I'm analyzing results from an experiment where 56 samples were tested for a specific response to electrical stimulation. Electrical stimulation was done at 2 different stimulation frequencies, and at 3 different time points. 9, 10, and 9 samples were stimulated on day 7, 10, and 14 respectively at frequency 1, and the same was done at frequency 2. Each of the 56 measurements was done on an independent sample.

2-way ANOVA analysis indicated that both frequency and time point had a significant effect on the response variable. However, Levene's test indicated the assumption of homoscedasticity was violated. Additionally the data seem non-normal. Here is the output: Normality and homoscedasticity output Y residual vs Y predicted

Based on the standard deviations, Levene's test, and the residuals vs predicted plot, it seems as though the data are quite heteroscedastic.

My question is what should I do to analyze these data in the 2-way ANOVA?

My first thought was to transform them to make the variances equal, but this answer seems to suggest when sample sizes are equal, heteroscedasticity can be ignored? I've read there are other problems with transforming data as well. Also, I only have SPSS and Graphpad Prism at my disposal, so please no suggestions that require using R, Stata, Python, etc.

  • $\begingroup$ Welcome to our site. Your plot is almost tantamount to a spread-vs-level plot. The near-linear increase of the SD with the predicted value strongly indicates performing an ANOVA of the log responses. Would that be meaningful for your study and have you tried it? $\endgroup$
    – whuber
    Commented Sep 6, 2018 at 21:56
  • $\begingroup$ @whuber Thanks for the welcome. I did try a natural logarithm transformation, and the residuals vs predicted look like this: i.imgur.com/yQ1BV2D.png. Whether it is meaningful is a good question. I believe so? The response variable is change in fluorescence, so a greater/lesser ln(Y) would still mean a greater/lesser change in fluorescence. $\endgroup$
    – user219200
    Commented Sep 6, 2018 at 22:56
  • 1
    $\begingroup$ You might do even better by predicting the reciprocal of the response. $\endgroup$
    – whuber
    Commented Sep 6, 2018 at 23:41
  • $\begingroup$ This transformation worked quite well. Levene's test is no longer significant, and more importantly the residual plot looks much more homogeneous i.imgur.com/pPgMJvx.png. I'll explore this and the answer I accepted and see which one works best for my purposes. Thanks. $\endgroup$
    – user219200
    Commented Sep 7, 2018 at 15:20
  • $\begingroup$ For an account of the technique, see stats.stackexchange.com/a/74594/919. $\endgroup$
    – whuber
    Commented Sep 7, 2018 at 15:42

1 Answer 1


Thanks for showing key output. For several reasons, I think the conclusions from your ANOVA are valid, even though some assumptions are not met.

Frequency effect: First, just from the information you gave, I can see that at Time 7 Frequencies 1 & 2 differ. Because this is just a two-sample t test, I can choose the Welch version which does not assume equal variances. This can be done by hand from formulas, but I did it in Minitab. You can see that this is the Welch (separate-variances) test because DF = 8 (instead of 16). [Welch has DF between 8 and 16, lower to the extent that sample SDs differ.] Even with this "loss" of DF, the P-value is very small.

Two-Sample T-Test and CI 

Sample  N    Mean   StDev  SE Mean
1       9  1.0000  0.0840    0.028
2       9   1.730   0.342     0.11

Difference = μ (1) - μ (2)
Estimate for difference:  -0.730
95% CI for difference:  (-1.001, -0.459)
T-Test of difference = 0 (vs ≠): 
    T-Value = -6.22  P-Value = 0.000  DF = 8

Similarly frequencies differ very significantly at Time 10:

Two-Sample T-Test and CI 

Sample  N   Mean  StDev  SE Mean
1       9  1.420  0.198    0.066
2       9  2.030  0.407     0.14

Difference = μ (1) - μ (2)
Estimate for difference:  -0.610
95% CI for difference:  (-0.942, -0.278)
T-Test of difference = 0 (vs ≠): 
    T-Value = -4.04  P-Value = 0.002  DF = 11

Again at Time 14, frequencies differ very significantly. P-value < 0.0005.

Summarizing the Frequency-effect, there is no problem declaring results for Frequencies 1 and 2 to differ across time points (consistently smaller for Frequency 1) because all three P-values are so small.

If you want to compare results for several different treatment combinations, you could look at 99% confidence intervals for pairs of the six cell means. They would be of the form $\bar X_{ij} \pm 2.306 S_{ij}/\sqrt{9}$ or $\bar X \pm 0.77\,S_{ij}.$ For example the 99% CI for the population mean in the cell for Frequency 1 at Time 7 is $1.0 \pm 0.06.$ You could make several Bonferroni comparisons at levels around 95%.

qt(.975, 8)
[1] 2.306004
[1] 0.7666667

Time effect: Now for the time effect, you can do a two one-way ANOVAs with all of your data: first 27 observations for Frequency 1, then second 27 for Frequency 2. Preferably, you would use the version that allows for unequal variances. I hope that version is available in SPSS.

However, because this is an balanced design, the standard ANOVA is easy to do based on information you provide.

