It is also called as Distribution Free Test. This means population may not be normal. Hence, the null hypothesis should be taken as-
H0 = Observations are in good agreement with hypothetical distribution.
Suppose Oi be observed frequency and ei be expected frequency where, i = 1, 2, …… n. the statistic will be written as-
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-
ri and ci = Attributes
Oij = Observed Frequencies
Hence, expected frequency can be calculated by using the formula,
eij= , 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-
H0 😛ij = P12 ….. = Pic, i = 1, 2, 3….r
Pijis the probability of obtaining observation that belongs to i-th row and j-th column.
If X2cal> X2tab, then H0 will be rejected and
IF Xcal< X2tab, then H0 will be accepted.
8B. A 2 x 2 Table (Simplified Form)
|Attribute B||Attribute A||Total|
|a||b||R1 = a + b|
|c||d||R2 = c + d|
|Total||c1 = a + c||c2 = b + d||Grand Total = N|
= R1 + R2
= c1 + c2
Here, statistics is calculated as-
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-
X2 (Corrected) =
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