Hypothesis test for the parameter of a geometric distributionEdexcel A-Level Further Maths: Revision notes

A geometric distribution $X\sim\mathrm{Geo}(p)$ counts the number of independent trials up to and including the first success. In a hypothesis test for $p$, you observe one value of $X$, the number of trials needed, and decide whether it is surprising if $p$ has its stated value. The key facts are $\mathrm{P}(X=x)=p(1-p)^{x-1}$, $\mathrm{P}(X\geq x)=(1-p)^{x-1}$ and $\mathrm{P}(X\leq x)=1-(1-p)^x$. These let you find any tail probability with a single power, with no tables needed.

The null hypothesis is $\mathrm{H}_0:p=p_0$, where $p_0$ is the stated probability. The alternative hypothesis is $\mathrm{H}_1:p<p_0$, $\mathrm{H}_1:p>p_0$ or $\mathrm{H}_1:p\neq p_0$, depending on the suspicion described. Hypotheses are always about the parameter $p$, never about $X$. Define $p$ in context, for example 'the probability that a card wins a prize'. Then state that under $\mathrm{H}_0$, $X\sim\mathrm{Geo}(p_0)$. The test is one-tailed when the alternative says $p$ is lower or higher, and two-tailed when it says only that $p$ is different.

Assume $\mathrm{H}_0$ is true. Find the probability of a result at least as extreme as the observed $x$ and compare it with the significance level. For $\mathrm{H}_1:p<p_0$ a lower $p$ means a longer wait, so use $\mathrm{P}(X\geq x)=(1-p_0)^{x-1}$. For $\mathrm{H}_1:p>p_0$ a higher $p$ means a shorter wait, so use $\mathrm{P}(X\leq x)=1-(1-p_0)^x$. Worked example: $\mathrm{H}_0:p=0.2$, $\mathrm{H}_1:p<0.2$, first win on card $12$. $\mathrm{P}(X\geq12)=0.8^{11}=0.0859>0.05$, so do not reject $\mathrm{H}_0$. There is insufficient evidence, at the $5\%$ level, that the probability of winning is less than $0.2$. Worked example: $\mathrm{H}_0:p=0.02$, $\mathrm{H}_1:p>0.02$, first defective at screen $2$. $\mathrm{P}(X\leq2)=1-0.98^2=0.0396<0.05$, so reject $\mathrm{H}_0$.

The critical region is the set of values of $X$ that lead to rejecting $\mathrm{H}_0$. For $\mathrm{H}_1:p<p_0$ it has the form $X\geq c$, where $c$ is the smallest value with $(1-p_0)^{c-1}\leq$ the significance level. Solve $(1-p_0)^{c-1}<0.05$ with logarithms, reversing the inequality because $\ln(1-p_0)<0$, then check the neighbouring values. For $p_0=0.1$: $c-1>\frac{\ln0.05}{\ln0.9}=28.4$, and $0.9^{28}=0.0523>0.05$ but $0.9^{29}=0.0471<0.05$, so the critical region is $X\geq30$. For $\mathrm{H}_1:p>p_0$ it has the form $X\leq c$ with $1-(1-p_0)^c<0.05$. For $p_0=0.02$, $X\leq2$ because $1-0.98^2=0.0396$ and $1-0.98^3=0.0588$. The actual significance level is the probability of the critical region under $\mathrm{H}_0$, for example $0.0471$ for $X\geq30$.

For $\mathrm{H}_1:p\neq p_0$, put half the significance level in each tail. At $10\%$ use $5\%$ in each tail. The lower tail $X\leq a$ needs $1-(1-p_0)^a\leq0.05$ and the upper tail $X\geq b$ needs $(1-p_0)^{b-1}\leq0.05$. For $p_0=0.3$: $\mathrm{P}(X\leq1)=0.3>0.05$, so there is no lower critical region (the smallest possible value of $X$ is not extreme enough). Upper tail: $0.7^{8}=0.0576$ and $0.7^{9}=0.0404$, so $X\geq10$ and the actual significance level is $0.0404$. When the critical region has only one part, say so.

State whether $\mathrm{H}_0$ is rejected, then give a conclusion in context: 'There is sufficient evidence, at the $5\%$ level, that the probability of a defective screen is greater than $0.02$.' If $\mathrm{H}_0$ is not rejected, say there is insufficient evidence against the claim. The test relies on independent trials with a constant probability of success. If the probability changes, for example because a player improves with practice, the geometric model is not suitable. A test based on a single observed value is also fairly weak, so comment on this if asked to evaluate.