It is also called as Distribution Free Test. This means population may not be normal. Hence, the null hypothesis should be taken as-

H_{0} = Observations are in good agreement with hypothetical distribution.

Suppose O_{i} be observed frequency and e_{i} be expected frequency where, i = 1, 2, …… n. the statistic will be written as-

X^{2}=

The above equation follows chi-square distribution. It will be made with (n – 1) degree of freedom. When this calculated value becomes greater than tabulated value, null hypothesis will not be accepted (at level of significance ‘a’).

**8A: Chi-Square Test of Independence**

Consider r x c table with following data to name it as contingency table-

c_{1} | c_{2} | c_{3} | Total | |

r_{1} | O_{11} | O_{12} | O_{13} | rt_{1} |

r_{2} | O_{21} | O_{22} | O_{23} | rt_{2} |

r_{3} | O_{31} | O_{32} | O_{33} | rt_{3} |

Total | ct_{1} | ct_{2} | ct_{3} | Grand Total |

Where,

r_{i} and c_{i} = Attributes

O_{ij} = Observed Frequencies

Hence, expected frequency can be calculated by using the formula,

e_{ij}=

e_{ij}= , etc.

The above equation will follow the chi-square distribution with (c – 1)(r – 1) degree of freedom. As attributes are independent, the null hypothesis will be-

H_{0} 😛_{ij} = P_{12} ….. = P_{ic}, i = 1, 2, 3….r

Where,

P_{ij}is the probability of obtaining observation that belongs to i-th row and j-th column.

Thus,

If X^{2}_{cal}> X^{2}_{tab}, then H_{0} will be rejected and

IF X_{cal}< X^{2}_{tab}, then H_{0} will be accepted.

**8B. A 2 x 2 Table (Simplified Form)**

Attribute B | Attribute A | Total | |

a | b | R_{1} = a + b | |

c | d | R_{2} = c + d | |

Total | c_{1} = a + c | c_{2} = b + d | Grand Total = N = R = c |

Here, statistics is calculated as-

X^{2} =

The degree of freedom will be (2 – 1) (2 – 1) = 1

**8C. Yate’s Correction**

Suppose the degree of freedom is one for four cell frequencies. If the total rows and columns are fixed, the Yate’s Correction has suggested the following-

- ad >bc, reduce a and d by the value 0.5 and increase the other two (b and c) by 0.5
- ad <bc, increase a and d by the value 0.5 and reduce the other two (a and d) by 0.5

X^{2} (Corrected) =