For Frequency 1: MS(Factor) is the variance of the three sample means multiplied by 9. Also, MS(Error) is the mean of the three sample variances. Using R mainly as a calculator, I got the following results. MS(Factor) = 0.6517 with DF = 2 and MS(Error) = 0.0231 with DF = 24, and F = 36.29, which greatly exceeds the critical value 5.61 for a test at the 1% level; the P-value is less than 0.0005.

msf = 9*var(c(1.1, 1.42, 1.71)); msf
[1] 0.8379
mse = mean(c(.084, .193, .158)^2); mse
[1] 0.02308967
f = msf/mse; 
[1] 36.28896
1 - pf(f, 2, 24)
[1] 5.546266e-08
qf(.99, 2, 24)
[1] 5.613591

You can do the same for Frequency 2.

Comments on assumptions: Heteroscasticity. From your residual plot we can see that variance differ considerably for among the $2 \times 3 = 6$ treatment combinations. Thus it is a good idea to look at procedures that do not assume equal variances. However, because this is a balanced design (9 replications in each cell) and effects seem convincingly large, it seems reasonable to assume significance, if not to rely on specific difference in post-hoc tests.

As you say, you might seek a transformation to make cell variances more nearly equal, but that can make post hoc comparisons awkward.

Normality. I am not sure how SPSS does the Shapiro-Wilk test for ANOVA designs. It would be incorrect to expect the 45 original data values to be collectively normally distributed because, even if individually normal, they would have a mixture distribution of normals with six potentially vary different means. (A mixture distribution of normals with different means is need not be even nearly normal.)

The correct way to assess normality is by looking at residuals. Looking at your residual plot I don't see departures from normality serious enough to ruin your ANOVA. You might want to test the residuals formally, but with nine observations per cell and differences in cell mean as large as yours I don't think the significant results of your ANOVA are in question.

Addendum: Curious to see what results for a Welch-type one-way ANOVA of various Times for Frequency 1 would give, I did a nonparametric bootstrap with data for the three times modeled after sample means and SDs in your printout. (In R, the procedure for a one-way ANOVA that does not assume equal variances is oneway.test.) Sampling 100,000 datasets roughly like yours for Frequency 1, I got an average P-value of $4.1 \times 10^{-7}$ and all P-values < 0.0005.

m = 10^5;  pv = numeric(m);  gp = as.factor(rep(c(7,10,14), each=9))
for(i in 1:m) {
  t.7 = rnorm(9, 1, .014); t.10 = rnorm(9, 1.4, .198);  t.14 = rnorm(9, 1.7, .158)
  all = c(t.7,t.10,t.14); pv[i] = oneway.test(all~gp)$p.val }
mean(pv);  mean(pv < .0005) 
[1] 4.122535e-07
[1] 1

Note: I know you asked for no recommendations to use R, but this gives a promising indication that an ANOVA you do in SPSS will give a significant result.

I would just point out that R is very good software, available without cost at r-project.org. Some people hesitate to use R because they see so many complicated things done with it. But there is no obligation to learn more about R than you need for the task at hand (probably much like you're doing with SPSS), and there are numerous help sessions on line.

  • $\begingroup$ Thanks for your response. Interesting, so you broke up the single 2-way ANOVA into 3 t-tests and a one-way ANOVA with unequal variances. I hadn't thought of this approach but it does the job quite well. Should the t-tests also be Bonferroni corrected? (although this will make little difference to significance). Re: your addendum, I have been feeling my lack of proper statistical software acutely, and I am planning to complete a Python data analysis tutorial this weekend. I've hesitantly settled on Python instead of R as I already have a fair amount of experience with Python. $\endgroup$
    – user219200
    Commented Sep 6, 2018 at 23:43
  • 1
    $\begingroup$ Good questions. Actually, broke 2-way ANOVA into 3 t-tests and two one-way ANOVAs to handle possibly different variances. // I would do separate Bonferroni "families" for Frequency (3) and time (2). // Congratulations on getting feet wet with Python. Wishing you success with that and other statistical adventures. $\endgroup$
    – BruceET
    Commented Sep 6, 2018 at 23:52
  • $\begingroup$ If you want to try transformations, perhaps see this section of the NIST handbook. Note that the vertical scale of the 'Plot of Predicted Values' should be marked as log scale. $\endgroup$
    – BruceET
    Commented Sep 7, 2018 at 3:42
  • $\begingroup$ Thanks for the resource. The reciprocal transformation was successful, so I'll explore transforming the data and your method and see which one I end up preferring. Your method has the advantage that I don't have to graph the reciprocal of my response variable and confuse readers. $\endgroup$
    – user219200
    Commented Sep 7, 2018 at 15:24
  • $\begingroup$ Maybe it's OK to graph the original data and explain that the reciprocal transformation was a technical device to facilitate data analysis. Your differences are large enough that no one is going to doubt they're real. // Might show residual plots of orig'l and transf data side-by-side to show why transf was useful. // One major drawback of transformations is difficulty of explaining transformed scale to general audiences. $\endgroup$
    – BruceET
    Commented Sep 7, 2018 at 17:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.