geometric-hypothesis-testing

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Geometric hypothesis testing

How do I test for the parameter p of a Geometric distribution?

If $X ~ Geo (p)$ , test for the probability of success, $p$ , using the following hypotheses
- $H_{0} : p = . . .$
- $H_{1} : p \neq . . .$ or $p < . . .$ or $p > . . .$
- with significance level $α$
  - For example, $α = 0.05$ for 5%
You will be given an observed value, $x$ , in the question
- This is the number of trials it takes to see the first success
  - For example, “They thought the coin was fair ( $p = \frac{1}{2}$ ), but last week it took 5 flips to get the first tail ( $x = 5$ )”
- It can help to compare $x$ with the expected number of trials to see the first success, $E (X) = \frac{1}{p}$
  - For example, “they expected a fair coin ( $p = \frac{1}{2}$ ) to take $\frac{1}{p} = 2$ attempts to see the first tail”
Assuming $H_{0} : p = . . .$
- Find the probability that $X$ is the observed value $x$ , or more extreme than that
- For $H_{1} : p < . . .$ the extreme values are $X \geq x$
  - Note the “change in inequality direction”
  - A lower probability of success means a higher number of attempts to first reach that success
- For $H_{1} : p > . . .$ the extreme values are $X \leq x$
  - A higher probability of success means a lower number of attempts to first reach that success
- For $H_{0} : p \neq . . .$ compare $x$ with $E (X) = \frac{1}{p}$
  - If $x$ is less than $\frac{1}{p}$ , then extreme values are $X \leq x$
  - If $x$ is more than $\frac{1}{p}$ , then the extreme values are $X \geq x$
- If the total probability of these values is $< α$ (or $< \frac{α}{2}$ for two-tailed tests)
  - Write that “there is sufficient evidence to reject $H_{0}$ “
- If not, write that “there is insufficient evidence to reject $H_{0}$ ”
Write a conclusion in context
- For example
  - “the probability of success is less than $\frac{1}{2}$ “
  - or “the probability of success has not changed from $\frac{1}{2}$ “

How do I find the critical region for a Geometric hypothesis test?

If $H_{1} : p < . . .$
- Assume that $H_{0} : p = . . .$
- Then test different integer values, $c$ , to get $P (X \geq c)$ as close to $α$ as possible, without exceeding it
  - Use the formula $P (X \geq c) = {(1 - p)}^{c - 1}$ to help
  - The integer that’s the nearest is called the critical value
  - Checking one integer lower should show that $P (X \leq c - 1)$ is $> α$
- The critical region is $X \geq c$
  - Note that the inequality is the opposite way round to $p < . . .$
- Instead of testing integers, you can also use logarithms to solve the critical region inequalities
  - Beware when dividing both sides by $\log (p)$
  - $\log (p) < 0$ so the inequality must be “flipped“
If $H_{1} : p > . . .$
- It’s the same process, but with $P (X \leq c)$ as close to $α$ as possible, without exceeding it
  - Use the formula $P (X \leq c) = 1 - {(1 - p)}^{c}$ to help
- The critical region is $X \leq c$
If $H_{1} : p \neq . . .$
- The critical region is $X \leq c_{1}$ or $X \geq c_{2}$
  - $P (X \leq c_{1})$ is as close to $\frac{α}{2}$ as possible, without exceeding it
  - $P (X \geq c_{2})$ is as close to $\frac{α}{2}$ as possible, without exceeding it
Your calculator may have an ‘Inverse Geometric Distribution’ function that can help with finding critical values
- But always check those values against the requirements of the question
- The calculator may not always give the exact answer you are looking for

What is the actual significance level?

As the geometric model is discrete, it’s not possible to get a critical region whose probability sums to $α$ exactly
- That’s because $X$ can only take integer values
Whatever it does sum to is called the actual significance level
- The actual amount of probability in the tail (or tails)
For example, if $H_{1} : p < . . .$ has the critical region $X \geq c$
- Then $P (X \geq c)$ will be just less than $α$
  - It’s value is the actual significance level
  - It represents the probability of rejecting $H_{0}$ incorrectly (when $H_{0}$ was actually true)
Some questions want a critical region that’s as close to $α$ as possible, even if that means probabilities that exceed $α$
- For example, if $P (X \geq 12) = 0.0511$ and $P (X \geq 13) = 0.0299$ where $α = 0.05$
  - Then $X \geq 12$ is the critical region that’s as close to $α$ as possible
  - The actual significance level is 0.0511

Examiner Tips and Tricks

Remember that, for geometric hypothesis testing, the inequalities for $p$ (in $H_{1}$ ) are the opposite way round to those used for the critical regions

Worked Example

Palamedes constructs a large spinner with the numbers 1 to 40 marked on it. He claims that it is fair, and in particular that the probability of the spinner landing on a ‘1’ is exactly $\frac{1}{40}$ . Odysseus is suspicious about this claim. They decide to conduct a two-tailed hypothesis test to test Palamedes’ claim, by having Odysseus spin the spinner and counting how many spins it takes until the spinner lands on a ‘1’ for the first time.

a) Write down the null and alternative hypotheses for the test

b) Using a 10% level of significance, find the critical regions for this test, where the probability of rejecting either tail should be as close as possible to 5%.