Which route through a binomial test usually needs fewer calculations?

Finding the probability of a value at least as extreme as the observed one, which takes a single cumulative calculation. Finding the critical region means testing candidate values one at a time until the boundary is located, so it costs more unless the question asks for the region.

What is different about finding the critical region for a two-tailed Poisson test?

Two critical values are needed, one in each tail, each found against half the significance level. The lower tail is a direct cumulative probability, while the upper tail still has to be reached by subtracting from one.

A LevelMathsCambridge (CIE)Probability & Statistics 2FlashcardsHypothesis TestingHypothesis Testing (Binomial & Poisson Distributions)

Hypothesis Testing (Binomial & Poisson Distributions) (Cambridge (CIE) A Level Maths: Probability & Statistics 2): Flashcards

Exam code: 9709

1/14

0Still learning

Know0

FrontHypothesis Testing for the Proportion of a Binomial Distribution
In a binomial hypothesis test, what is the population parameter and what does the random variable count?

Was this flashcard helpful?

Cards in this collection (14)

In a binomial hypothesis test, what is the population parameter and what does the random variable count?
The parameter is the probability of success, $p$ , in a binomial distribution $B (n, p)$ .
The random variable counts the number of successes in a fixed number of trials $n$ , and its value for the sample is the observed value.
Why is the distribution written as $X ~ B (12, p)$ rather than $X ~ B (12, 0.6)$ ?
Because $p$ is the quantity being tested, so it must stay as a letter when the distribution is first defined.
The assumed value 0.6 enters only through the null hypothesis, and is used from that point on to calculate probabilities assuming $H_{0}$ is true.
Complete the conditions that make $c$ the critical value of an upper-tailed binomial test at the $α %$ level.
$P (X \geq c)______α %$
$P (X \geq c - 1)______α %$
The completed conditions are:
$P (X \geq c) \leq α %$
$P (X \geq c - 1) > α %$
The critical value is the first value inside the region, so the value one step nearer the middle has to fall outside it.
How do the critical value conditions change for a lower-tailed test?
The tail is reversed, so the conditions become $P (X \leq c) \leq α %$ and $P (X \leq c + 1) > α %$ .
The second condition still steps towards the middle of the distribution, which for a lower tail means going up rather than down.
True or False?
In a two-tailed binomial test the two critical regions are the same size.
False.
The binomial distribution is only symmetrical when $p = 0.5$ , so in general one tail reaches its half of the significance level sooner than the other.
Often one of the two regions is much larger than the other.
How is $P (X \geq 10)$ found for $X ~ B (12, 0.6)$ ?
Either add the three individual probabilities for $X = 10$ , $X = 11$ and $X = 12$ , or subtract the cumulative probability from one as $1 - P (X \leq 9)$ .
Both give $0.0834$ to three significant figures.
Which route through a binomial test usually needs fewer calculations?
Finding the probability of a value at least as extreme as the observed one, which takes a single cumulative calculation.
Finding the critical region means testing candidate values one at a time until the boundary is located, so it costs more unless the question asks for the region.
What must be done when the observed value covers a different time interval from the stated mean?
Scale the mean in proportion so that it applies to the same interval as the observation.
A mean of 8 per 24 hours becomes a mean of 2 per 6 hours, and the calculation then uses $X ~ Po (2)$ .
The random variable counts occurrences in that interval, so the two have to match before any probability is worked out.
When the mean has been rescaled, which value of $λ$ do the hypotheses use?
The value for the interval $λ$ was defined in, so a mean defined as likes per day keeps $H_{0} : λ = 8$ .
The rescaled mean of 2 belongs to the distribution used for the calculation, not to the hypotheses.
Defining $λ$ with its interval in words is what keeps the two apart.
Why is an upper-tail Poisson probability found by subtracting from one?
A Poisson variable has no upper limit, so $P (X \geq c)$ cannot be reached by adding terms.
Use $P (X \geq c) = 1 - P (X \leq c - 1)$ , which turns it into a finite cumulative sum.
The upper tail is where mistakes are most often made, usually by dropping or adding one to the limit.
How is the rejection region found for an upper-tailed Poisson test at the 5% level?
Work out $P (X \geq c)$ for candidate values of $c$ until it first drops below 0.05.
Check the value one step lower is still above 0.05, which confirms $c$ is the boundary.
With $λ = 2$ this gives $P (X \geq 5) = 0.0527$ and $P (X \geq 6) = 0.0166$ , so the region is $X \geq 6$ .
True or False?
For $X ~ Po (2)$ tested at the 5% level for an increase, an observed value of 5 is not enough to reject $H_{0}$ .
True.
The rejection region is $X \geq 6$ , and 5 falls outside it.
A Poisson distribution has a long right tail, so a value two and a half times the mean can still be unremarkable.
What is different about finding the critical region for a two-tailed Poisson test?
Two critical values are needed, one in each tail, each found against half the significance level.
The lower tail is a direct cumulative probability, while the upper tail still has to be reached by subtracting from one.
A test on $X ~ Po (2)$ has rejection region $X \geq 6$ . What is the probability of a Type I error?
It is $P (X \geq 6)$ worked out with $λ = 2$ , since a Type I error means landing in the rejection region while $H_{0}$ is true.
That gives $1 - P (X \leq 5) = 1 - 0.9834 = 0.0166$ to three significant figures.

Sign up to unlock flashcards

By signing up you agree to our Terms and Privacy Policy

Previous:Hypothesis TestingNext:Hypothesis Testing (Normal Distribution)