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WCU ECO 252 - Classroom Example for Tests of Hypotheses Involving a Poisson Distribution

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252onesx2 2/12/03(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 5:H0against 5:H1. Assume .05.The simplest way to test this is to use a p- value. We wish to test 5:H0against 5:H1. Count the number of service calls for a month. Let's say 7x. Since 7 is on the high side of 5, find  7xP. Use the Poisson table with a parameter (mean) of 5.    .23782.76218.1617  xPxPSince this is 2-sided test, double the p-value.    .47564.23782.272  xPpval Since this is above our significance level, do not reject 0H.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 cvlx with 025.cvlxxPAgain look at the Poisson table with a parameter (mean) of 5. It says that .00674.0 xP Since  04043.1 xPis above 2.5%, our lower reject zone is only .0x The upper reject zone will be the numbers above and including the smallest number cvux with 025.cvuxxP. Again look at the Poisson table with a parameter (mean) of 5. It says that   .01370.98630.110111  xPxP Since   03183.96817.19110  xPxPis above 2.5%, our upper reject zone is .11xFor a longer, more powerful test, observe the number of service calls over a year. ,x the numberof calls, will have the Poisson distribution with parameter  .60512 Our hypotheses are now60:H0against .60:H1 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 xz. If we are still working with ,05. we will reject the null hypothesis if 960.1025. zz or 960.1025. zz. For example if  ,84712 x098.3606084xz and we will reject 0H.If we want a reject zone for ,x use 2zxcv , in this case 6096.160 cvx. These critical values work out to 45.82 and 75.18. Again if  ,84712 xx is above 75.18 and we will reject 0H.1Classroom Example for Tests of Hypotheses Involving a Variance (Quoted from document is the Syllabus supplement)The formula table has the following:Interval for Confidence IntervalHypotheses Test Ratio Critical ValueVariance- Small Sample  25.5.2221sn202:12020:H:H 20221sn  12025.5.22nscvVariance- Large Sample  DFzDFs22220212020:H:H 1222DFz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 64:20H and 64:21H. The test ratio is the only method that iscommonly used. Our data is a sample of 17n workers, and our computations give us a sample variance of .1002s. Let us set our significance level at 2%  02. . The test ratio has the chi-squared distribution with 161 ndegrees of freedom. We will not reject the null hypothesis if the test ratio lies between    162.991-n2-1 2 and    16201.122n. If we look up these two values we find on the column in the chi-squared table for 16 degrees of freedom that   812.5 162.99 and  000.32 16201.. We can then compute the test ratio    .25641001612022sn 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 64:20H and 64:21H., but that this time our data is a sample of 73n workers, and our computations give us a sample variance of .1002s Letus set our significance level again at 2%  02. . We once again compute our test ratio of   .5.112641007212022snThe test ratio has the chi-squared distribution with 721 ndegrees of freedom. We cannot find an appropriate 2 on our table because of the high number of degrees of freedom, so we use the z formula in the table excerpt above.    04.396.1100.1514322517225.11221222 DFz.Since this statistic is  1,0N, we will not reject the null hypothesis if z is between 2z and 2z. Since ,02. we use .327.201.2 zz. Since 3.04 is above the upper critical value, 2.327, we reject 0H. In fact, if we use a p- value,    0024.4988.5.204.32  zPpval.© 2002 R. E.


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