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 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 2 P 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 x cvu 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 H 1 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 x 84 60 z 3 098 and we will reject H 0 60 If we want a reject zone for x use x cv z in this case 2 These critical values work out to 45 82 and 75 18 Again if x 12 7 84 x cv 60 1 96 60 x is above 75 18 and we will reject H 0 1 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 Variances 2 DF Large Sample z 2 DF 2 Hypotheses Test Ratio H 0 2 02 H 1 2 02 2 H 0 02 H 1 2 02 2 Critical Value n 1 s 2 02 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 chisquared distribution with n 1 16 degrees of freedom We will not reject the null hypothesis if the test ratio lies between 2 16 2 n 1 12 n 1 99 201 16 If we look up these two values we find on the and 2 2 2 16 5 812 and 201 16 32 000 column in the chi squared table for 16 degrees of freedom that 99 2 We can then compute the test ratio n 1 s 2 02 16 100 25 Since this value is not below the 64 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 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 64 degrees 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 z 2 2 2 DF 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 Since 02 we use z 2 z 01 2 327 z is between z 2 and z 2 Since 3 04 is above the upper critical value 2 327 we reject H 0 In fact if we use a p value pval 2 P z 3 04 2 5 4988 0024 2002 R E Bove 2
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