The Poisson probability law gives the probability distribution of the number of events occurring in a given interval of time or space. There are several different tests to compare the rates from two Poisson populations.
When you start the tests, the app will ask you to input some values:
The following is an example:
TESTS FOR TWO POISSON RATES
G X T S R
----------------------------------
1 209 1 40 5.225
2 230 2 30 3.8333
----------------------------------
G: Group
X: Number of events observed
T: Time interval
S: Size or space
R: Rate = X / (TS)
Difference of rates:
r1-r2 = 5.225 - 3.833333
= 1.391667
95% Confidence Interval
λ1 = 5.225 ± 0.708371
= (4.516629 5.933371)
λ2 = 3.833333 ± 0.495405
= (3.337928 4.328738)
λ1-λ2 = 1.391667 ± 0.864417
= (0.52725 2.256083)
Two-Tailed Test for Population Rates
Ho: λ1 = λ2
Ha: λ1 ≠ λ2
Significance level = 0.05
Using Normal Distribution
Critical Value = ±1.96
Test Statistic
Z = (cx - y) / √(c2x + y)
= (1.5·209) / √(1.52·209 + 230)
= 3.1554
where c = (t2 / t1) * (s2 / s1)
P(>|Z|) = 0.001603 (1.60E-3)
Reject the Ho at α = 0.05
∴ λ1 ≠ λ2
Using Chi-Square Distribution
G X TS E
------------------------------
1 209 40 175.6
2 230 60 263.4
------------------------------
439 100
G: Group
X: Number of events observed
TS: T * S
T: Time interval
S: Size or space
E: Expected value
= 439 * (TS / 100)
Critical Value:
χ2(1, 0.05) = 3.8415
Test Statistic:
χ2 = ∑(O - E)2 / E
= (209 - 175.6)2 / 175.6
+ (230 - 263.4)2 / 263.4
= 10.5881
P(>χ2) = 0.001138 (1.14E-3)
Reject the Ho at α = 0.05
∴ &lambda1 ≠ &lambda2
Using Binomial Distribution
(Exact Test, C-Test)
P-value = 0.001466 (1.47E-3)
Reject the Ho at α = 0.05
∴ &lambda1 ≠ &lambda2
One-Tailed Test for Population Rates
Ho: λ = λ2
Ha: λ1 > λ2
Significance level = 0.05
Using Normal Distribution
Critical Value = 1.6449
Test Statistic
Z = (cx - y) / √(c2x + y)
= (1.5·209) / √(1.52 · 209 + 230)
= 3.1554
where c = (t2 / t1) * (s2 / s1)
P(>Z) = 0.000801 (8.01E-4)
Reject the Ho at α = 0.05
∴ λ1 > λ2
Using Binomial Distribution
(Exact Test, C-Test)
P-value = 0.000733 (7.33E-4)
Reject the Ho at α = 0.05
∴ λ1 > λ2
When you input more than two input for each category, the app will test only the χ2-test:
TESTS FOR 4 POISSON RATES
G X T S R
----------------------------------
1 4 1 1,260 0.0032
2 1 1 2,080 0.0005
3 7 1 1,425 0.0049
4 10 1 1,650 0.0061
----------------------------------
G: Group
X: Number of events observed
T: Time interval
S: Size or space
R: Rate = X / (TS)
Two-Tailed Test for Population Rates
Ho: All groups have the same rates.
Ha: Not Ho
Significance level = 0.05
Using Chi-Square Distribution
G X TS E
------------------------------
1 4 1,260 4.3211
2 1 2,080 7.1333
3 7 1,425 4.887
4 10 1,650 5.6586
------------------------------
22 6,415
G: Group
X: Number of events observed
TS: T * S
T: Time interval
S: Size or space
E: Expected value
= 22 * (TS / 6,415)
Critical Value:
χ2(3, 0.05) = 7.8147
Test Statistic:
χ2 = ∑(O - E)2 / E
= (4 - 4.3211)2 / 4.3211
+ (1 - 7.1333)2 / 7.1333
+ (7 - 4.887)2 / 4.887
+ (10 - 5.6586)2 / 5.6586
= 9.5417
P(>χ2) = 0.022892
Reject the Ho at α = 0.05
∴ At least one group has different rate.