Up to this time, we have said that if you do NOT have homogeneity of variance and you do NOT have equal Ns, you could not do ANOVA:




What we have said is that if you do not have homogeneity AND you do not have equal n's, then you should do a nonparametric analysis.

If you have 2 means - Mann-Whitney U.

If you have more than 2 means - Kruskal-Wallis H.

HOWEVER, there IS another alternative.

You can try transforming the data by doing something mathematical to each score.


Sometimes, but not always, such a transformation will bring about homogeneity.

Another good thing is that if you find a transformation that will bring about homogeneity, it will probably also bring about normality, if the distribution lacks normality, which is another assumption of ANOVA.

In fact, usually, if a transformation causes an unmet assumption to be met, it usually will cause us to meet all the other unmet assumptions.


Remember that normality is also an assumption of ANOVA, t-tests, and other parametric tests.

If a transformation can't be found to bring about homogeneity or normality, then your only alternative may be to go to a nonparametric test that does not require homogeneity.


You already know how to check for homogeneity of variance.

We do Levene's test - if it is significant, there is no homogeneity of variance. Even a significance level of .10 is cause for concern.

But, how can we check for normality?

There are statistical tests and there are visual methods involving plots.

There are several statistical tests.

You have probably already seen a couple of simple ones - skew and kurtosis.

When a distribution is normal, the values for skewness and kurtosis are both zero.

If a distribution is positively skewed (cases cluster to the left and right tail is extended with only a few cases), then the statistic for skew will be greater than zero.

If it has negative skew, skewness will be less than zero.

If a distribution is too peaked (leptokurtic), the kurtosis value will be positive.

If it is too flat, with many cases in the tails (platykurtic), the kurtosis value will be negative.

There are tables of critical values for skew and kurtosis, and if you use these, do it at the .01 or .001 for small to moderate size samples.

If not, a rule of thumb is that if both kurtosis and skew is between +1 and -1, then we have normality.

If you have a really large sample, don't worry too much - ANOVA is robust to violation if the sample is large.

SPSS will give you kurtosis and skew, and it will also give you some better statistics

For normality, you can get the Kolmogorov-Smirnov test and the Shapiro-Wilks test.

The Kilmogorov-Smirnov test will have the Lilliefors modification with small samples.

Shapiro-Wilks is for distributions with fewer than 50 cases.

It tests the null hypothesis that the population is normally distributed.

So, if the test is significant, it means the variable is probably not normally distributed.


In SPSS we check both homogeneity and normality using the EXPLORE routine followed by a PLOT.

Here are some scores:

          Group 1     Group 2     Group 3

               3                 6                12

               0                 4                  6

               4                 2                  6

               2                 4                10

               2                 7                  6

Mean     2.2             4.6               8.0

Var        2.2             4.4               8.0

This is a POISSON distribution we'll talk more about that later.

 Note that in each group, the mean is just about exactly equal to the variance - typical in POISSON distributions.

We can check this for normality and for homogeneity with the EXPLORE routine in SPSS:

_____ 1. Put in the scores as usual, with all in the same column and with another variable as a dummy variable to identify the group:

_____ 2. Now click on ANALYZE - DESCRIPTIVE STATISTICS and EXPLORE. The EXPLORE box will open. Make it look like this::

_____ 3. Make sure that under DISPLAY in the lower left of the screen, BOTH is marked.

_____ 4. Click the PLOTS button at the bottom of the box. The following EXPLORE: PLOTS box will open:

_____ 5. Make the box look like the one pictured above. Then, click Continue and then OK. The EXPLORE analysis will run. The output will appear. Here is a copy:

HANDOUT - transformations1.spo

There are several things we can look at here - all having to do with the assumption of normality.

With regard to normality, look at page 1.

Both skewness and kurtosis for the SCORES variable fall between -1 and +1.

So we suspect it is from a normal population.

They are both positive, suggesting that it is slightly positively skewed and slightly leptokurtic.


In the middle of page 1, you will find the statistical test for normality.

Look under Kilmogorov-Smirov and you will see that the normality is .200 and under Shapiro-Wilk it is .392.

