Hypothesis Tests for a Poisson Mean

A2 · S2 · 15 min

Has a new road layout reduced the number of accidents? Did an advertising campaign increase the rate of calls to a help desk? When events happen at random at some average rate, the count in a fixed period has a Poisson distribution, and a single observed count can test a claim about the rate. This is one of the most common full questions on Paper 6. It combines everything from the Poisson unit (scaling the mean to the right interval, summing Poisson variables, approximating a binomial) with the logic of a hypothesis test, and it often continues into Type I and Type II errors.

The set-up

Events occur singly, independently and at random, at a constant average rate. The null hypothesis states a value for that rate. The test statistic is the number of events $X$ observed in a period of a stated length, and under $H_0$

X \sim \text{Po}(\lambda_0),

where $\lambda_0$ is the mean number of events for the period actually observed. If the claimed rate is $1.2$ per minute and calls are counted over $5$ minutes, then $\lambda_0 = 6$ .

The hypotheses can be written for the rate in the question's units or for the mean over the observed period, as long as you are consistent and clear:

H_0: \lambda = 1.2 \text{ per minute} \quad\text{or}\quad H_0: \lambda = 6 \text{ (per 5 minutes)}.

Many students find it safest to write the hypotheses with the mean for the observed period, so that the same number appears in the distribution.

Large counts are evidence of a higher rate, and small counts of a lower rate. Everything else works as in binomial tests: compare a tail probability with the significance level, or find a critical region. There are no Poisson tables in the formula booklet, so every probability comes from

P(X = r) = e^{-\lambda}\frac{\lambda^r}{r!}.

Poisson test

Find the mean for the period observed, scaling the claimed rate in proportion to the length of the period.
Define the parameter and state $H_0$ and $H_1$ , using the wording of the question to choose the tail.
State "under $H_0$ , $X \sim \text{Po}(\lambda_0)$ ".
For an increase, find $P(X \ge x) = 1 - P(X \le x - 1)$ ; for a decrease, find $P(X \le x)$ . Write the sum out with $e^{-\lambda_0}$ factored out.
Compare with the significance level (half of it, per tail, for a two-tailed test).
Decide about $H_0$ and conclude in context, without certainty.

Routine: has the rate increased?

A help desk receives calls at random at an average rate of $1.2$ per minute. After an advertising campaign, $11$ calls are received in a randomly chosen $5$ -minute period. Test at the $5\%$ significance level whether the rate of calls has increased.

Solution

For $5$ minutes the mean under the old rate is $5 \times 1.2 = 6$ .

Let $\lambda$ be the mean number of calls in $5$ minutes. $H_0: \lambda = 6$ , $\quad H_1: \lambda > 6$ .

Under $H_0$ , $X \sim \text{Po}(6)$ , where $X$ is the number of calls in $5$ minutes.

P(X \ge 11) = 1 - P(X \le 10) = 1 - e^{-6}\left(1 + 6 + \frac{6^2}{2!} + \frac{6^3}{3!} + \cdots + \frac{6^{10}}{10!}\right) = 1 - 0.9574 = 0.0426

$0.0426 < 0.05$ , so reject $H_0$ .

There is evidence at the $5\%$ level that the rate of calls has increased.

Critical regions and the actual significance level

Exactly as for the binomial, the counts are discrete, so the critical region is chosen to have probability as close as possible to, but not more than, the significance level.

Poisson critical regions

Upper tail ( $H_1: \lambda > \lambda_0$ ): the smallest $c$ with $P(X \ge c) \le \alpha$ ; critical region $X \ge c$ .
Lower tail ( $H_1: \lambda < \lambda_0$ ): the largest $c$ with $P(X \le c) \le \alpha$ ; critical region $X \le c$ .
Two-tailed: each tail has probability at most $\tfrac{1}{2}\alpha$ .
The actual significance level is the probability of the critical region under $H_0$ .

The graph shows $\text{Po}(6)$ with the upper-tail critical region at the $5\%$ level, $X \ge 11$ , shaded. Its probability is $0.0426$ . Adding $X = 10$ would raise it to $0.0839$ , which is too large.

y = 6^floor(x + 0.5) exp(-6) / fact(floor(x + 0.5)) + 0*sqrt(x + 0.5) fill 10.5 17 y = 6^floor(x + 0.5) exp(-6) / fact(floor(x + 0.5))

A two-tailed critical region

The number of goals scored per match in a league has mean $2$ . A journalist wants to test whether the scoring rate has changed this season, using the total number of goals in $4$ randomly chosen matches and a $10\%$ significance level.

