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Poisson & Geometric Hypothesis Testing (Edexcel A Level Further Maths: Further Statistics 1): Flashcards

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FrontPoisson Hypothesis Testing
For a test of whether the mean of a Poisson distribution has increased from 8, complete the hypotheses:
$H_{0} : λ =______and H_{1} : λ______8$

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Cards in this collection (15)

For a test of whether the mean of a Poisson distribution has increased from 8, complete the hypotheses:
$H_{0} : λ =______and H_{1} : λ______8$
The completed hypotheses are:
$H_{0} : λ = 8 and H_{1} : λ > 8$
The null hypothesis always fixes the parameter at its assumed value, and the alternative carries the direction being tested.
Define the critical region of a hypothesis test.
The critical region is the set of values of the test statistic for which $H_{0}$ would be rejected.
Its boundary value is the critical value, and the region is built so that the probability of landing in it, assuming $H_{0}$ , is at most the significance level.
Having assumed $H_{0}$ , what probability do you actually calculate?
The probability of getting the observed value or something more extreme, in the direction the alternative hypothesis points.
That total is then compared with $α$ , or with $\frac{α}{2}$ for a two-tailed test.
For $H_{1} : λ > 8$ , how do you find the critical region?
Find the smallest integer $c$ for which $P (X \geq c)$ falls below $α$ , and the critical region is then $X \geq c$ .
Take care with the discrete inequality, since $P (X \geq c)$ is $1 - P (X \leq c - 1)$ and it is easy to end up having found $c - 1$ instead.
True or False?
In a two-tailed test, each tail is compared with the full significance level $α$ .
False.
Each tail is compared with $\frac{α}{2}$ , so that the two tails together carry a total probability of $α$ .
Using $α$ in each tail would double the chance of wrongly rejecting $H_{0}$ .
A blog receives 8 likes per 24 hours, and 7 likes are observed in a 12-hour period. What must you do before testing?
Rescale $λ$ to the same interval as the observation, so the test uses $λ = 4$ for a 12-hour period rather than 8.
The observed value and the parameter must always refer to the same length of time or space, or the comparison means nothing.
True or False?
In a test at the 5% level on a discrete variable, the probability of wrongly rejecting $H_{0}$ is usually less than 5%.
True.
A discrete variable takes only whole number values, so the tail probability jumps from one integer to the next and cannot be tuned to land on $α$ exactly.
The critical region is normally chosen to stay below $α$ , and whatever its probability actually comes to is the actual significance level.
The tail probability comes out below the significance level. How do you write the conclusion?
Say that there is sufficient evidence to reject $H_{0}$ , then state what that means in context.
You never say that $H_{0}$ is false or that $H_{1}$ has been proved: a test weighs evidence, so the wording stays about evidence.
Why is the critical region always calculated assuming that $H_{0}$ is true?
Because $H_{0}$ fixes the parameter at a single value, which is the only assumption specific enough to give you a distribution to calculate with.
The alternative hypothesis names a range rather than a value, so no probabilities could be worked out from it at all.
What parameter does a geometric hypothesis test examine, and what is the observed value?
The test is on $p$ , the probability of success in one trial, and the observed value $x$ is the number of trials taken to reach the first success.
Comparing $x$ with the expected number of trials $\frac{1}{p}$ shows at once which way the evidence points.
True or False?
For $H_{1} : p < 0.5$ , the critical region of a geometric test takes the form $X \leq c$ .
False.
A lower probability of success means it takes more trials to reach the first success, so the extreme values are large ones and the critical region is $X \geq c$ .
The inequality in the critical region is always the opposite way round to the one in $H_{1}$ .
For $X ~ Geo (p)$ , complete the formula that gives the upper tail directly, filling in the missing index:
$P (X \geq c) = {(1 - p)}^{______}$
The completed formula is:
$P (X \geq c) = {(1 - p)}^{c - 1}$
The first success falls on trial $c$ or later exactly when the first $c - 1$ trials all fail, which is why the index is one less than $c$ .
You solve ${(1 - p)}^{c - 1} < α$ using logarithms. What must you watch out for?
Dividing both sides by $\log (1 - p)$ reverses the inequality, because that logarithm is negative whenever $1 - p$ lies between 0 and 1.
Forgetting the flip gives a critical region pointing the wrong way, which is easy to miss because the algebra otherwise looks perfectly correct.
In a two-tailed geometric test, how do you decide which tail the observed value belongs to?
Compare $x$ with the expected number of trials $\frac{1}{p}$ : if $x$ is smaller the extreme values are $X \leq x$ , and if $x$ is larger they are $X \geq x$ .
The observed value tells you which side of the expectation it has fallen, and only that tail is tested.
What form does the critical region take for a two-tailed geometric test?
The critical region has two parts, $X \leq c_{1}$ or $X \geq c_{2}$ , one drawn from each tail of the distribution.
Each critical value is found separately, and neither is simply the mirror of the other.

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