<>11]/P 17 0 R/Pg 47 0 R/S/Link>> e CP i! The Wald interval can be repaired by using a different procedure (Geyer, 2009, Electronic Journal of Statistics, 3, 259--289). \] or \[ \], \[ This same fix can be used for any confidence interval, not just the Wald interval. 2020-06-26T15:30:45-07:00 The test statistic is \[ <>/Metadata 2 0 R/Outlines 5 0 R/Pages 3 0 R/StructTreeRoot 6 0 R/Type/Catalog/ViewerPreferences<>>> \\ \mu^2 - (2 x + z_{\alpha / 2}^2) \mu + x^2 < 0 43 0 obj 6 0 obj alpha <- 1 - conf.level c(0, - log(alpha)) \], the section on likelihood-based confidence intervals below. endobj H_1: & \mu \neq \mu_0 \hat{\mu} \pm z_{\alpha / 2} \cdot \sqrt{\hat{\mu}} <>9]/P 26 0 R/Pg 59 0 R/S/Link>> endobj endobj endobj <>2]/P 6 0 R/Pg 47 0 R/S/Link>> Call that \(\mu_0\) and we have \[ endobj \], \[ For the Poisson distribution, this recipe uses the interval \([0, - \log(\alpha)]\) for coverage \(1 - \alpha\). x^2 - 2 x \mu + \mu^2 < \mu z_{\alpha / 2}^2 endstream %PDF-1.7
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And we can also use this signed likelihood ratio test statistic for the two-tailed test because its square is the original likelhood ratio test statistic. \]. \] where in both places \(\mu\) is now the value of \(\mu\) hypothesized under the null hypothesis. <>34]/P 22 0 R/Pg 47 0 R/S/Link>> <>21]/P 19 0 R/Pg 47 0 R/S/Link>> \hat{\mu} \pm z_{\alpha / 2} \cdot \sqrt{\hat{\mu}} 32 0 obj uuid:e6f70d20-aeb8-11b2-0a00-80860c4efe7f This turns out to also be the maximum likelihood estimator. 5 0 obj \\ x��VۊG}��HMI�+4=7�`C�f�`�]�1��e���QUOO�d�!K /Xj��J:ґ�j5>>=ܿ}������ϻ~�����_���~��v�D˶��g](���C�X��Hٰ�?.��{������������ۗ��NV7���g#! \] so the interval is the set of \(\mu\) such that \[ endobj endobj \], \[ 59 0 obj The log likelihood goes to minus infinity as \(\mu \to 0\) or \(\mu \to \infty\). 35 0 obj It will turn out that the only interesting part of the log likelihood is the region near the maximum. Nishantha Janith Chandrasena Poddiwala Hewage 65 0 obj I This formula, we call Wald method, it is easy to present and compute but it has poor coverage properties for small n. <>41 0 R]/P 44 0 R/S/Link>> posed an approximate confidence interval for a Poisson mean, called adding the tail probability of the Wald in- terval (AWC) as follows, 2 2 2. z X Xz 2n n (8) For any nominal (1 )100% confidence interval for mean ( ), the coverage probability at a fixed value of is given by i Li Ui i0. H_1: & \mu < \mu_0 Then use that variable elsewhere. \], \[ \] or \[ Thus we cannot try to draw the curve from zero to infinity but rather from a little bit above zero to some big number. endobj This interval is \[ Confidence interval (CI) of Poisson means that popular is 2 ˆ znˆ and z 2 is the 100 1 2 percentile of the standard normal distribution. 44 0 obj \\ z_{\alpha / 2} \sqrt{x + \frac{z_{\alpha / 2}^2}{4}} Thus when we observe \(x = 0\) and want 95% confidence, the interval is. 2020-06-26T15:30:45-07:00 There is no exact two-tailed because the exact (Poisson) distribution is not symmetric, so there is no reason to us \(\lvert X - \mu_0 \rvert\) as a test statistic. Also it really doesn't matter, since we are just using this plot to get some idea what is going on. endobj {�}��}�:POR�ﻁi��cX�_�wof��7�僌�9����u�ao���o��ǟa˛s\$7����/"�8�|e�ўGxJ�_}e�"���.r0!Kj, Nishantha Janith Chandrasena Poddiwala Hewage, Wald Confidence Intervals for a Single Poisson Parameter and Binomial Misclassification Parameter When the Data is Subject to Misclassification. \] where \(z_{\alpha / 2}\) is the \(1 - \alpha / 2\) quantile of the standard normal distribution, because the variance of the distribution is \(\mu\). 39 0 obj z_{\alpha / 2} \sqrt{x + \frac{z_{\alpha / 2}^2}{4}} 42 0 obj H_1: & \mu > \mu_0 To derive this inverval we invert the score test. endobj 2020-06-26T15:30:45-07:00 Wald Confidence Intervals for a Single Poisson Parameter and Binomial Misclassification Parameter When the Data is Subject to Misclassification Set the confidence level. <> So nobody recommends this (for Poisson), and we won't even illustrate it. interval or region of space. Hence we include in our plot only the part of the curve in which the log likelihood is within 10 of the maximum. \], \[ The dashed vertical line shows where the MLE is, and it does appear to be where the log likelihood is maximized. APPROVED: Kent Riggs, Ph.D., Thesis Director Robert K. Henderson, Ph.D., Committee Member Jacob Turner, Ph.D., Committee Member Prince 12.5 (www.princexml.com) It is best programming practice to never hard code numbers like this, that is, the number 0.95 should only occur in your document once where it is used to initialize a variable. 1 0 obj \pm <> t = 2 [ l(\hat{\mu}) - l(\mu_0) ] 27 0 obj 26 0 obj 45 0 obj endobj <>1]/P 12 0 R/Pg 47 0 R/S/Link>> The 10 was pulled out of the air. Hence the ylim optional argument to R function curve. For the Poisson distribution, this recipe uses the interval \([0, - \log(\alpha)]\) for coverage \(1 - \alpha\). <> 1 \left\lvert \frac{x - \mu}{\sqrt{\mu}} \right\rvert < z_{\alpha / 2} \], \[ \[ But that gives us a plot in which it is hard to see what is going on. There is a fuzzy test, but there is no computer implementation available. \frac{2 x + z_{\alpha / 2}^2 \pm \sqrt{(2 x + z_{\alpha / 2}^2)^2 - 4 x^2}}{2} application/pdf t = \frac{x - \mu_0}{\sqrt{\hat{\mu}}} <>17]/P 18 0 R/Pg 47 0 R/S/Link>> uuid:e6f70d1f-aeb8-11b2-0a00-70d268010000 endobj \] and has a chi-square distribution on one degree of freedom under the null hypothesis. Thus when we observe \(x = 0\) and want 95% confidence, the interval is. 37 0 obj We don't really want to get scientific about this yet (but do in the section on likelihood-based confidence intervals below). endobj <>stream
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