Geometric Distribution - How to show that a certain event is unplausible? We have given a geometric distribution with parameter $p$ as well as some result $r$, which we doubt is an outcome of the given distribution. What is the best way to show that $r$ is indeed not a plausible outcome the given distribution?
In general, any single event has a quite small probability for geometric distributions. So it does not make sense to test for the probability of the result $r$ given our geometric distribution. It seems possible to check for the probability that the outcome is within the interval $[c r, (1+c) r]$ for some $0 < c < 1$.
Is this a reasonable approach or are there more appropriate methods?
 A: Let me change the notation slightly:  We have one observation $r$ from some geometric distribution with unknown parameter $p.$ We want to perform the hypothesis test:
$$H_0:p=p_0 $$
$$H_1:p\ne p_0$$
at significance level $\alpha=0.05$ and $p_0$ is a specified number.
I assume we have the geometric distribution defined for $k\ge 1$: $P_0(k)=(1-p_0)p_0^{k-1}.$
If we have the simpler case with $H_1:p=p_1$ the first thing we would think about (due to the Neyman-Pearson lemma) is the Likelihood ratio $$L(r)=\frac{P_0(r)}{P_1(r)}   $$
with Rejection region $L(r)\le w$ for some $w$. If $p_0>p_1$ then $L(r)\le w$ is:
$$ (r-1)ln(\frac{p_0}{p_1})\le \frac{w(1-p_0)}{1-p_1}  $$
$$ r\le 1+\frac{w(1-p_0)}{(1-p_1)ln(\frac{p_0}{p_1})} := r^{**} $$
So this says if $r \in \{1,2,...,r^{**}\}  $ then we Reject $H_0$ because $r$ is "too extreme" or "too unlikely" to have come from $H_0.$ Now if we change the alternative hypothesis to $H_1:p<p_0  $ then we wind up with the same type of Rejection region: from $1$ up to some number $r^{**}$. And if we have the alternative $H_1:p>p_0$ then repeating the above steps gives us a rejection region of $\{r^*,r^*+1,...   \}.$ (The inequality in $L(r)\le w$ flips since the $ln$ is negative.) We splice together these two "one-sided tests" to get the test for $H_1:p\ne p_0.$  Namely Reject the null hypothesis if the observed $r$ value is "too large" or "too small."
Now all we need to do is find the values of $r^*, r^{**}.$ Since we set the Type I error at $\alpha=0.05,$ we set: 
$$ P_0(1)+P_0(2)+,...,+P_0(r^{**})=\alpha/2=0.025   $$
$$ P_0(r^*)+P_0(r^*+1),+..=\alpha/2=0.025   $$
Summing up these probabilities gives us: 
$r^{**}=ln(0.975)/ln(p_0)$ and $r^*=1+ln(0.025)/ln(p_0)$
Example: $p_0=0.7.$ Then $r^{**}=0.07$ and $r^*=11.3$. Since $0.07$ is smaller than the minimum value of the geometric we will ignore the left tail of the Rejection region. So we will Reject the null hypothesis if $r\ge 12.$
