The chi-square test for trend

We have already discussed the chi-square (χ 2 ) test and 2 × 2 tables (known as cross-tabulation) to test the association between 2 categorical variables where the distribution of 1 varies according to the values of the other. In this article, we will describe the chi-square test for trend where 1 variable is binary and the other is ordered categorical, and we want to assess whether the association between the variables follows a trend.

As an example, we are interested in whether the frequency of using oral mouth rinse is associated with the presence or absence of gingivitis in orthodontic patients. In other words, in the following example, we will compare the distribution of a binary variable (presence or absence of gingivitis) across the levels of an ordered categorical variable (frequency of mouth rinse use). For the sake of simplicity, we assume that all other measures of oral hygiene are the same for all patients. The distribution of the presence or absence of gingivitis according to the mouth rinse use schedule is shown in Table I .

Table I
Presence or absence of gingivitis according to the mouth rinse schedule
Presence of gingivitis Mouth rinse usage Total
Once a week Twice a week Seven days a week
Yes 16 9 2 27
No 29 36 39 104
Total 45 45 41 131

The question that we want to answer is whether the proportion of orthodontic patients with gingivitis increases or decreases across mouth rinse groups. To answer this question, we will first assign a numeric score across the mouth rinse use groups starting from number 1 for the group with the lowest frequency and 3 for the group with the highest frequency of mouth rinse use. See last row of Table II .

Table II
Presence or absence of gingivitis according to the mouth rinse schedule and assigned score per mouth rinse use category
Presence of gingivitis Mouth rinse usage Total
Once a week Twice a week Seven days a week
Yes 16 9 2 27
No 29 36 39 104
Total 45 45 41 131
Score 1 2 3

The next step is to calculate the mean score for each gingivitis category and overall using the following procedure.

Xyes=([116]+[29]+[32])27=1.48
X yes = ( [ 1 ∗ 16 ] + [ 2 ∗ 9 ] + [ 3 ∗ 2 ] ) 27 = 1.48
Xno=([129]+[236]+[339])104=2.10
X no = ( [ 1 ∗ 29 ] + [ 2 ∗ 36 ] + [ 3 ∗ 39 ] ) 104 = 2.10
Xoverall=([145]+[245]+[341])131=1.97
X overall = ( [ 1 ∗ 45 ] + [ 2 ∗ 45 ] + [ 3 ∗ 41 ] ) 131 = 1.97
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Apr 4, 2017 | Posted by in Orthodontics | Comments Off on The chi-square test for trend
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