endobj So we don't cover that either. \] or \[ \], \[ x + \frac{z_{\alpha / 2}^2}{2} endobj e CP i! 27 0 obj \mu^2 - (2 x + z_{\alpha / 2}^2) \mu + x^2 < 0 <>41 0 R]/P 44 0 R/S/Link>> x��VۊG}��HMI�+4=7�`C�f�`�]�1��e���QUOO�d�!K /Xj��J:ґ�j5>>=ܿ}������ϻ~�����_���~��v�D˶��g](���C�X��Hٰ�?.��{������������ۗ��NV7���g#! endobj \] and this has the problem of blowing up when \(\hat{mu} = 0\). <><>19 20]/P 18 0 R/Pg 47 0 R/S/Link>> This interval is \[ \\ \hat{\mu} \pm z_{\alpha / 2} \cdot \sqrt{\hat{\mu}} \] From the quadratic formula this interval has end points \[ This is just like the score test except that the variance is estimated under the alternative rather than under the null. endobj 13 0 obj 39 0 obj We can plot the log likelihood function using the following code. The log likelihood goes to minus infinity as \(\mu \to 0\) or \(\mu \to \infty\). x^2 - 2 x \mu + \mu^2 < \mu z_{\alpha / 2}^2 \end{align*}\], \[\begin{align*} <>stream t = \frac{x - \mu_0}{\sqrt{\mu_0}} <>45 0 R]/P 25 0 R/S/Link>> <>11]/P 17 0 R/Pg 47 0 R/S/Link>> endobj \pm endobj 44 0 obj H_1: & \mu \neq \mu_0 Appligent AppendPDF Pro 6.3 2020-06-26T15:30:45-07:00 endobj <> H_1: & \mu > \mu_0 Nishantha Janith Chandrasena Poddiwala Hewage As always the test statistic for a two-tailed test is \[ <> 35 0 obj H_0: & \mu = \mu_0 \] so the interval is the set of \(\mu\) such that \[ alpha <- 1 - conf.level c(0, - log(alpha)) H_0: & \mu = \mu_0 z_{\alpha / 2} \sqrt{x + \frac{z_{\alpha / 2}^2}{4}} Wald Confidence Intervals for a Single Poisson Parameter and Binomial Misclassification Parameter When the Data is Subject to Misclassification \frac{x - \mu}{\sqrt{\mu}} z_{\alpha / 2} \sqrt{x + \frac{z_{\alpha / 2}^2}{4}} \] where \(z_{\alpha / 2}\) is the \(1 - \alpha / 2\) quantile of the standard normal distribution, because the variance of the distribution is \(\mu\). <> \frac{2 x + z_{\alpha / 2}^2 \pm \sqrt{(2 x + z_{\alpha / 2}^2)^2 - 4 x^2}}{2} \]. <>2]/P 6 0 R/Pg 47 0 R/S/Link>> endobj 1 \[ Hence we include in our plot only the part of the curve in which the log likelihood is within 10 of the maximum. t = \frac{x - \mu_0}{\sqrt{\mu_0}} 1 0 obj endobj The 10 was pulled out of the air. \\ 64 0 obj There is a fuzzy test, but there is no computer implementation available. Prince 12.5 (www.princexml.com) The Wald interval can be repaired by using a different procedure (Geyer, 2009, Electronic Journal of Statistics, 3, 259--289). x + \frac{z_{\alpha / 2}^2}{2} \], \[ Thus when we observe \(x = 0\) and want 95% confidence, the interval is. AppendPDF Pro 6.3 Linux 64 bit Aug 30 2019 Library 15.0.4 [58 0 R 60 0 R 61 0 R 62 0 R 63 0 R] \], \[\begin{align*} \[ On could also there are many possible two-tailed exact tests and no reason to prefer any one of them. uuid:e6f70d1f-aeb8-11b2-0a00-70d268010000 \] or \[ posed an approximate confidence interval for a Poisson mean, called adding the tail probability of the Wald in- terval (AWC) as follows, 2 2 2. z X Xz 2n n (8) For any nominal (1 )100% confidence interval for mean ( ), the coverage probability at a fixed value of is given by i Li Ui i0. endobj t = \frac{x - \mu_0}{\sqrt{\hat{\mu}}} \], the section on likelihood-based confidence intervals below. 2 0 obj endobj endstream t = 2 [ l(\hat{\mu}) - l(\mu_0) ] 34 0 obj %PDF-1.7 %���� \], \[ It is best programming practice to never hard code numbers like this, that is, the number 0.95 should only occur in your document once where it is used to initialize a variable. Confidence interval (CI) of Poisson means that popular is 2 ˆ znˆ and z 2 is the 100 1 2 percentile of the standard normal distribution. \], \[ \] where in both places \(\mu\) is now the value of \(\mu\) hypothesized under the null hypothesis. x^2 - 2 x \mu + \mu^2 < \mu z_{\alpha / 2}^2 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. endobj \] or \[ endobj \hat{\mu} \pm z_{\alpha / 2} \cdot \sqrt{\hat{\mu}} 37 0 obj We don't use this plot for statistical inference. <><>5 6]/P 25 0 R/Pg 59 0 R/S/Link>> \end{align*}\], \[\begin{align*} Hence the ylim optional argument to R function curve. <>/Metadata 2 0 R/Outlines 5 0 R/Pages 3 0 R/StructTreeRoot 6 0 R/Type/Catalog/ViewerPreferences<>>> endobj \], The test statistic is (repeating what was said in the section on the score interval above) \[ 36 0 obj endobj <>1]/P 12 0 R/Pg 47 0 R/S/Link>> <>15]/P 18 0 R/Pg 47 0 R/S/Link>> endobj \frac{2 x + z_{\alpha / 2}^2 \pm \sqrt{(2 x + z_{\alpha / 2}^2)^2 - 4 x^2}}{2} H_1: & \mu < \mu_0 \end{align*}\], \[ 43 0 obj <>9]/P 26 0 R/Pg 59 0 R/S/Link>> 45 0 obj The usual estimator of the parameter \(\mu\) is \(\hat{\mu} = x\). So nobody recommends this (for Poisson), and we won't even illustrate it. 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. \], \[ 42 0 obj endobj <>/MediaBox[0 0 612 792]/Parent 67 0 R/Resources<>/Font<>/ProcSet[/PDF/Text/ImageC]/XObject<>>>/StructParents 1/Tabs/S/Type/Page>> 28 0 obj t = \frac{x - \mu}{\sqrt{\mu}} 32 0 obj <> endobj Then use that variable elsewhere. <>17]/P 18 0 R/Pg 47 0 R/S/Link>> t = 2 [ l(\hat{\mu}) - l(\mu_0) ] The test statistic is \[ The reason for the x - 1 is the discreteness of the Poisson distribution (that's the way lower.tail = FALSE works). But that gives us a plot in which it is hard to see what is going on. \mu^2 - (2 x + z_{\alpha / 2}^2) \mu + x^2 < 0 \left\lvert \frac{x - \mu}{\sqrt{\mu}} \right\rvert < z_{\alpha / 2} \], \[ To derive this inverval we invert the score test.

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