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CHI-SQUARE TEST ( X2 )
 It is the most popular test of significance in social
science research
 It is used to make comparison between two or
more nominal variables
 It is used to make comparison between
frequencies rather than between the mean
 This test evaluated whether the difference
between the observed frequencies and the
expected frequencies under the null hypothesis
A chi square value is obtained by the formula:
X2=∑(f0-fe)2/fe
 fo= the observed frequency
 fe=the expected frequency in terms of the null
hypothesis
 Suppose we want to examine the relationship between
education and income.
What will be,
 The Null hypothesis ?
 Alternative / Research hypothesis?
Income
Education
Low High Total
Low 30 12 42
High 10 28 40
Total 40 40 80
 H0= there is a no relation between educational
and income level
 Ha=there is a relation between educational and
income level
Procedure,
X2=∑(f0-fe)2/fe
 We can compute the expected frequencies (fe) with the
observed once. (f0)
 To compute the (fe) for any cell, the following formula is
used:
fe= (row total * column total)/n
 n= total number
 To compute X2 , the expected frequency of each cell is
subtracted from the observed one, squared, divided by the
expected frequency of the cell and then all quotients are
added up.
 The expected frequencies computed are shown in
brackets.
Income
Education
Low High Total
Low 30 (21) 12 (21) 42
High 10 (19 ) 28 (19) 40
Total 40 40 80
 21= (42*40)/ 80
 19=(38*40)/ 80
X2= X2=∑(f0-fe)2/fe
= ((30-21) 2/ 21)
+ ((12-21)2/ 21)
+ ((10-19)2/19 )
+((28-19)2/19)
= 3.86 + 3.86 + 4.26 + 4.26
X2 =16.24
To reject or accept the null hypothesis, we need to compare
X2 value to the critical value of sampling distribution of
X2
The sampling distribution of X2 is determined by
1. The level of significance
2. The number of degree of freedom
1. Level of significance = 0.01 or 0.05
2. The number of degrees of freedom of the X2 distribution is
set by the number of cells for which the expected
frequencies can be selected freely.
The formula used is:
df= (r-1)*(c-1)
r= the number of rows
C=the number of columns
Thus
in a 2*2 table df=(2-1)*(2-1)=1
In a 3*3 table df=(3-1)*(3-1)=4 and so on.
 In our example with 1 degree of freedom and 0.01
level of significance the table value of X2 is 6.64.
 Our obtained X2 value 16.24 is much larger than
the table value
 Hence the null hypothesis is rejected and
research hypothesis is supported
 i.e. we may conclude that there is significant
relationship between education and income.
 In a survey of a brand preference of high school
students for the soft drink, the following result was
obtained:
 Find out is there any relationship between brand
preference and the gender of the consumer? Use chi
square test at the 0.05 level of significance
Gender
Brand
A B
Boys 25 30
Girls 46 22
Total 71 52
Chi square test ( x2  )

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Chi square test ( x2 )

  • 2.  It is the most popular test of significance in social science research  It is used to make comparison between two or more nominal variables  It is used to make comparison between frequencies rather than between the mean  This test evaluated whether the difference between the observed frequencies and the expected frequencies under the null hypothesis
  • 3. A chi square value is obtained by the formula: X2=∑(f0-fe)2/fe  fo= the observed frequency  fe=the expected frequency in terms of the null hypothesis
  • 4.  Suppose we want to examine the relationship between education and income. What will be,  The Null hypothesis ?  Alternative / Research hypothesis? Income Education Low High Total Low 30 12 42 High 10 28 40 Total 40 40 80
  • 5.  H0= there is a no relation between educational and income level  Ha=there is a relation between educational and income level
  • 6. Procedure, X2=∑(f0-fe)2/fe  We can compute the expected frequencies (fe) with the observed once. (f0)  To compute the (fe) for any cell, the following formula is used: fe= (row total * column total)/n  n= total number  To compute X2 , the expected frequency of each cell is subtracted from the observed one, squared, divided by the expected frequency of the cell and then all quotients are added up.
  • 7.  The expected frequencies computed are shown in brackets. Income Education Low High Total Low 30 (21) 12 (21) 42 High 10 (19 ) 28 (19) 40 Total 40 40 80
  • 8.  21= (42*40)/ 80  19=(38*40)/ 80
  • 9. X2= X2=∑(f0-fe)2/fe = ((30-21) 2/ 21) + ((12-21)2/ 21) + ((10-19)2/19 ) +((28-19)2/19) = 3.86 + 3.86 + 4.26 + 4.26 X2 =16.24
  • 10. To reject or accept the null hypothesis, we need to compare X2 value to the critical value of sampling distribution of X2 The sampling distribution of X2 is determined by 1. The level of significance 2. The number of degree of freedom
  • 11. 1. Level of significance = 0.01 or 0.05 2. The number of degrees of freedom of the X2 distribution is set by the number of cells for which the expected frequencies can be selected freely. The formula used is: df= (r-1)*(c-1) r= the number of rows C=the number of columns Thus in a 2*2 table df=(2-1)*(2-1)=1 In a 3*3 table df=(3-1)*(3-1)=4 and so on.
  • 12.  In our example with 1 degree of freedom and 0.01 level of significance the table value of X2 is 6.64.  Our obtained X2 value 16.24 is much larger than the table value  Hence the null hypothesis is rejected and research hypothesis is supported  i.e. we may conclude that there is significant relationship between education and income.
  • 13.  In a survey of a brand preference of high school students for the soft drink, the following result was obtained:  Find out is there any relationship between brand preference and the gender of the consumer? Use chi square test at the 0.05 level of significance Gender Brand A B Boys 25 30 Girls 46 22 Total 71 52