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-

 c1 c2 c3 Total r1 O11 O12 O13 rt1 r2 O21 O22 O23 rt2 r3 O31 O32 O33 rt3 Total ct1 ct2 ct3 Grand 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 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-

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) = ### Customer Reviews

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