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326 Fundamentals of Probability and Statistics for Engineers
Table 10.6 Table for 2 test for Example 10.4
2
Interval, A i n i p i np i n /np i
i
x 56 13 0.161 11.27 15.00
56 < x 72 10 0.178 12.46 8.03
72 < x 88 20 0.224 15.68 25.51
88 < x 104 13 0.205 14.35 11.78
104 < x 120 8 0.136 9.52 6.72
120 < x 6 0.096 6.72 5.36
70 1.0 70 72.40
With intervals A i defined as shown in the first column of Table 10.6, theoretical
probabilities P(A i ) now can be calculated with the aid of Table A.3. For
example, the first two of these probabilities are
56 83:6
P
A 1 P
X 56 PU p F U
0:990
777:4
1 F U
0:990 1 0:8389 0:161;
P
A 2 P
56 < X 72 P
0:990 < U 0:416
1 F U
0:416 1 F U
0:990
0:339 0:161 0:178:
The information given above allows us to construct Table 10.6. Hence,
we have
k 2
X n i
d n 72:40 70 2:40:
np i
i1
r
The number of degrees of freedom in this case is k 1 6 1 3.
2
Table A.5 thus gives
2 3; 0:05 7:815:
Since d < 2 3, 0 05 normal distribution N(83 6, 777 4) is acceptable at the 5%
.
.
significance level.
Before leaving this section, let us remark again that statistic D in the 2 test is
>
2 -distributed only when n !1 . It is thus a large sample test. As a rule, n 50
is considered satisfactory for fulfilling the large-sample requirement.
TLFeBOOK