These are not significant, so the variable is normal, although less so than ideal..

On page 2 is the histogram, and you will see the visual evidence for this - slightly skewed positively and slightly peaked, but not too far from normal.


Look at page 2 of the printout.

There, you will find Q-Q Plots for each group.

These are called normal probability plots, or normal Q-Q plots.

The observations are listed from low to high and plotted against the expected values if the distribution were normal.

Observed values are along the x axis and the predicted values from a normal dist. are on the y axis.

If the distribution is normal, the plot should resemble a straight line.

The detrended normal Q-Q plot on page 3 is a plot of the amount each score varies from the straight line.

In these, a normal distribution would have every score falling on a horizontal line that passes through 0.


Page 3 has a boxplot of the data.

The median is a dark line.

The length of the box indicates the variability, because the 75th percentile is the top of the box, and the 25th percentile is the bottom of the box.

Therefore, the length of the box shows the interquartile range (the middle 50% of the scores).

Lines are drawn from the edge of the box to the largest and smallest values that are not outliers (outliers are cases between 1.5 and 3 boxlengths from the edge of the box, and extremes are more than 3 boxlengths away.)

The length of the box is a visual depiction of one measure of variability - the interquartile range.

The longer the box, the more variability.


This variable does not depart significantly from normality.

If it does, there are a large number of different transformations that are possible to try to get the distribution to be normal.

There is not a whole lot that can be done with normality with so few scores, but there is a possibility of transforming even these few scores to get more homogeneity of variances.

Thee are many different transformations that are possible.


One of the simplest transformations is a square root transformation.

To do this one, you simply take the square root of each of your raw data scores, then do an ANOVA on the transformed scores.

This is a transformation that is often also used when a distribution lacks normality or homogeneity.

It is best for bringing about normality when a distribution departs only moderately from normality.

And, use if the skew is positive.

In addition, this transformation is almost always the best one to use when dealing with counts of rare events - insects on a leaf for example.

Such a distribution is called a POISSON distribution.

For example, suppose that you know that a seed manufacturer is selling seed for which 0.1% is assumed will be dead.

That means that the probability of any one seed being dead is 1 in 1000 - that is quite rare.

If you take 100 samples of 1000 seeds each, you will get:

37 samples with no dead seeds

37 samples with 1 dead seeds

18 samples with 2 dead seeds

  6 samples with 3 dead seeds

  2 samples with 4 dead seeds

If your raw data is the number of dead seeds in each 1000-seed sample, then that distribution will probably be a poisson distribution.

The above is not a normal distribution by any stretch.

It is a POISSON distribution, in which, by definition, the variance is equal to the mean (or a little larger).

You can often normalize such a distribution, and make the variances homogeneous, by transforming each score to square roots.

Just take the root of each score.

If you have any scores less than 10, then use the root of X +1 or root of X + .5.


But, so far, we have not tested the variances for homogeneity.

That is because in step #3 above, we did not identify a FACTOR so that SPSS could break the scores into the three groups of five scores each.

When we were checking for normality, we were evaluating the entire distribution.

Let's go back now and identify a factor in step #3 above. Repeat the first few steps, then here is the EXPLORE box again:

Note that we moved group into the FACTOR LIST field.

_____ 4. Now, click on the PLOTS button. The PLOTS box will open:

_____ 5. This time, make the PLOTS box look like the above. Notice that you should check the NORMALITY PLOTS WITH TESTS checkbox and the UNTRANSFORMED radio button should be chosen.

_____ 6. Click CONTINUE and OK. The output will be produced:

HANDOUT - transformations2.spo

This will give us statistics on each group separately, and it will also give us Levene's test for homogeneity of variance.

You will find it on page 3.

It is .101, which is in the "nervous" category, although not significant.

This is exactly what you would have found had you run a one-way ANOVA.


Now, we can consider some different transformations.

Let's see what the effect would be on homogeneity of taking the root of each score.

If you have any scores less than 10, then use the root of X +1 or root of X + .5.

Since we do have scores less than 10, we should do that.

Let's use the root of X + .5.

How do we do that?

