MTB > # The information with the data indicates that the mean is 10097/2608.
MTB > # I'll enter the data and check this.
MTB > set c1
DATA> 0:14
DATA> end
MTB > set c2
DATA> 57 203 383 525 532 408 273 139 45 27 10 4 0 1 1
DATA> end
MTB > let k1 = sum( c1*c2 )
MTB > let k2 = sum( c2 )
MTB > let k3 = k1/k2
MTB > name k1 'scintill' k2 'interval' k3 'mean'
MTB > print k1-k3
scintill 10097.0
interval 2608.00
mean 3.87155
MTB > # The supplied info checks out. The estimated mean is stored in k3.
MTB > # I'll put the probabilities for the categories 0, 1, 2, ..., 12, 13, 14+ in c3
MTB > # (where 14+ is 14 or more).
MTB > pdf c1 c3;
SUBC> pois k3.
MTB > # Right now, c3(15) is just P( exactly 14 ). I'll alter it to be P( 14 or more ).
MTB > dele 15 c3
MTB > let c3(15) = 1 - sum( c3 )
MTB > # I'll put the estimated expected frequencies in c4.
MTB > let c4 = k2*c3
MTB > name c1 'Count' c2 'Freq' c3 'prob' c4 'expected'
MTB > print c1-c4
ROW Count Freq prob expected
1 0 57 0.020826 54.314
2 1 203 0.080629 210.281
3 2 383 0.156080 407.057
4 3 525 0.201424 525.313
5 4 532 0.194955 508.444
6 5 408 0.150956 393.693
7 6 273 0.097406 254.034
8 7 139 0.053873 140.501
9 8 45 0.026071 67.994
10 9 27 0.011215 29.249
11 10 10 0.004342 11.324
12 11 4 0.001528 3.986
13 12 0 0.000493 1.286
14 13 1 0.000147 0.383
15 14 1 0.000054 0.142
MTB > # I'll do a check to make sure the sum of estimated expected values is 2608.
MTB > let k4 = sum( c4 )
MTB > name k4 'check' k5 'q' k6 'Q p-val' k7 'GLR stat' k8 'GLRp-val'
MTB > print k4
check 2608.00
MTB > # If I combined the last 4 categories to correspond to 11 or more,
MTB > # all of the estimated expected values will exceed 5. But I think
MTB > # the accuracy should be okay by just combining the last 3 categories
MTB > # (to have all of the estimated expected values be at least 1, and
MTB > # 11 out of 13 of them be at least 5).
MTB > dele 13:15 c2 c4
MTB > let c2(13) = 2608 - sum( c2 )
MTB > let c4(13) = 2608 - sum( c4 )
MTB > print c2 c4
ROW Freq expected
1 57 54.314
2 203 210.281
3 383 407.057
4 525 525.313
5 532 508.444
6 408 393.693
7 273 254.034
8 139 140.501
9 45 67.994
10 27 29.249
11 10 11.324
12 4 3.986
13 2 1.811
MTB > let k5 = sum( (c2 - c4)*(c2 - c4)/c4 )
MTB > let k7 = 2*sum( c2*loge( c2/c4 ) )
MTB > cdf k5 k6;
SUBC> chis 11.
MTB > cdf k7 k8;
SUBC> chis 11.
MTB > let k6 = 1 - k6
MTB > let k8 = 1 - k8
MTB > print k5-k8
q 12.9740
Q p-val 0.294982
GLR stat 14.0214
GLRp-val 0.231767
MTB > save 'Geiger'
Saving worksheet in file: Geiger.MTW
Above I got an approximate p-value of 0.29 using Pearson's chi-square test,
and an approximate p-value of 0.23 using an asymptotic GLR test. (Usually
the two tests give results in closer agreeement, especially when the sample
size is large.) But it can be noted that in both cases no strong evidence
was found against the Poisson model based on scientific theory. (So in that
sense, the two tests are in agreement.)