Page 372 - DMTH404_STATISTICS

Statistics

Notes 2 2 2 2
(46 40- ) (68 60- ) (62 60- ) (24 40- )
2
Thus, we have  = + + + = 8.44.
40 60 60 40
Since this value is greater than the tabulated value, 7.81, for 3 d.f. and 5% level of significance,
H is rejected.
0

Example 42: A survey of 320 families with 5 children each revealed the following
distribution:

No.of boys : 5 4 3 2 1 0
No.of girls : 0 1 2 3 4 5
No.of families : 14 56 110 88 40 12

Is the result consistent with the hypothesis that male and female births are equally probable?
Solution.
Assuming that H (i.e., male and female births are equally probable) is true, the expected number
0
5
æ
1 ö
of families having r boys (or equivalently 5 - r girls) is given by E = 320´ 5 C r ç ÷ = 10 ´ 5 C . On
r
r
è 2 ø
substituting r = 5, 4, 3, 2, 1, 0, the respective expected frequencies are 10, 50, 100, 100, 50 and 10.
(14 10- ) 2 (56 50- ) 2 (110 100- ) 2 (88 100- ) 2 (40 50- ) 2 (12 10- ) 2
2
  = + + + + + = 7.16.
10 50 100 100 50 10
The value from table for 5 d.f. at 5% level of significance is 11.07, which is greater than the
calculated value. Thus, there is no evidence against H .
0
Example 43:
The record for a period of 180 days, showing the number of electricity failures per day in Delhi
are shown in the following table :

No .of failures : 0 1 2 3 4 5 6 7
No .of days : 12 39 47 40 20 17 3 2

Determine, by using c - test, whether the number of failures can be regarded as a Poisson
2
variate?
Solution.
We have to test H : No. of failures is a Poisson variate against H : No. of failures is not a Poisson
0 a
variate.
The mean of the Poisson distribution is
+
´
+
+
´
´
´
+
+
0 12 1 39 2 47 + 3 40 4 20 5 17 6 3 7 2
´
´
+
´
´
m = = 2.5
180
364 LOVELY PROFESSIONAL UNIVERSITY