_____ 1. Return to the data and click on TRANSFORM and COMPUTE. The following box will appear:

What we have to do is name a new "TARGET VARIABLE" in the upper left of the box and write an equation to produce that new variable in the NUMERIC EXPRESSION field.

Since it will be the root of each score after adding .5, we will call it rootplh, or root plus half a point.

_____ 2. Type rootplh in the TARGET VARIABLE FIELD.

_____ 3. Scroll down the list of functions in the FUNCTIONS box until you find the one for square roots. It is called SQRT(numexpr). (Right click on any function to get a definition.) Click on it to highlight it and then click the UP ARROW button just above this list of functions. This will move the function up into the NUMERIC EXPRESSION field. Now the COMPUTE VARIABLE box should look like this:

Notice the question mark inside the function you have chosen. SPSS is asking what you want to take the square root of.

_____ 4. Since we want to take the square root of whatever number is in the SCORES variable, but only after adding .5 to each score, we will highlight SCORES in the bottom left field and click on the RIGHT ARROW button to move it in to replace the question mark.

_____ 5. Now, type a plus sign (or select the symbol on the pictured keypad) followed by .5 (or select .5 on the pictured keyboard). The box should look like this:

_____ 6. Now, click OK. You will be returned to the data and you will see the new variable:

_____ 7. Now, run a one-way ANOVA on this data:

Here are the results:

Note that Levene's sig. is now .917 - much better than before, when it was .101


How do you write it up?

Explain what you did.

You can either report the untransformed means, but the analysis of transformed means, or you can de-transform the means to get weighted means. Do this by squaring the transformed means and subtracting .5. You won't get the same means you got from raw scores, however. These are weighted means and you should refer to them as such.

There are many other transformations. Mertler and Vennatta (2001) (Advanced and Multivariate Statistical Methods) have a useful chart showing which transformations are often best for distributions with varying skew.


 Moderate positive skew, as we have already seen, should first try a square root transformation as we did above. The SPSS compute function is SQRT(var), or, if any values are below 10, SQRT(var + .5)

Substantial positive skew should cause us to try a logarithm transformation. This is usually best if the means and standard deviations tend to be proportional. The SPSS function is LG10(var). If any values are negative or 0, add 1 to each data point (or any constant to bring the smallest value to at least 1) before transforming. LG10(var + 1). If any data points are less than one, you can avoid negative logarithms by multiplying all data points by a constant. So, if the lowest value is 0.13, you could multiply by 10: LG10(var*10).

Severe positive skew should cause us to try an inverse, or reciprocal transformation. This will work well in distributions where the square of treatment means and standard deviations are proportional: 1/var will work if there are no zeroes in the data. If there are zeroes use 1/var + 1 or 1/(var + C), where C is a constant added to each score to make the smallest value be at least 1.

Moderate negative skew should cause us to try a transformation in which we "reflect" the variable and then try one of the transformations above. To "reflect" means find the largest score, add 1 to it, and the result is a constant that is larger than any score in the distribution. Now, subtract each score from the constant. (This will make the distribution be distributed positively, and so one of the above transformations would then work.) The SPSS function for reflect followed by square root transformation is SQRT(K - var) when Kis the constant formed as shown above. Using the constant will make it so that the smallest score equals 1.

For substantial negative skew, try reflect followed by logarithm. Reflect as shown in the previous case, then take the log of the new data. SPSS function is LG10(K - var)

Severe negative skew should cause us to try reflect followed by inverse transformation. Reflect as usual, then take the inverse of the new data. SPSS function is 1/K - var)

There are many other transformations that are possible.  Tabachnick and Fidell (1996) have a good section on transformations with lots of examples.

At this point, we should probably review logarithms.


What are logarithms?

A logarithm is a number that represents an exponent.

You know what an exponent is - it tells us how many times to multiply a number times itself.

So 103 = 1000

The abbreviation for a logarithm is LOG.

The common logs are base 10 logs, but logs can be to any base.

What does that mean?

The base 10 log of 1000 is the number of times 10 must be multiplied by itself to equal 1000 - so the base 10 log of 1000 is 3.


