    # Chi-Square Test Introduction

Statistics mostly deals with quantitative data that is data which is expressed numerically. But we can also analyse qualitative data using statistics. In real life situations, the data may not always be quantitative in nature, hence we need a method to analyse qualitative data. Chi-square is one such method that can be used to analyse qualitative data. To do this test we consider two attributes and try to check if there are associated or not. The number of attributes may be more also. This test is called "chi-square test for independence of attributes".

Definition

Consider two attributes A and B. let each attribute have only two outcome. Then the respective frequencies can be represented in the form of a table as shown

 B/A A1 A2 Total B1 a b a + c B2 c d c + d total a + c b + d N

Here we want to test the hypothesis

H0: there is no association between the two attributes

H1: there is association between he two attributes

The test statistic here is Under H0 follows distribution with n-1 degrees of freedom. Here since n=2, the degrees of freedom is always 1. We use this information and the chi-square table values to take the decision.

Example:

Following is the data collected on two characters namely the cigarette smoking and literacy and the data is given as follows

 Literacy/Smoking Smoker Non-smoker Total literates 40(a) 50(b) 90(a+b) illiterates 60(c) 35(d) 95(c+d) total 100(a+c) 85(b+d) 185(N)

Here we want to test the hypothesis

H0 : cigarette smoking and literacy are not associated

H1: cigarette smoking and literacy are associated

Lets use the level of significance Hence =6.51

Now to take the decision, we use the critical value. We can find this from excel, using the function CHISQ.INV.RT(0.05,1). The value we get is 3.8414.

Since the test statistic value is greater than the critical value, we reject H0 at 5% l evel of significance and conclude that there is association between cigarette smoking and literacy.

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