This is a 2 by 3 table, which shows the different schools and their different performance in an exam. We use a χ2 test to determine whether the two factors are independent, or whether there is an association between them. According to the table above, we want to know whether the school affects their exam performance. Or in other words, since the amount of students of school A and school B. are different (80 and 70 respectively), we know that, if they have the same ratios of credit, pass and fail, it means that whichever the school it is, also it doesn’t affect the grades.
This kind of test is known as the test for independence. As usual, we shall find the expected frequency, find the degree of freedom ν and find the test statistic X2 which has the same formula as the previous section.
Let’s take the above example. The degree of freedom for a h × k contingency table can be found using the formula
ν = (h – 1)(k – 1)
From here, we proceed to find X2 by making use of the 6 values of O and E that we just calculated. Now let me give you an example:
A research worker studying the ages of adults and the number of credit cards they posses obtained the results shown in the table.
Use the χ2 statistic and a significance test at the 5% level to decide whether or not there’s an association between age and number of credit cards possessed.
H0: There’s no association between age and number of credit cards possessed.
H1: There’s an association between age and number of credit cards possessed.
ν = (2 – 1)(2 – 1) = 1, the Yates’ Correction is used.
Use the χ2 (1) distribution, perform the test at 5% level.
Since χ2(5%) (1) = 3.841, reject H0 if X2 > 3.841.
Easy? That’s all for this chapter. ☺