LibraryPirate

606 CHAPTER 12 THE CHI-SQUARE DISTRIBUTION AND THE ANALYSIS OF FREQUENCIES
TABLE 12.3.8 Observed and Expected Frequencies and Components
of X 2 for Example 12.3.3
Number of
Number of Days this Expected
Emergency Number Relative Expected
Admissions Occurred, O i Frequency Frequency
1O i E i 2 2
E i
0 5 .050 4.50 .056
1 14 .149 13.41 .026
2 15 .224 20.16 1.321
3 23 .224 20.16 .400
4 16 .168 15.12 .051
5 9 .101 9.09 .001
6 3 .050 4.50 .500
7 3 .022 1.98 .525
8 1 .008 .72
9 1 r 2 .003 .27 r .784
10 or more 0 .001 .09
Total 90 1.000 90.00 3.664
is multiplied by 90 to obtain the corresponding expected frequencies.
These values along with the observed and expected frequencies and the
components of X 2 , 1O i - E i 2 2 >E i , are displayed in Table 12.3.8, in which
we see that
X 2 = a c 1O i - E i 2 2
E i
d =
15 - 4.5022
4.50
+ . . . +
12 - 1.0822
1.08
= 3.664
We also note that the last three expected frequencies are less than 1, so that
they must be combined to avoid having any expected frequencies less than 1.
This means that we have only nine effective categories for computing
degrees of freedom. Since the parameter, l, was specified in the null
hypothesis, we do not lose a degree of freedom for reasons of estimation,
so that the appropriate degrees of freedom are 9 - 1 = 8. By consulting
Appendix Table F, we find that the critical value of x 2 for 8 degrees of freedom
and a = .05 is 15.507, so that we cannot reject the null hypothesis at
the .05 level, or for that matter any reasonable level, of significance
1p 7 .102. We conclude, therefore, that emergency admissions at this hospital
may follow a Poisson distribution with l = 3. At least the observed
data do not cast any doubt on that hypothesis.
If the parameter l has to be estimated from sample data, the estimate
is obtained by multiplying each value x by its frequency, summing these
products, and dividing the total by the sum of the frequencies.
■