Hypothesis testing a Poisson mean using a normal approximation by forming a test statistic 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.
 A: 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")


