\[ \frac{2 x + z_{\alpha / 2}^2 \pm \sqrt{(2 x + z_{\alpha / 2}^2)^2 - 4 x^2}}{2} t = \frac{x - \mu_0}{\sqrt{\mu_0}} \\ Set the confidence level. The dashed vertical line shows where the MLE is, and it does appear to be where the log likelihood is maximized. t = \frac{x - \mu_0}{\sqrt{\hat{\mu}}} It will turn out that the only interesting part of the log likelihood is the region near the maximum. \], the section on likelihood-based confidence intervals below. 45 0 obj endobj \], \[ <> The usual estimator of the parameter \(\mu\) is \(\hat{\mu} = x\). H_0: & \mu = \mu_0 APPROVED: Kent Riggs, Ph.D., Thesis Director Robert K. Henderson, Ph.D., Committee Member Jacob Turner, Ph.D., Committee Member This turns out to also be the maximum likelihood estimator. endobj 2020-06-26T15:30:45-07:00 t = \frac{x - \mu}{\sqrt{\mu}} \hat{\mu} \pm z_{\alpha / 2} \cdot \sqrt{\hat{\mu}} endobj 2020-06-26T15:30:45-07:00 Nishantha Janith Chandrasena Poddiwala Hewage \], \[ endobj <> 64 0 obj The test statistic is \[ 27 0 obj endobj \left\lvert \frac{x - \mu}{\sqrt{\mu}} \right\rvert < z_{\alpha / 2} Then use that variable elsewhere. Also it really doesn't matter, since we are just using this plot to get some idea what is going on. <>21]/P 19 0 R/Pg 47 0 R/S/Link>> This interval is \[ <>1]/P 12 0 R/Pg 47 0 R/S/Link>> <>45 0 R]/P 25 0 R/S/Link>> This formula, we call Wald method, it is easy to present and compute but it has poor coverage properties for small n. <> x + \frac{z_{\alpha / 2}^2}{2} Wald Confidence Intervals for a Single Poisson Parameter and Binomial Misclassification Parameter When the Data is Subject to Misclassification <><>5 6]/P 25 0 R/Pg 59 0 R/S/Link>> 2 0 obj \end{align*}\], \[\begin{align*} For the Poisson distribution, this recipe uses the interval \([0, - \log(\alpha)]\) for coverage \(1 - \alpha\). <><>19 20]/P 18 0 R/Pg 47 0 R/S/Link>> The log likelihood goes to minus infinity as \(\mu \to 0\) or \(\mu \to \infty\). t = 2 [ l(\hat{\mu}) - l(\mu_0) ] endobj 2020-06-26T15:30:45-07:00 The reason for the x - 1 is the discreteness of the Poisson distribution (that's the way lower.tail = FALSE works). <>9]/P 26 0 R/Pg 59 0 R/S/Link>> As with the binomial the Wald interval is zero width, hence ridiculous when \(\hat{\mu} = 0\). <>30]/P 21 0 R/Pg 47 0 R/S/Link>> There is a fuzzy test, but there is no computer implementation available. The Wald interval can be repaired by using a different procedure (Geyer, 2009, Electronic Journal of Statistics, 3, 259--289). \end{align*}\], \[\begin{align*} z_{\alpha / 2} \sqrt{x + \frac{z_{\alpha / 2}^2}{4}} \], \[ \] or \[ Thus when we observe \(x = 0\) and want 95% confidence, the interval is. \frac{2 x + z_{\alpha / 2}^2 \pm \sqrt{(2 x + z_{\alpha / 2}^2)^2 - 4 x^2}}{2}

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