Hypothesis testing without standard deviation Given $X_1,\dots,X_{100}$, test $H_0: \lambda=1$ against $H_a: \lambda=4$. The mean $\bar{X}_{100}=1.5$
(1) Take the decision on 3% level.
(2) Find the p-value
I'm considering to use $t=\frac{\bar{x}-\mu}{s/\sqrt{n}}$.
I see that the level of significance is $0.03$. The p-value I need to evaluate $\frac{\bar{x}-\mu}{s/\sqrt{n}}$, but I am missing $s$. Is it still possible to have a solution without $s$?
 A: On the hunch that the population from which the $X_i$'s are sampled is
Poisson, and that you are supposed to use a normal approximation to
Poisson probabilities, you would test $H_0: \lambda = 1$ vs. $H_a: \lambda = 4$
using the $Z$ statistic:
$$Z = \frac{\bar X_{100} - \lambda_0}{\sqrt{\lambda_0/100}} = 
\frac{1.5 - 1}{\sqrt{1/100}} = 5,$$
where $Z$ is approximately standard normal.
The P-value is $P(Z \ge 5) \approx 2.9 \times 10^{-7}.$
So you would reject $H_0$ at any reasonable level of significance,
including $\alpha = 3\%.$
From printed tables, you can see that the z-value 1.88 cuts about 3% from the upper tail of the standard normal
distribution. So you would reject at the 3% level for any value of $Z \ge 1.88.$

Note: For an exact test one would note that $\bar X_{100} = 1.5$ implies the total
$T = 150,$ where $T \sim \mathsf{Pois}(100)$ under $H_0$ (the green dots in the figure below) and
$T \sim \mathsf{Pois}(400)$ under $H_a$ (dots connected by a blue curve). The observed $T = 150$ is nearer to 100 than to 400, but it is clearly beyond
likely values of  $\mathsf{Pois}(100)$. Exact Poisson probabilities can be
found using software.