(a) State the hypotheses and find the critical region.

(b) Find the actual significance level.

Solution

(a) Let $\lambda$ be the mean number of goals in $4$ matches. $H_0: \lambda = 8$ , $\quad H_1: \lambda \ne 8$ . Under $H_0$ , $X \sim \text{Po}(8)$ , and each tail can have probability at most $0.05$ .

Lower tail:

P(X \le 3) = e^{-8}\left(1 + 8 + 32 + \frac{256}{3}\right) = 0.0424 \le 0.05, \qquad P(X \le 4) = 0.0996 > 0.05.

Upper tail:

P(X \ge 14) = 1 - P(X \le 13) = 0.0342 \le 0.05, \qquad P(X \ge 13) = 0.0638 > 0.05.

The critical region is $X \le 3$ or $X \ge 14$ .

(b) Actual significance level $= 0.0424 + 0.0342 = 0.0766$ .

(c) $13$ is not in the critical region. Do not reject $H_0$ : there is insufficient evidence at the $10\%$ level that the scoring rate has changed.

Treating the total for $4$ matches as $\text{Po}(8)$ assumes goals in different matches are independent, using the fact that a sum of independent Poisson variables is Poisson.

Combining periods and approximating a binomial

Two Poisson results feed directly into tests.

Combining several periods. If accidents are $\text{Po}(4)$ per month, the total over three months is $\text{Po}(12)$ , provided months are independent. A test based on the total uses $\text{Po}(12)$ . The same applies to two different kinds of event (cars and lorries, say): their total is Poisson with the sum of the means.

Approximating a binomial. If the test statistic is binomial with large $n$ and small $p$ , such as the number of defective items in a sample of $600$ when $p = 0.005$ , the Poisson approximation $B(n, p) \approx \text{Po}(np)$ is used (see approximating a binomial by a Poisson). The test then proceeds as a Poisson test, with the hypotheses still written for the proportion $p$ .

Combining months: has the rate decreased?

The number of accidents per month at a junction has had the distribution $\text{Po}(4)$ . New warning signs are installed, and in the following $3$ months there are $6$ accidents in total. Test at the $5\%$ significance level whether the rate of accidents has decreased.

Solution

Over $3$ months, the mean under the old rate is $3 \times 4 = 12$ .

Let $\lambda$ be the mean number of accidents in $3$ months. $H_0: \lambda = 12$ , $\quad H_1: \lambda < 12$ .

Under $H_0$ , $X \sim \text{Po}(12)$ .

P(X \le 6) = e^{-12}\left(1 + 12 + \frac{12^2}{2} + \frac{12^3}{6} + \frac{12^4}{24} + \frac{12^5}{120} + \frac{12^6}{720}\right) = 0.0458

$0.0458 < 0.05$ , so reject $H_0$ .

There is evidence at the $5\%$ level that the rate of accidents has decreased since the signs were installed.

A binomial test via the Poisson approximation

On a production line, $0.5\%$ of items are defective. After a machine is serviced, a random sample of $600$ items contains $7$ defectives. Use a suitable approximation to test at the $5\%$ significance level whether the proportion of defective items has increased.

Solution

Let $p$ be the proportion of defective items. $H_0: p = 0.005$ , $\quad H_1: p > 0.005$ .

Under $H_0$ , $X \sim B(600, 0.005)$ . Since $n = 600 > 50$ and $np = 3 < 5$ , use $X \approx \text{Po}(3)$ .

P(X \ge 7) = 1 - e^{-3}\left(1 + 3 + \frac{9}{2} + \frac{27}{6} + \frac{81}{24} + \frac{243}{120} + \frac{729}{720}\right) = 1 - 0.9665 = 0.0335

$0.0335 < 0.05$ , so reject $H_0$ .

There is evidence at the $5\%$ level that the proportion of defective items has increased.

Exam-hard: how long must the observation be?

Faults occur at random in a type of cable at an average rate of $0.3$ per $100$ m. A new manufacturing process is introduced, and it is hoped that the fault rate has decreased. A length of $L$ metres of cable from the new process will be examined, and a test carried out at the $5\%$ significance level.

(a) Show that $H_0$ can be rejected only if $L$ is greater than about $1000$ .

(b) A length of $1500$ m is examined. Find the critical region and the actual significance level.

