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 5:H0against 5:H1. Assume .05.The simplest way to test this is to use a p- value. We wish to test 5:H0against 5:H1. Count the number of service calls for a month. Let's say 7x. Since 7 is on the high side of 5, find 7xP. 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. {poiss}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.cvlxxP. Again 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 .0x 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 11xP .01370.98630.1101 xP Since 9110 xPxP03183.96817.1 is above 2.5%, our upper reject zone is .11xFor 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 .60512 Our hypotheses are now 60:H0against .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 x098.3606084xz and we will reject 0H.If we want a reject zone for ,x use 2zxcv , in this case the critical value is 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.1252oneslx1 1/13/05Classroom Example for Tests of Hypotheses Involving a Variance (Quoted from 252conf in the Syllabus supplement)The formula table has the following:Interval for Confidence IntervalHypotheses Test Ratio Critical ValueVariance- Small Sample 25.5.2221sn202:12020:H:H 20221sn 12025.5.22nscvVariance- Large Sample DFzDFs22220212020:H:H 1222DFzDFzDFscv222Assume that we want to see if the variance of the ages of a group of workers is 64 (Chiswick and Chiswick). Our hypotheses are 64:20H and 64:21H. The test ratio is the only method that is commonly used. Our data is a sample of 17n workers, and our computations give us a sample variance of .1002s 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 16201.122n. 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{chiSq} and 000.32 16201.. We can then compute the test ratio 20221sn .256410016 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 64:20H and 64:21H., but that this time our data is a sample of 73n workers, and our computations give us a sample variance of .1002s Let us set our significance level again at 2% 02. . We once again compute our test ratio of .5.112641007212022snThe 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 2zand 2z. 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, 4988.5.204.32 zPpval 0024.4988.5.2 .© 2002 R. E.
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