Hypothesis Testing Question (Poisson Random Variable)

Question

$\text{The question:}$ A new web page design is intended to increase the rate at which customers place orders. Prior to the new design, the number of orders in an hour was a Poisson random variable with mean $\lambda=30$. Eight one-hour measurements with the new design find an average 32 orders completed per hour.

a) At a $5\%$ significance level, do the data support the claim that the order placement has increased? b) For $1\%$?

$\text{My Solution:}$ We model this Poisson random variable as a Gaussian (with variance being equal to $\lambda$), so: $$\alpha = 0.05 = P(\hat{X}_8>30+c)$$ $$P\Bigg[\frac{\hat{X}_{8}-30}{\sqrt{\frac{30}{8}}}>\frac{c}{\sqrt{\frac{30}{8}}}\Bigg]=Q\Bigg[\frac{c}{\sqrt{\frac{30}{8}}}\Bigg]$$ $$\implies c= Q^{-1}(0.05)\cdot\sqrt{\frac{30}{8}}\approx3.18$$ Now, since $32<33.18$, we can't reject the null hypothesis that the mean is $30$.

Now with $\alpha=0.01$ we get $c= Q^{-1}(0.01)\cdot\sqrt{\frac{30}{8}}\approx4.496$, which yields the same conclusion as before: the null hypothesis can't be rejected. Is this just a dumb question setup, and I did it right? It seems strange that $a)$ and $b)$ would give the same answer.

Thanks!

Answer 1

Exact Poisson computation: If $X \sim Pois(240),$ then what is $P(X \ge 256).$ This is the P-value of your test of $H_0: \lambda = 240$ against $H_a: \lambda > 240.$ Notice that (per @barrycarter) everything has been re-scaled to an 8-hour period.

The answer from R statistical software is that $P(X \ge 256) \approx 0.15,$ so we do not reject at any level below 15%. In particular, no rejection at 5% or 1% level. Simply put, an average of 32 orders per hour over an 8-hour period would not be really surprising if the true Poisson rate were still 30 per hour.

 1 - ppois(255, 240)  # in R, 'ppois' is the Poisson CDF
 ## 0.1584904

Because it is easy to get an exact Poisson probability using software, it is not really necessary to use a normal approximation. However, in case the point of your exercise is drill using the normal distribution, then it goes like this:

$$P(X \ge 256) \approx P\left(Z \ge \frac{256 - 240}{\sqrt{240}} = 1.033 \right) \approx 0.151,$$

where the first approximation sign is because of the normal approximation to the binomial and the second is for approximations in evaluating the normal probability. (If you use printed normal tables, you may get a slightly different answer.) Generally speaking, one cannot expect better than about two place accuracy using a normal approximation to a Poisson distribution.