2.1 Numerical and graphical summaries

The methods for exploring data discussed so far are most applicable when at least one of the variables you are examining is numerical. However, graphical illustrations for examining potential dependencies between two categorical variables are not useful as, for example, a scatter plot will only show a point at each distinct pair of events with multiple occurrences being drawn on top of one another.

The relationship between two categorical variables are best examined using a contingency table, or cross tabulation, that displays the frequency between all possible pairs of events from the two categorical variables of interest.

	X=1	X=2	$\mathrm{\cdots }$	X=I	Total
Y=1	$n_{11}$	$n_{21}$	$\mathrm{\cdots }$	$n_{I1}$	$r_1$
Y=2	$n_{12}$	$n_{22}$	$\mathrm{\cdots }$	$n_{I2}$	$r_2$
$\mathrm{⋮}$	$\mathrm{⋮}$	$\mathrm{⋮}$	$\mathrm{\ddots }$	$\mathrm{⋮}$	$\mathrm{⋮}$
Y=J	$n_{1J}$	$n_{2J}$	$\mathrm{\cdots }$	$n_{IJ}$	$r_J$
Total	$c_1$	$c_2$	$\mathrm{\cdots }$	$c_I$	$N$

Here, $n_{ij}$ denotes the frequency of the pair $(X,Y)=(i,j)$ appears in the data set that contains a total of $N$ records. The column and row totals, denoted by $c_i$ and $r_j$ respectively, represents the marginal tables for each of the categorical variables.

For example, the data set APP contains information from 480 customers who took part in a consumer satisfaction survey regarding three smartphone applications. The contingency table below present the consumer’s responses to the question “Would you recommend the app to your friends?” against which of the three apps that they have downloaded.

Here, the response ‘Yes’ is more frequent across all three apps, and the frequency increases from App1 to App2 and again for App3 for both Yes/No responses. Since the marginal relationship between events for each categorical variable is reflected within the contingency table, is this sufficient to state that there is no relationship between these categorical variables?

It is not sufficient to compare values based on magnitude but rather comparisons should be performed on proportions. For example, about half of the users of App3 responded ‘Yes’ to the question whereas about three-quarters of App1 users gave the same response. Further investigation is needed to distinguish whether this difference in proportions represents some form of dependency or natural sampling variation.

Recall that if two random variables are independent, the joint probability mass/density function can be expressed as the product of the marginal probability mass/density function. For the general contingency table, the probability of a record possesses the event pair $(i,j)$, denoted by $p_{ij}$, can be written under the assumption of independence as:

$$p_{ij}=p_i{p}_j,\mathrm{for}i=\{1,\mathrm{\dots },I\}\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}j\in \{1,\mathrm{\dots },J\}$$

where $\{p_i\}$ and $\{p_j\}$ are the marginal probabilities for each categorical variable. The true probabilities are unknown, but the proportion of the column/row totals relative to the overall total provides unbiased estimates.

$${\widehat{p}}_i=\frac{c_i}{N}\mathit{\hspace{1em}}\mathrm{and}\mathit{\hspace{1em}}{\widehat{p}}_j=\frac{r_j}{N}$$

Given these marginal probabilities, we can then estimate the expected number of samples to appear in each cell of the contingency table under the assumption of independence by:

$$E_{i,j}=N{p}_{ij}=N{p}_i{p}_j\approx N{\widehat{p}}_i{\widehat{p}}_j=N\frac{c_i}{N}\frac{r_j}{N}=\frac{c_i{r}_j}{N}.$$

 
Exercise 2.8
For the app example, calculate the expected number of consumers there should be for each application and Yes/No pair under the assumption of independence.

 

Now that we know the expected number of events for each event pair $(X,Y)$ under the assumption of independence, we need to assess whether the differences between the observed and expected tables are truly an indication of some relationship or sampling variation. For this, we use the Pearson’s chi-squared test for independence.

Informal proof: Consider the simplest table with $I=1$ and $J=2$ where the table consists of only two cells. Let $n$ be the number of events in the first cell and $N$ denote the overall total number or records, so that there are $N-n$ events for the second cell. It is clear that the distribution for the count in the first cell is $n\sim \mathrm{Binom}(N,p)$ for some unknown probability $p$. The expected number for the first cell is $E_1=Np$ whist that for the second cell is $E_2=N(1-p)$. Substituting these values into the chi-squared test statistic gives:

	$X^2$	$={\displaystyle \frac{{(n-Np)}^2}{Np}}+{\displaystyle \frac{{(N-n-N(1-p))}^2}{N(1-p)}}$
		$={\displaystyle \frac{{(n-Np)}^2}{Np}}+{\displaystyle \frac{{(-n+Np)}^2}{N(1-p)}}$
		$={\displaystyle \frac{(1-p){(n-Np)}^2+p{(n-Np)}^2}{Np(1-p)}}$
		$={\displaystyle \frac{{(n-Np)}^2}{Np(1-p)}}$
		$={\left({\displaystyle \frac{n-𝔼[n]}{\sqrt{\mathrm{var}(n)}}}\right)}^2$

For large $N$, the distribution for $n$ is approximately normal $N(np,np(1-p))$. Thus $X^2\sim {\chi }_1^2$. When generalising to a $I\times J$ table, the distribution of the test statistic corresponds a sum of chi-squares, but we must take into account the fact that the last row and column of the table can be derived from the marginal totals and the values in the other cells for that row or column. Therefore, the degree-of-freedom for a general table is $(I-1)(J-1)$.

 
Exercise 2.9
Perform a chi-squared test for independence for the app study. What do you conclude?

 

When using the Pearson’s chi-squared test, it is important to be aware of the following assumptions.

• •

  The data are assumed independent of each other and are drawn from a population where every member has an equal probability of selection.

• •

  The overall sample size $N$ must be sufficiently large such that the count in any one cell is not too small.

• •

  The expected count for any cell must not be zero or too small.