critical
value for P(0.05)but not any greater level of significance. Note also that SPSS does not list what that critical value is (it is not necessary).
To discover what the critical value is, we would still have to use tables.

From the manual working above you will recall that for P(0.05), it was 10.1.

We also have the necessary values to complete the regression formula for this particular line:

y = - 3.256 + 1.233x or re-arranged as y = 1.233x - 3.256

As a check, we know that any 'line of best fit' must pass through the mean of 'x' and the mean of 'y'.
The mean of all the 'x' values is 5.4 so substitute this value in the formula.

y(mean) = 6.658 - 3.256 = 3.402, the mean of 'y' was also found (from the data set) to be 17÷ 5 = 3.4

Correct!

Let us next carry out a regression ANOVA fully using SPSS...

The Daily Mail, 'money page' has carried out a comparison survey of Life Assurance premiums. They asked 6 leading providers to give quotations for healthy males aged 40, 50, 60 and 70, all living in the same location and with similar jobs. How can we best compare the quotations?

We can quickly produce a 'line of best fit' but just how valid is the inference? You might say that if all the points fall close to the line then this will give us a good indication (and this is true) but it is not very scientific. ANOVA will give a more precise result.

Task:Construct your null and alternative hypotheses.

The data set is:SPex59 Life Assurance ANOVA

Open SPSS in the usual way, the two variables are 'age' and '£quoted' , produce a simple scattergraph and then superimpose the regression line...

Note that we are treating all the quotations as though they belong to one group.

Q. What effect do you think this might have on the result?

Hint: A one-way ANOVA would be even more accurate (as described later).

Go to 'Analyse', 'Regression', 'Linear' & tick 'Statistics' (this gives the descriptives i.e. means and s.d)

The output involves 5 frames but here are the two most important that you should have generated....

The formula for the line is: y = 20.85 + 3.38x

Task: Check that this is correct (hint: you will need to obtain the mean of x and y )

Task:The F- ratio is shown to be 43.571, how would you find the critical values for P at both 5% and 1%.

Explain the SPSS output for P(sig)

Q. Why can we not simply use an Independent samples t-test on this dataset?

(Method 2) One-way ANOVA's

If we now carry out the alternative one-way ANOVA test on the same data set ....

Open the dataset as above: Go to 'Analyse', 'Compare Means', 'One-way ANOVA'.

Transfer 'premium quoted' to the dependent box and 'age of client' to the factor box

Click 'options' and tick the 'Descriptives' and 'means plot' boxes.... also tick the 'Homogeneity-of-variance' box (see note
below concerning the Levene statistic)

Click 'continue' and then 'OK'.....

We get the same result in terms of the level of significance but a much lower (more stringent) F-value of 13.308. This is because we are now being more precise and treating the data as 4 separate (age) groups rather than one large one as we were with Regression ANOVA.

A One-Factor Independent-Groups ANOVA example..

Multigym International have many fitness clubs throughout the UK but 110 branches are situated in busy city centres and cater for lunchtime users mainly. They are currently testing 3 three new makes of 'executive' rowing machines ('Beacon','Prima', and 'Chalice') to see if there is any real differences between the three. Before buying these very expensive machines it is important to know which allows the fastest 'turnaround' time because this is a crucial element in the selection process. Timers were attached to each machine and the results analysed. Of course, there may be no discernible differences between the three.

6 branches were to chosen to participate and each had each of the three machines installed in turn.Thus there are 18 results to consider...

Here is the basic data set... SPex 75 Multigym

(figures in brackets are that value squared)

These results present the company with a dilemma because the mean turnaround times seem so similar but can we be sure?
Can we say that statistically, there is no real difference between the three or is there a significant difference?

Task: Construct your own Ho and H1.

The calculations should be done manually to follow the progress of the calculation:

Grand total time: 23.59 mins.......grand mean (divide by 18) = 1.311 mins

Now calculate the square of each of the 18 values (already shown in brackets above) and sum them. 1.613 + 1.769 + 2.132 etc etc =.... this gives a total of 31.110

Thus SS (total) would appear to be 31.110 but we will first need to incorporate a correction factor (see second box below)..

Now we need to find the final expression for SS (total) and this is explained below:

At this point we can say that the sum of squares (between) + the sum of squares (within) must add up to 0.194

Now for the SS (between-groups)

Square the sum of each group and then divide the answer by the n for that group...

'Beacon' .......7.86 squared = 61.780 divided by 6 = 10.2966

'Prima' ...........8.35 squared = 69.723 divided by 6 = 11.6205

'Chalice' ........7.38 squared = 54.464 divided by 6 = 9.0774

all summed = 30.9945 (i.e 30.995)

Once again, subtract the (already calculated) correction factor (30.916) = 0.079

Finally, the SS(within-groups) must be the difference i.e. 0.194 - 0.079 = 0.115

Let us follow this same method in a mathematical layout...

