Page 375 - DMTH404_STATISTICS

2
Unit 26:  - Test Hypothesis

Computation of Expected Frequencies Notes

Cities 
A B C D Total
Income (Rs 
)
Under 3000 10.00 13.33 13.33 13.33 50
3000-5000 8.00 10.67 10.67 10.67 40
Over 5000 12.00 16.00 16.00 16.00 60
Total 30 40 40 40 150

Note that the expected frequencies for city A, under various income groups, are computed as
´
30 50 30 40 30 60
´
´
= 10.00, = 8.00 and = 12.00. Other frequencies have also been computed in
150 150 150
a similar manner.
Using the observed and expected frequencies, the value of c = 8.28.
2
Further, the value of X from tables for 6 d.f. at 5% level of significance is
2
Since the calculated value is less than the tabulated value, there is no evidence against H .
0
2
The value of  for a 2 ´ 2 Contingency table
For a 2 ´ 2 contingency table,

a b a b
+
c d c + d , the value of  can be directly computed with the use of the
2
+
d
+
+
a c b d a b c + = N
+
following formula :
N (ad bc- ) 2
2
 =
(a b a c+ )( + )(b d+ )(c d+ )
Yate's correction for continuity
We know that c is a continuous random variate but the frequencies of
2
(O E- ) 2
various cells of a contingency table are integers. When N is large, the distribution of å
E
2
is approximately c . However, the corrections for continuity are required when N is small. Yates
has suggested the following corrections for continuity in a 2 × 2 contingency table :
1 1
If ad > bc, reduce a and d by and increase b and c by . Similarly, If ad < bc, increase a and
2 2
1 1
d by and decrease b and c by .Thus, the contingency tables in the two situations become
2 2
1 1 1 1
a - b + a + b -
2 2 2 2
and respectively.
1 1 1 1
c + d - c - d +
2 2 2 2

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