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-

X2=

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-

c1c2c3Total
r1O11O12O13rt1
r2O21O22O23rt2
r3O31O32O33rt3
Totalct1ct2ct3Grand Total

 

Where,

ri and ci = Attributes

Oij = Observed Frequencies

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

eij=

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

Where,

Pijis the probability of obtaining observation that belongs to i-th row and j-th column.

Thus,

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 BAttribute ATotal
abR1 = a + b
cdR2 = c + d
Totalc1 = a + cc2 = b + dGrand Total = N

= R1 + R2

= c1 + c2

 

Here, statistics is calculated as-

X2 =

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

X2 (Corrected) =

 

 

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