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One of my study materials claims that when you need to do a hypothesis test for a poisson mean $\lambda$, if the statistics table does not have the value for the hypothesized $\lambda$ value, then a hypothesis test should be carried out by calculating a test statistic $Z$ where $$Z=\frac{\hat\lambda-\lambda_0}{\hat\lambda/n}$$

$\hat\lambda$ is the point estimate for the mean from a sample of size $n$, and

$\lambda_0$ is the hypothesised mean.

Then, reject/fail to reject $H_0$ accordingly.

Could someone please confirm both the test statistic and the approach, as I have found few to no sources that agrees/disagrees with this.

EDIT: Some weeks back I did suspect that the denominator should have a square root, as suggested by @gung in the comments. The study material I used came from this site.

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  • $\begingroup$ Does the denominator have a square root? $\endgroup$ May 9, 2017 at 13:24
  • $\begingroup$ @gung no it doesn't, at least not according to the material $\endgroup$
    – tsp216
    May 9, 2017 at 13:31
  • $\begingroup$ You should add the self-study tag, and add a reference to your study material. $\endgroup$ May 9, 2017 at 15:46

1 Answer 1

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I'm not sure what that is. I wonder if there is a typo, or an error in the formula.

You can test if a parameter estimate, estimated by maximum likelihood, differs from a null value by performing a Wald test. Wald tests follow from the assumption that the maximum likelihood estimate would be distributed as a standard normal 'at infinity'. If your $N$ is sufficiently large, you could use it. Here are two forms of the Wald test:
\begin{align} z &= \frac{\hat\theta - \theta_0}{\sqrt{\frac{{\rm Var}(\hat\theta)}{N}}} \\[10pt] \chi^2 &= \frac{(\hat\theta - \theta_0)^2}{\frac{{\rm Var}(\hat\theta)}{N}} \end{align} Note that the second version is just the square of the first. Your formula looks like the top one, with the square root of the denominator missing. (Remember that for the Poisson distribution, $\lambda$ is both the mean and the variance.) The result is that the denominator will be too small and the standard deviation of the values will be too large for a standard normal. Note that in the quickie simulation below (coded in R) using the square root yields values that range over a sensible interval, but omitting it makes they range over a huge interval. Other than that, the resulting values are normally distributed either way:

library(car)                   # we'll use this package
set.seed(514)                  # this makes the example exactly reproducible
z.vect = vector(length=10000)  # these will store the results of the simulation
Z.vect = vector(length=10000)
for(i in 1:10000){             # we'll do this 10k times
  N = 500                      # there will be N=500 on each iteration
  l = 5                        # the null value of lambda is 5
  x = rpois(N, lambda=l)       # the true lambda is 5 (ie, the null is true)
  m = mean(x)                  # here we estimate the mean / lambda
  z.vect[i] = (m-l)/sqrt(m/N)  # these are the two formulae
  Z.vect[i] = (m-l)/    (m/N)
}
sd(z.vect)                     # the SD matches a standard normal
# [1] 1.001831
sd(Z.vect)                     # the SD is way too large for a standard normal
# [1] 10.02672
windows(height=4,width=7)
  layout(matrix(1:2, nrow=1))
  qqPlot(z.vect, main="With square root")
  qqPlot(Z.vect, main="Without square root")

enter image description here

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  • $\begingroup$ Then I guess the TS I proposed (with the sqrt in the denominator) is more or less accurate. $\endgroup$
    – tsp216
    May 10, 2017 at 3:53
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    $\begingroup$ With such a large sample size (500) and Poisson variates with mean 5 (whose averages will go to normal nice and fast), a standardized version of the square-rootless statistic will still be asymptotically normal (via Slutsky and the CLT) and for this case we'd expect the approximation to work well at fairly moderate sample sizes -- so your results here are exactly as we'd expect. If you want to see a difference in the two forms you'd need a considerable smaller $N$. Try N=20. see here (the blue line is the line for a standard normal) $\endgroup$
    – Glen_b
    May 10, 2017 at 10:18
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    $\begingroup$ An even better asymptotic test (in the sense that it works well at smaller $\lambda$ and $N$) is $Z = 2\left(\sqrt{\hat{\lambda}+3/8}-\sqrt{\lambda_0+3/8}+\frac{1}{4\sqrt {\lambda_0}}\right)$. $\endgroup$
    – Glen_b
    May 10, 2017 at 10:38

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