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Classroom Example for Tests of Hypotheses Involving a Poisson distribution.
In statistics, ‘rare events’ is a code word for the Poisson distribution.
The example given in class was a situation in which a service contract business asserts that it only
gets 5 calls per month. We wish to test H 0 :   5 against H 1 :   5 . Assume   .05 . The simplest way to
test this is to use a p- value. We wish to test H 0 :   5 against H 1 :   5 . Count the number of service
calls for a month. Let's say x  7 . Since 7 is on the high side of 5, find Px  7 . Use the Poisson table
with a parameter (mean) of 5. Px  7  1  Px  6  1  .76218  .23782 . Since this is 2-sided test,
double the p-value. pval  2Px  7  2.23782   .47564 . Since this is above our significance level, do
not reject H 0 .
A harder way to do this is to set up 'accept' and reject zones. If we are still working with   .05,
the lower reject zone will be the numbers below and including the largest number x cvl with
Px  x cvl   .025 Again look at the Poisson table with a parameter (mean) of 5. It says that
Px  0  .00674. Since Px  1  .04043 is above 2.5%, our lower reject zone is only x  0. The
upper reject zone will be the numbers above and including the smallest number x cvu with
Px  xcvu   .025 . Again look at the Poisson table with a parameter (mean) of 5. It says that
Px  11  1  Px  10  1  .98630  .01370 . Since Px  10  1  Px  9  1  .96817  .03183 is above
2.5%, our upper reject zone is x  11 .
For a longer, more powerful test, observe the number of service calls over a year. x, the number of
calls, will have the Poisson distribution with parameter   125  60 . Our hypotheses are now
H 0 :   60 against H1 :   60 . We do not have an appropriate Poisson table, but we know that this
distribution is approximated by a Normal distribution with a mean equal to the Poisson mean of 60 and a
standard deviation equal to the square root of the Poisson mean. This means that if we use a test ratio, for
x
example, we will use z 
. If we are still working with   .05, we will reject the null hypothesis if

z   z.025  1.960 or z  z.025  1.960 . For example if x  12 7   84, z 
x


84  60
 3.098 and
60
we will reject H 0 .
If we want a reject zone for x, use xcv    z  , in this case x cv  60  1.96 60 . These
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critical values work out to 45.82 and 75.18. Again if x  12 7   84, x is above 75.18 and we will reject
H0 .
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Classroom Example for Tests of Hypotheses Involving a Variance (Quoted from document is the Syllabus
supplement)
The formula table has the following:
Interval for
Confidence
Interval
Variancen  1s 2
2  2
Small Sample
.5 .5 2 
s 2DF 
Variance 
Large Sample
 z  2DF 
Hypotheses
Test Ratio
H 0 :  2   02
2 
H1: :  2   02
H 0 :    02
H1 :  2   02
2
2
n  1s 2
 02
Critical Value
2
s cv

 .25 .5 2  02
n 1
z 
2  2  2DF   1
Assume that we want to see if the variance of the ages of a group of workers is 64 (Chiswick and Chiswick)
Our hypotheses are H 0 :  2  64 and H 1 :  2  64 . The test ratio is the only method that is commonly
used. Our data is a sample of n  17 workers, and our computations give us a sample variance of
s 2  100 . . Let us set our significance level at 2%   .02  . The test ratio has the chi-squared distribution
with n  1  16 degrees of freedom. We will not reject the null hypothesis if the test ratio lies between
216
and  2n1   .20116 . If we look up these two values we find on the column in the chi 12-n -1   .99
2
2
216
squared table for 16 degrees of freedom that  .99
 5.812 and  .20116  32 .000 . We can then compute
the test ratio  2 
n  1s 2
 02

16 100 
 25 . Since this value is not below the first number from the table or
64
above the second number from the table, we cannot reject the null hypothesis.
Assume again that our hypotheses are H 0 :  2  64 and H 1 :  2  64 ., but that this time our data
is a sample of n  73 workers, and our computations give us a sample variance of s 2  100 . Let us set
our significance level again at 2%   .02  . We once again compute our test ratio of
2 
n  1s 2
 02

72 100 
 112 .5. The test ratio has the chi-squared distribution with n  1  72 degrees of
64
freedom. We cannot find an appropriate  on our table because of the high number of degrees of freedom,
so we use the z formula in the table excerpt above.
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z  2 2  2DF  1  2112 .5  272   1  225  143  15 .00  11 .96  3.04 .
Since this statistic is N 0,1 , we will not reject the null hypothesis if z is between  z and z . Since
2
2
  .02 , we use  z   z.01  2.327. . Since 3.04 is above the upper critical value, 2.327, we reject
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H 0 . In fact, if we use a p- value, pval  2Pz  3.04  2.5  .4988   .0024 .
© 2002 R. E. Bove
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