Solution

(a) The most extreme possible result for "decreased" is no faults at all. In $L$ metres the mean is $\dfrac{0.3L}{100} = 0.003L$ , so we need

P(X = 0) = e^{-0.003L} \le 0.05 \;\Rightarrow\; 0.003L \ge \ln 20 = 2.9957 \;\Rightarrow\; L \ge 998.6.

For any shorter length, even zero faults would not be significant, so $L$ must be about $1000$ m or more.

(b) For $1500$ m, $\lambda_0 = 4.5$ . $H_0: \lambda = 4.5$ , $\quad H_1: \lambda < 4.5$ , and under $H_0$ , $X \sim \text{Po}(4.5)$ .

P(X = 0) = e^{-4.5} = 0.0111 \le 0.05, \qquad P(X \le 1) = e^{-4.5}(1 + 4.5) = 0.0611 > 0.05.

The critical region is $X = 0$ , and the actual significance level is $0.0111$ , that is $1.11\%$ .

(c) $1$ is not in the critical region. Do not reject $H_0$ : there is insufficient evidence at the $5\%$ level that the fault rate has decreased.

Common mistakes

Not scaling the mean. A rate of $1.2$ per minute tested on a $5$ -minute count needs $\text{Po}(6)$ , not $\text{Po}(1.2)$ . This is the most common error.
Off-by-one in the upper tail. $P(X \ge 11) = 1 - P(X \le 10)$ . Using $1 - P(X \le 11)$ throws away the observed value.
Dropping the $r = 0$ term. $P(X \le 6)$ has seven terms, starting with $e^{-\lambda}$ .
Using $P(X = x)$ instead of the tail probability.
Forgetting to halve the significance level in a two-tailed test.
Taking a critical region with probability just above $\alpha$ because it is "closer". The probability must not exceed $\alpha$ .
Writing hypotheses about the sample count, such as $H_0: X = 11$ . Hypotheses are about the population mean $\lambda$ (or the proportion $p$ if the underlying model is binomial).

Exam tip

Write the scaled mean explicitly: "mean for $5$ minutes $= 5 \times 1.2 = 6$ ". It is often worth a mark and makes the rest of your work clear.
Use $\lambda$ or $\mu$ for the Poisson mean in hypotheses; both are accepted. State what it is the mean of.
Show the Poisson sum with $e^{-\lambda}$ factored out, even when you get the value from a calculator.
If a binomial is approximated by a Poisson, say so and justify it: " $n > 50$ and $np < 5$ ".
Many questions ask you to state an assumption, such as "accidents occur independently and at random" or "the rate is constant". Give one in context.
For critical region questions, show the probabilities on both sides of the boundary and state the region as an inequality in $X$ .

Summary

Test statistic: the count $X$ in the observed period; under $H_0$ , $X \sim \text{Po}(\lambda_0)$ with $\lambda_0$ scaled to that period.
Increase: compare $P(X \ge x)$ with $\alpha$ . Decrease: compare $P(X \le x)$ with $\alpha$ . Two-tailed: compare the relevant tail with $\tfrac{1}{2}\alpha$ .
Critical regions have probability as close as possible to, but not above, $\alpha$ ; the actual significance level is that probability.
Totals over several independent periods, or of independent event types, are Poisson with the sum of the means.
$B(n, p)$ with large $n$ and small $p$ is tested via $\text{Po}(np)$ .
If $e^{-\lambda_0} > \alpha$ , a lower-tail test can never reject $H_0$ : the observation period is too short.
Conclude in context with "evidence" language.

Practice questions

Question

A test of $H_0: \lambda = 6$ against $H_1: \lambda > 6$ is carried out at the $5\%$ significance level, using a single observation of $X \sim \text{Po}(\lambda)$ . Find the critical region and the actual significance level.
A person receives emails at random at an average rate of $5$ per hour. On the first day back after a holiday, $10$ emails arrive in a randomly chosen hour. Test at the $5\%$ significance level whether the rate has increased.
A shop sells a particular item at random at an average rate of $2.4$ per day. After a price rise, the shop sells $3$ of these items in $3$ days. Test at the $5\%$ significance level whether the rate of sales has decreased.
The number of breakdowns per week of a machine is $\text{Po}(5)$ . After maintenance, there are $11$ breakdowns in one week. Test at the $5\%$ significance level whether the rate of breakdowns has changed.
State two conditions needed for the number of calls in a period to be modelled by a Poisson distribution in a hypothesis test about the rate of calls.
It is known that $2\%$ of plates made in a factory are faulty. After new equipment is installed, a random sample of $200$ plates contains no faulty plates. Use a suitable approximation to test at the $5\%$ significance level whether the proportion of faulty plates has decreased.
Typing errors occur at random at an average rate of $0.5$ per hour of typing. A trainer wants to test, at the $1\%$ significance level, whether a new keyboard reduces the error rate. Find the least whole number of hours of typing that must be observed for the test to be able to reject $H_0$ .
Cars pass a point on a road at random at an average rate of $3$ per minute, and lorries pass independently at random at an average rate of $1$ per minute. After a new bypass opens, $15$ vehicles pass the point in a randomly chosen $2$ -minute period. (a) Test at the $5\%$ significance level whether the total rate of vehicles has increased. (b) Find the critical region for the same test at the $1\%$ significance level, and state the conclusion at this level.