Because 10 times 10 times 10, or 103 = 1000.

We can write it like this:

log10 1000 = 3 or, sometimes log10 (1000) = 3

Remember, logs can be to any base.

The base 5 log of 25 is the number of times 5 must be multiplied by itself to equal 25 - so the base 5 log of 25 is 2.

log5 25 = 2 or, sometimes log5 (25) = 2

2 is the logarithm of 25 to base 5.

What is the log of 81 to base 3? 4

What is the log of 125 to base 5?  3

What is the log of 64 to base 8? 2

What is the log of 8 to base 2? 3

When doing transformations, you can use any base, but base 10 is probably the most convenient.


HANDOUT - transformations.first.lec.after.factorial.homework1.doc



Often, you will have no alternative but to give up on a parametric test such as ANOVA.

If there is no homogeneity and you can't achieve it through data transformation, that is about your only choice.

You will have to find a nonparametric test that will work for your data.

Some people call these tests "assumption-free," although that is really not accurate.

Unfortunately, these tests are, in general, not as powerful as are parametric tests.

Many nonparametric tests (but not all) work by RANKING THE DATA.

What we often do is give the lowest score a rank of one, etc.

As you can appreciate, a low score will have a low rank, a high score a high rank.

Then, we do our analysis on the RANKS, not on the raw data.

You can probably see why nonparametric tests have less power - we lose some data on the magnitude of the differences between scores when we rank them.

However, it gets us around the controversy about not being able to use parametric tests with data that is not true interval or ratio data.

No one argues that nonparametric tests are inappropriate for ordinal data.


The Mann-Whitney U is the nonparametric test we often use if we want to run a t-test but cannot because we lack normality, homogeneity, or because the data is not interval or ratio data.

This test is also sometimes called the Wilcoxon W Test, and even sometimes the Mann-Whitney-Wilcoxon.

Let's use it on the first two groups we looked at in your homework: (transformations.first.lec.after.factorial.sav)

Exp.     Control 

1.00        1.00

4.00        2.00

7.00        3.00

10.00      4.00

13.00      5.00

16.00      6.00

19.00      7.00

22.00      8.00

25.00      9.00


We will first run a t-test to investigate homogeneity of variance and so you can compare the Mann-Whitney results to independent t-test results.

Here is the t-test:

Note the means. It appears that the experimental group outperformed the control group means are 14.5 to 5.00. But note LEVENE's Test. There is no homogeneity and no equal n's. Notice the significance is .008 or .009 depending on which line you read.

Now the Mann-Whitney U. You will find it by clicking ANALYZE, then NONPARAMETRIC TESTS, then 2 INDPENDENT SAMPLES.

When the next box opens, make it look like this:

Be sure that Mann-Whitney U is checked.

Click the OPTIONS button and choose DESCRIPTIVE STATISTICS.

Click OK to run the analysis.

Here is the output:

Look at the RANKS box. Remember that Mann-Whitney ranks data from low to high. So, the group with the highest mean rank has the highest scores and vice versa.

The Test Statistics box has the result. Mann-Whitney U and Wilcoxon W yield different statistics, but the same results. Both appear here. The asymptotic significance assumes large samples and corrects for ties in ranks. The exact significance does not correct for ties, but is for small groups. They will usually be very close, as they are here. If they aren't the same, use Exact Sig., because our samples are small and because a correction for ties will always make it MORE SIGNIFICANT, not less.

So, the results here are the same as with the t-test - the experimental group has higher scores. (Null relates to median.) But, note that the t-test sig. was .008 while the Mann-Whitney sig. was .017, illustrating the greater power of the t-test.

BUT, if critics are right that t-test requires interval or ratio data, and if our data is ordinal, then the Mann-Whitney is the better and more appropriate test regardless.

Homework: copy these scores and do both a t-test and a Mann-Whitney. Draw conclusions.

experimental: 28, 35, 35, 24, 39, 32, 27, 29, 36, 35

     control: 05, 24, 06, 14, 09, 07, 17, 06, 03, 10

END - Kruskal-Wallis Next Time