252Oneslx2 9/20/07(Open This Document in 'Page Layout' View!)

252oneslx2 9/20/07(Open this document in 'Page Layout' view!)

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 against . Assume

The simplest way to test this is to use a p- value. We wish to test

against . Count the number of service calls for a month. Let's say .

Since 7 is on the high side of 5, find . Use the Poisson table with a parameter

(mean) of 5. Since this is 2-sided test,

double the p-value. Since this is above our

significance level, do not reject . {poiss}

A harder way to do this is to set up 'accept' and reject zones. If we are still

working with the lower reject zone will be the numbers below and including

the largest number with . Again look at the Poisson table with a

parameter (mean) of 5. It says that Since is

above 2.5%, our lower reject zone is only The upper reject zone will be the

numbers above and including the smallest number with . Again

look at the Poisson table with a parameter (mean) of 5. It says that Since

is above 2.5%, our upper reject zone is

For a longer, more powerful test, observe the number of service calls over a

year. the number of calls, will have the Poisson distribution with parameter

Our hypotheses are now against 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 example, we will use . If we are still working with we will

reject the null hypothesis if or . For example if and we will reject .

If we want a reject zone for use , in this case the critical

value is . These critical values work out to 45.82 and 75.18. Again

if is above 75.18 and we will reject .

252oneslx1 1/13/05

Classroom Example for Tests of Hypotheses Involving a Variance (Quoted from

252confin the Syllabus supplement)

The formula table has the following:

Interval for / Confidence Interval / Hypotheses / Test Ratio / Critical Value
Variance-
Small Sample / / / /
Variance-
Large Sample / / / /

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 and . The

test ratio is the only method that is commonly used. Our data is a sample of

workers, and our computations give us a sample variance of Let us set our

significance level at 2% . The test ratio has the chi-squared distribution with

degrees of freedom. We will not reject the null hypothesis if the test ratio lies

between and . If we look up these two values we find

on the column in the chi-squared table for 16 degrees of freedom that {chiSq}

and . We can then compute the test ratio

Since this value is not below the first number from the table or above

the second number from the table, we cannot reject the null hypothesis.

Assume again that our hypotheses are and ., but

that this time our data is a sample of workers, and our computations give us a

sample variance of Let us set our significance level again at 2% .

We once again compute our test ratio of The test

ratio has the chi-squared distribution with degrees of freedom. We cannot find

an appropriate on our table because of the high number of degrees of freedom, so we

use the formula in the table excerpt above.

Since this statistic is , we will not reject the null hypothesis if is between

and . Since we use Since 3.04 is above the upper

critical value, 2.327, we reject . In fact, if we use a p- value, .