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How do you carry out hypothesis tests on a Poisson mean and on a population mean, and what do Type I and Type II errors mean?

Hypothesis tests for the mean of a Poisson distribution, tests for a population mean using the normal distribution, one-tailed and two-tailed tests, and the meaning of Type I and Type II errors.

A focused answer to the AQA A-Level Further Mathematics hypothesis testing content, covering tests for the mean of a Poisson distribution, tests for a population mean using the normal distribution, one-tailed and two-tailed tests, and the meaning of Type I and Type II errors.

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  1. What this dot point is asking
  2. Setting up a test
  3. Test for a Poisson mean
  4. Test for a population mean
  5. Type I and Type II errors

What this dot point is asking

AQA wants you to set up and carry out hypothesis tests for the mean of a Poisson distribution and for a population mean using the normal distribution, to distinguish one-tailed and two-tailed tests, to compare a test statistic with a critical value or a probability with the significance level, and to explain Type I and Type II errors.

Setting up a test

Every test follows the same skeleton, and marks are lost by skipping a step rather than by hard calculation. State the null hypothesis H0H_0 (the value being assumed) and the alternative H1H_1 (what you are testing for). Choose the significance level, usually 5%5\% or 1%1\%. Decide on one tail or two: a claim that a parameter has increased or decreased is one-tailed, while a claim that it has merely changed is two-tailed, splitting the significance level between the two tails. Then either find the critical region in advance and check whether the observation falls in it, or compute the probability of a result as extreme as observed and compare it with the significance level. Finally, state the conclusion in the context of the original problem, not just as accept or reject.

Test for a Poisson mean

For a Poisson test the test statistic is the observed count itself, and you work with exact Poisson probabilities or cumulative tables rather than a continuous approximation. For an upper-tail test the relevant probability is P(Xobserved)P(X \geq \text{observed}), computed as 1P(Xobserved1)1 - P(X \leq \text{observed} - 1) from cumulative tables; for a lower-tail test it is P(Xobserved)P(X \leq \text{observed}) directly.

Test for a population mean

This ZZ test relies on the sampling distribution of the mean. If the population is normal with known variance, the sample mean Xˉ\bar{X} is normally distributed with mean μ\mu and standard error σn\frac{\sigma}{\sqrt{n}}, so standardising gives the ZZ statistic. By the central limit theorem the same test is approximately valid for a large sample from any population, even one that is not normal, which is why it appears so widely. A large absolute value of ZZ means the observed sample mean is many standard errors away from the hypothesised mean, which is unlikely if H0H_0 is true, so H0H_0 is rejected.

Type I and Type II errors

The two error types pull against each other, which is the heart of choosing a significance level. A very small significance level makes you reluctant to reject H0H_0, so you rarely raise a false alarm (low Type I rate) but more often miss a genuine effect (high Type II rate). Increasing the sample size is the only way to reduce both at once, because a larger sample sharpens the sampling distribution and separates the competing hypotheses more cleanly. In context, a Type I error is a false positive (acting on a change that is not real) and a Type II error is a false negative (missing a change that is real); which is worse depends on the practical consequences.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20207 marksThe number of flaws in rolls of fabric follows a Poisson distribution with historic mean 22 flaws per roll. After a change to the process, a single roll is inspected and found to contain 66 flaws. Test at the 5%5\% significance level whether the mean number of flaws has increased. Use P(X5)=0.9834P(X \leq 5) = 0.9834 for XPo(2)X \sim \text{Po}(2).
Show worked answer →

State the hypotheses. H0:λ=2H_0: \lambda = 2, H1:λ>2H_1: \lambda > 2 (one-tailed, because the question asks about an increase).

The test is one-tailed at 5%5\%. Find the probability of a result as extreme as observed, assuming H0H_0 is true: P(X6)=1P(X5)=10.9834=0.0166P(X \geq 6) = 1 - P(X \leq 5) = 1 - 0.9834 = 0.0166.

Compare with the significance level: 0.0166<0.050.0166 < 0.05.

Since the probability of the observed result (or more extreme) is less than 5%5\%, the result lies in the critical region, so reject H0H_0.

Conclusion in context: there is evidence at the 5%5\% level that the mean number of flaws per roll has increased.

Markers reward the one-tailed hypotheses, computing P(X6)P(X \geq 6) as 1P(X5)1 - P(X \leq 5), the comparison with 0.050.05, and the contextual conclusion.

AQA 20226 marksThe weights of bags filled by a machine are normally distributed with known standard deviation σ=4\sigma = 4 g and supposed mean 500500 g. A random sample of 1616 bags has mean weight 497.5497.5 g. Test at the 5%5\% level whether the mean weight differs from 500500 g, and explain what a Type I error would mean in this context.
Show worked answer →

State the hypotheses for a two-tailed test. H0:μ=500H_0: \mu = 500, H1:μ500H_1: \mu \neq 500.

Compute the test statistic Z=xˉμσ/n=497.55004/16=2.51=2.5Z = \frac{\bar{x} - \mu}{\sigma / \sqrt{n}} = \frac{497.5 - 500}{4 / \sqrt{16}} = \frac{-2.5}{1} = -2.5.

For a two-tailed 5%5\% test the critical values are ±1.96\pm 1.96. Since 2.5<1.96-2.5 < -1.96, the test statistic lies in the critical region.

Reject H0H_0: there is evidence at the 5%5\% level that the mean weight differs from 500500 g.

A Type I error here would be concluding the mean weight has changed (rejecting H0H_0) when in fact the machine is still correctly set at 500500 g. Its probability equals the significance level, 5%5\%.

Markers reward the two-tailed hypotheses, the ZZ statistic, comparison with ±1.96\pm 1.96, the conclusion, and a correct contextual Type I error description.

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