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.)