Answers

Under $H_0$ , $X \sim \text{Po}(6)$ . $P(X \ge 11) = 1 - P(X \le 10) = 0.0426 \le 0.05$ ; $P(X \ge 10) = 0.0839 > 0.05$ . Critical region $X \ge 11$ ; actual significance level $0.0426$ .
$H_0: \lambda = 5$ , $H_1: \lambda > 5$ (per hour). Under $H_0$ , $X \sim \text{Po}(5)$ . $P(X \ge 10) = 1 - P(X \le 9) = 1 - 0.9682 = 0.0318$ . $0.0318 < 0.05$ : reject $H_0$ . There is evidence at the $5\%$ level that the rate of emails has increased.
Mean for $3$ days $= 7.2$ . $H_0: \lambda = 7.2$ , $H_1: \lambda < 7.2$ . Under $H_0$ , $X \sim \text{Po}(7.2)$ . $P(X \le 3) = e^{-7.2}\left(1 + 7.2 + \tfrac{7.2^2}{2} + \tfrac{7.2^3}{6}\right) = 0.00075 + 0.00538 + 0.01935 + 0.04644 = 0.0719$ . $0.0719 > 0.05$ : do not reject $H_0$ . There is insufficient evidence at the $5\%$ level that sales have decreased.
$H_0: \lambda = 5$ , $H_1: \lambda \ne 5$ . Under $H_0$ , $X \sim \text{Po}(5)$ . $11 > 5$ , so use the upper tail. $P(X \ge 11) = 1 - P(X \le 10) = 1 - 0.9863 = 0.0137$ . Two-tailed, so compare with $0.025$ : $0.0137 < 0.025$ . Reject $H_0$ : there is evidence at the $5\%$ level that the rate of breakdowns has changed.
Any two of: calls occur independently of each other; calls occur at random; calls occur at a constant average rate; calls occur singly (two calls cannot arrive at exactly the same instant).
$H_0: p = 0.02$ , $H_1: p < 0.02$ . Under $H_0$ , $X \sim B(200, 0.02)$ ; $n > 50$ and $np = 4 < 5$ , so use $X \approx \text{Po}(4)$ . $P(X = 0) = e^{-4} = 0.0183$ . $0.0183 < 0.05$ : reject $H_0$ . There is evidence at the $5\%$ level that the proportion of faulty plates has decreased.
In $t$ hours, $\lambda_0 = 0.5t$ . The most extreme result is $0$ errors, so we need $e^{-0.5t} \le 0.01$ , giving $0.5t \ge \ln 100 = 4.605$ and $t \ge 9.21$ . The least whole number of hours is $10$ . (Check: $e^{-4.5} = 0.0111 > 0.01$ for $9$ hours; $e^{-5} = 0.00674 \le 0.01$ for $10$ hours.)
(a) The total number of vehicles per minute is $\text{Po}(3 + 1) = \text{Po}(4)$ , so in $2$ minutes it is $\text{Po}(8)$ . $H_0: \lambda = 8$ , $H_1: \lambda > 8$ (per $2$ minutes). Under $H_0$ , $X \sim \text{Po}(8)$ . $P(X \ge 15) = 1 - P(X \le 14) = 0.0173 < 0.05$ . Reject $H_0$ : there is evidence at the $5\%$ level that the rate of vehicles has increased. (b) $P(X \ge 16) = 0.0082 \le 0.01$ and $P(X \ge 15) = 0.0173 > 0.01$ , so the critical region is $X \ge 16$ . $15$ is not in this region, so at the $1\%$ level $H_0$ is not rejected: there is insufficient evidence at the $1\%$ level that the rate has increased.

The set-up

Critical regions and the actual significance level

Combining periods and approximating a binomial

Practice questions

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