The chi-squared test is used to determine whether there is a significant difference between the expected frequencies and the observed frequencies in one or more categories.

Take for instance the data on the image rating (top2box score) of shampoo brands in Exhibit 33.23. To gauge the image profile of Organics shampoo, the brand manager needs to compare attribute ratings for the brand with competing brands.

You may recall from Chapter *Brand Sensing* that the basis for comparison is the *Expected Score*:
$$Expected Score=\frac{AVG(Brand Score)×AVG(Attribute Score)}{AVG(All Score)}$$
$$ = \frac{AVG(Column)×AVG(Row)}{AVG(All)}$$

*Note: Total scores may be used, instead of average score.*

The attribute *profile* (*Profile = Actual Score – Expected Score*)
reveals whether the brand’s association with the attribute is strong (*profile > 0*) or
weak (*profile < 0*). Organics shampoo for instance is strongly associated with the
attribute ‘Nourishes Roots’ (*Profile = 35 – 24 = 11*).

While big brands like Pantene and Organics are rated high on all attributes, image profiling mathematically eliminates influence of brand and attribute “size” to determine each brand’s strengths and weaknesses in relation to each other, expressed as variation from what one would expect if the brand were average.

Chi-Square (χ2) test of independence is a universal metric that standardizes the data so that it becomes comparable across data sets of different magnitude.

$$Chi˗square \,statistic:\,χ^2 = \sum\frac{(Observed-Expected)^2}{Expected}=\sum\frac{Profile^2}{Expected}$$For Organics on the attribute nourishes roots:

$$Chi˗square \,value \,of \,the \,cell=\frac{(Observed-Expected)^2}{Expected}=\frac{(35-24)^2}{24}=4.9$$For the entire data set, sum of these quantities over all the cells is the Chi˗square test statistic:

$$Chi˗square\, statistic \,χ2 = \sum\frac{(Observed-Expected)^2}{Expected}=230.4$$Under the null hypothesis there is no difference in the proportions across brands. This has approximately a chi-squared distribution whose number of degrees of freedom are: $$df =degrees \,of \,freedom=(attributes-1)(brands-1)=(11 - 1)(9 -1)=80$$

If the test statistic for the chi-squared distribution is improbably large, then the null hypothesis is rejected.

In our shampoo example, since the p-value (for χ2 = 230.4, df = 80) is extremely small (p-value << 0.05), the null hypothesis is rejected. There are significant differences in the attribute ratings across the brands.

The Chi-square test is often used in research studies to test the relationship between a variable pertaining to behaviour or attitude, with a variable pertaining to classification. For instance, the relationship between the consumption of a product with income level, location, or age. The variables are cross tabulated, and then tested. The test will reveal whether relationship exists between the two variables.

*Note: Chi-square function in Excel is CHISQ.TEST. In SPSS the analysis falls under
‘Descriptive statistics’ > ‘Crosstab’. Check the Chi-square box within the statistics pop-up page.*

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