We now have all the details necessary to construct and complete the standard ANOVA table....

n =18

From tables (table D7, p222 in Wheater & Cook) we find that P(0.05)(d.f.2,15) the critical value for F is 3.68 and at P(0.01)(d.f.2, 15) it is 6.36

Remember, results are significant if our calculated value (5.15) is greater than the tabulated value. Compare your conclusion to that shown in the SPSS output below....

Task: Write down the conclusion for these results and relate them back to H1 and H0.

The same calculation / output using SPSS / One-way ANOVA...

Enter the names and labels in the usual way and then enter the data or use: SPex 75 Multigym

Now go to 'Analyse', 'Compare Means', 'One-way ANOVA'.

Transfer 'time ' to the dependent box and 'make of machine' to the factor box

Click 'options' and tick the 'Descriptives' and 'means plot' boxes....
also tick the 'Homogeneity-of-variance' box (see note below concerning the Levene statistic)

Click 'continue' and then 'OK'.....

The difference in the F-ratio between the manual (5.15) and SPSS (5.128) methods is simply a rounding difference.

Once again, SPSS gives a direct P value (.020) for the significance of our F-statistic. Compare this with the result obtained via the manual method.

Task: Indicate which machine is 13% faster than which other machine.

Q. Which machine would you advise the Company to include in their buying policy?

A note about Levene's test for Homogeneity

Homogeneity could be defined as "the level of 'equality' between the samples".

The Levene test examines whether the variance (variation) of values for the groups is generally the same. It does not have anything to do with disparity of sample size. In SPSS the outcome of the test will determine if there is or is not evidence of homogeneity or the opposite state; heterogeneity. The P-value for the Levene statistic is significant if the output shows it to be less than 0.05 and in those circumstances; equal variance must not be assumed.

Here is the output from the multigym example, the Levene statistic is .824 and is therefore not significant, so equal variance can be assumed....

Tukey's test

Post hoc tests are carried out once the issue of significance has been established. After all, there is little point in checking on the 'fine tuning' of a significance output if the result is not significant in the first place! If there is no significant difference detected, then that is the end of the analysis.

The test is designed to make adjustments that help to lower the risk of making a Type 1 error.
(Type 1 and 2 errors are important and are fully explained in Focus 4)

If there are only two means involved and the results indicate a significant difference between the two; then the answer is simple, "the larger mean is significantly larger than the smaller one "!

Unfortunately, we cannot say this when there are 3 or more means involved. There will be several possible scenarios and to determine exactly where the differences lie, we need to carry out a comparison test.

Tukey's test assumes equal variance. The test will compare all possible combinations of pairs of means:

Be aware however, that the more tests we do,the more likely we are to falsely reject a null hypothesis and claim a significant difference when in fact there is none....a Type 1 error . If we do nothing (that is leave the calculation at the point where we have concluded an overall significant difference, we are in danger of committing a Type 2 error.

Tukey's test is often referred to as yielding the HSD (honestly significant difference) test and is a compromise between the two risks outlined above.

Let us return to the three makes of rowing machines ....

Go back to the SPSS menu and carry out the procedure as before but this time also click on the 'post hoc' box and select the 'Tukey test'...

and here is the output...

This chart shows that there is a significant difference between the 'Prima' and 'Chalice' machines
P= (0.015), but no other comparisons of the means showed a significant difference. This is made clearer in the chart below; each machine has been slotted into a group with any of its homogeneous partners and again it can be seen that because 'Chalice' and 'Prima' have been found to be significantly different (at least as far as 'set up time' is concerned), they have been placed in separate subsets by SPSS.....

Q. Are these findings borne out by the boxplot we produced earlier?

Task: Carry out the next analysis yourself....

The trial was conducted again the following year on a slightly larger scale and now 5 machines were tested including two new computerised 'hi tech' versions; 'Saturn' and 'Geneva 3'.

Use:SPex 76 Multigym 2 to carry this analysis >>>>

Remember to construct an H0 and H1 before you begin.

Task: produce your own boxplot as below........

Task: Carry out a One-way ANOVA and the post hoc (Tukey's) test

Task: Explain your results in such a way as to be comprehensible in a Powerpoint presentation to a group of branch managers.

Britmax are evaluating three new robot welding machines from three different manufactures for use in constructing their bike frames. The assembly team was asked to turn out 6 frames from each machine and to record exactly the time it took to complete each frame.

The results are recorded in SPex62 ANOVA bike frames. After carrying out the work, the results are analysed using a one-way ANOVA test (as above). There is found to be a significant difference (P-value 0.007) between the three machines in terms of construction times.

Remember to construct an H0 and H1 before you begin.

Q. Which is the fastest and which is the slowest machine ?

Task: carry out the SPSS One-way ANOVA analysis and include a post hoc Tukey's test to determine between which machines there is a significant difference and at what level.

Go back to Contents Page

Go on to Focus 13a

Go back to Focus 12