I have a common question on how to explain significant association between categorical variables in mosaic plot.

For example,in this plot,based on Pearson residuals, can we say that $[2.0, 5.1]$ and $[-3, -2.0]$ residuals values mean there is a statistically significant association in $40+$ age,with memory and moderate attitude? And how to consider Pearson residual value , we use $[2.0, 5.1]$ value or $[4.0,5.1]$ or $[-2.0 ,-3.0]$ also?

enter image description here


2 Answers 2


The formula for the standardized residuals is:

$$\begin{align}\text{Pearson's residuals}\,&=\,\frac{\text{Observed - Expected}}{ \sqrt{\text{Expected}}}\\ d_{ij}&=\frac{n_{ij}-m_{ij}}{\sqrt{m_{ij}}} \end{align}$$

where $m_{ij} = E( f_{ij})$ is the expected frequency of the $i$-th row and the $j$-th column.

The sum of squared standardized residuals is the chi square value.

From Extending Mosaic Displays: Marginal, Partial, and Conditional Views of Categorical Data by Michael Friendly

Under the assumption of independence, these values roughly correspond to two-tailed probabilities $p < .05$ and $p < .0001$ that a given value of $| d_{ij} |$ exceeds $2$ or $4$.

Notice the following footnote:

For exploratory purposes, we do not usually make adjustments (e.g., Bonferroni) for multiple tests because the goal is to display the pattern of residuals in the table as a whole. However, the number and values of these cutoffs can be easily set by the user.

We are dealing with a multi-way table, in reference to which the R documentation for the mosaicplot package states:

Extended mosaic displays show the standardized residuals of a loglinear model of the counts from by the color and outline of the mosaic's tiles. (Standardized residuals are often referred to a standard normal distribution.) Negative residuals are drawn in shaded of red and with broken outlines; positive ones are drawn in blue with solid outlines.

The fact that this is a three-way contingency table complicates the interpretation, which is very nicely explained in @roando2's answer.

Here is a simulation with a made-up table that resembles the OP to clarify the calculations:

tab_df = data.frame(expand.grid(
  age = c("15-24", "25-39", ">40"),
  attitude = c("no","moderate"),
  memory = c("yes", "no")),
  count = c(1,4,3,1,8,39,32,36,25,35,32,38) ) 
(tab = xtabs(count ~ ., data = tab_df))

, , memory = yes
age     no moderate
  15-24  1        1
  25-39  4        8
  >40    3       39
, , memory = no
age     no moderate
  15-24 32       35
  25-39 36       32
  >40   25       38

Call: xtabs(formula = count ~ ., data = tab)
Number of cases in table: 254 
Number of factors: 3 
Test for independence of all factors:
    Chisq = 78.33, df = 7, p-value = 3.011e-14

mosaic(~ memory + age + attitude, data = tab, shade = T)
expected = mosaic(~ memory + age + attitude, data = tab, type = "expected") 

# Finding, as an example, the expected counts in >40 with memory and moderate att.:

over_forty = sum(3,39,25,38)
mem_yes = sum(1,4,3,1,8,39)
att_mod = sum(1,8,39,35,32,38)
exp_older_mem_mod = over_forty * mem_yes * att_mod / sum(tab)^2

# Corresponding standardized Pearson's residual:

(39 - exp_older_mem_mod) / sqrt(exp_older_mem_mod) # [1] 6.709703

enter image description here

It is interesting to compare the graphical representation to the results of the Poisson regression, which illustrates perfectly the English interpretation in @rolando2 's answer:

fit <- glm(count ~ age + attitude + memory, data=tab_df, family=poisson())

glm(formula = count ~ age + attitude + memory, family = poisson(), 
    data = tab_df)

                 Estimate Std. Error z value Pr(>|z|)    
(Intercept)        1.7999     0.1854   9.708  < 2e-16 ***
age25-39           0.1479     0.1643   0.900  0.36794    
age>40             0.4199     0.1550   2.709  0.00674 ** 
attitudemoderate   0.4153     0.1282   3.239  0.00120 ** 
memoryno           1.2629     0.1514   8.344  < 2e-16 ***
  • 1
    $\begingroup$ @gung for this case,I use vcd package,and use mosaic(Freq ~ age +memory,data =test,colorize = T).What do you mean for switch from red & blue? $\endgroup$
    – Rya
    Mar 7, 2017 at 20:02
  • 1
    $\begingroup$ @AntoniParellada so this case,we can say one dark blue part and three small red part has statistically significant right? because absolute Pearson residuals value is greater than 1.96. $\endgroup$
    – Rya
    Mar 7, 2017 at 20:27

This is best interpreted using some specific language. Within the 40+ age group (in the plot, labeled "40-") there is a significant association between the variables memory and attitude. We cite associations between variables, not between values or categories within them (such as "moderate" or "no").

A more specific statement one could make is that, for those 40+ but not for other age groups, "yes" on memory is disproportionately paired with "moderate" on attitude.

We could also say there is an interaction between age and memory as they relate to attitude, or between age and attitude as they relate to memory. Only rarely would one put a variable like age at the end of such a sentence, since age is ordinarily a candidate to be a predictor or cause, not an effect.

All of the above is based on the plot's characterization of each cell using, via a color, a range of Pearson residuals. The plot does not give us sufficient information to further specify the values of each residual. Nor does any individual residual value determine significance in this context. The mosaic plot, being based on a Chi-square test, does not address significance except by yielding a single, overall, "omnibus" p-value.

  • $\begingroup$ I am not aware of a standard way of adjusting, and I guess the simplest way would be to apply a Bonferroni correction. There are three variables ~ memory + age + attitude, and the model doesn't explicitly include interactions... $\endgroup$ Mar 8, 2017 at 14:51

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