Page 373 - DMTH404_STATISTICS
P. 373
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Unit 26: - Test Hypothesis
2
The computations of are done in the following table : Notes
No .of Expected Observed (O - E i ) 2
i
families freq .( ) freq .(O i ) E i
E
i
0 180 e - 2.5 = 14.76 12 0.52
´
1 E ´ 2.5 = 36.94 39 0.11
0
2 E ´ 2.5/2 = 46.17 47 0.01
1
3 E ´ 2.5/3 = 38.48 40 0.06
2
4 E ´ 2.5/4 = 24.05 20 0.68
3
5 E ´ 2.5/5 = 12.02 17 2.06
4
6 or more by difference = 7.58 5 0.88
2
Total 180 = 4.32
The value of from table at 5% level of significance and 5 d.f. is 11.07. Since the calculated value
2
is less than the tabulated value, there is no evidence against H .
0
2
26.2.2 - test as a Test for Independence of Two Attributes
Let us assume that a population is classified into m mutually exclusive classes, A , A , ...... A ,
1 2 m
according to an attribute A and each of these m classes are further classified into n mutually
exclusive classes, like A B , A B , ...... A B , etc., according to another attribute B.
i 1 i 2 i n
If O is the observed frequency of A B , i.e., (A B) = O , the above classification can be expressed
ij i j i j ij
in form of following table, popularly known as contingency table.
B
B 1 B 2 B n Total
A
A
A 1 O 11 O 12 O 1n ( )
1
A O O O (A )
2 21 22 2n 2
A O O O (A )
m m 1 m 2 mn m
B
B
B
Total ( ) ( ) ( ) N
1
2
n
Assuming that A and B are independent, we can compute the expected frequencies of each cell,
( )( ) m n (O - E ) 2
B
A
2
i.e., E = i j . Thus, = åå ij ij would be a c - variate with (m - 1)(n - 1) d.f.
2
ij
N i= 1 j= 1 E ij
Remarks : The expected frequencies of some cells may be obtained by the application of the
above formula while the remaining cell frequencies can be obtained by subtraction. The minimum
number of cell frequencies, that must be computed by the use of the formula, is equal to the
2
degrees of freedom of the c statistic.
Example 44: The employees in 4 different firms are distributed in three skill categories
shown in the following table. Test the hypothesis that there is no relationship between the firm
and the type of labour. Let the level of significance be 5%.
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