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Type II errors and powerAQA A-Level Further Maths: Revision notes

Section 1

Type II errors and their probability

A Type II error is failing to reject H0\mathrm{H}_0 when H0\mathrm{H}_0 is false.

To find its probability you need a specific true value of the parameter. Then P(Type II error)=P(test statistic is outside the critical region∣true value).\mathrm{P}(\text{Type II error})=\mathrm{P}(\text{test statistic is outside the critical region}\mid\text{true value}). The distribution used is the one for the true value, not the value in H0\mathrm{H}_0. (For a Type I error it is the other way round.)

Key termsType II error
Common mistake

Using the H0\mathrm{H}_0 value of the parameter when finding P\mathrm{P}(Type II). Use the stated true value.

Section 2

The power of a test

The power of a test is the probability of rejecting H0\mathrm{H}_0 when it is false: power=1−P(Type II error)=P(test statistic in the critical region∣true value).\text{power}=1-\mathrm{P}(\text{Type II error})=\mathrm{P}(\text{test statistic in the critical region}\mid\text{true value}). Power depends on the true value, so a test has a different power for each possible alternative. A high power means the test is good at detecting a real change.

Key termspower

Section 3

Binomial and Poisson tests

Binomial. H0:p=0.5\mathrm{H}_0:p=0.5, H1:p>0.5\mathrm{H}_1:p>0.5, X∼B(10,p)X\sim\mathrm{B}(10,p), reject if X≥8X\ge8. True p=0.7p=0.7: P(Type II)=P(X≤7)=0.6172,power=P(X≥8)=0.3828.\mathrm{P}(\text{Type II})=\mathrm{P}(X\le7)=0.6172,\qquad\text{power}=\mathrm{P}(X\ge8)=0.3828. Poisson. H0:λ=5\mathrm{H}_0:\lambda=5, H1:λ>5\mathrm{H}_1:\lambda>5, reject if Y≥10Y\ge10. True λ=8\lambda=8: P(Type II)=P(Y≤9)=0.7166,power=0.2834.\mathrm{P}(\text{Type II})=\mathrm{P}(Y\le9)=0.7166,\qquad\text{power}=0.2834. Always write the distribution with the true parameter first.

Key termscritical region
Exam tip

The non-rejection region is the complement of the critical region. Take care with the boundary, for example X≤7X\le7 when the critical region is X≥8X\ge8.

Section 4

Normal tests

For a test of a mean with σ\sigma known, xˉ∼N(μ,σ2n)\bar x\sim\mathrm{N}\left(\mu,\frac{\sigma^2}{n}\right) with μ\mu the true mean.

One-tailed: H0:μ=500\mathrm{H}_0:\mu=500, H1:μ<500\mathrm{H}_1:\mu<500, σ=4\sigma=4, n=16n=16, reject if xˉ<498\bar x<498. True mean 497497: xˉ∼N(497,12)\bar x\sim\mathrm{N}(497,1^2), so power=P(xˉ<498)=P(Z<1)=0.841.\text{power}=\mathrm{P}(\bar x<498)=\mathrm{P}(Z<1)=0.841. Two-tailed: reject if xˉ<993\bar x<993 or xˉ>1007\bar x>1007, with true mean 10101010 and standard deviation 44: P(Type II)=P(993≤xˉ≤1007)=P(−4.25≤Z≤−0.75)=0.227.\mathrm{P}(\text{Type II})=\mathrm{P}(993\le\bar x\le1007)=\mathrm{P}(-4.25\le Z\le-0.75)=0.227.

Key termsnon-rejection region

Section 5

What affects power

Power is higher when:

  • the true value is further from the H0\mathrm{H}_0 value, because a real difference is easier to detect,
  • the sample size is larger, because xˉ\bar x has a smaller standard deviation,
  • the critical region is larger (a larger significance level), but this also raises the probability of a Type I error.

So there is a trade-off: enlarging the critical region lowers P\mathrm{P}(Type II) but raises P\mathrm{P}(Type I). Only a bigger sample can improve both.

Key termstrade-off
Exam tip

In explanations, link the cause to the distribution: 'closer to the H0\mathrm{H}_0 value, so more of the distribution lies in the non-rejection region'.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Type II errors and power

  1. A spinner lands on red with probability pp. To test H0:p=0.5\mathrm{H}_0:p=0.5 against H1:p>0.5\mathrm{H}_1:p>0.5, it is spun 10 times and H0\mathrm{H}_0 is rejected if it lands on red 8 or more times. Let XX be the number of reds, so X∼B(10,p)X\sim\mathrm{B}(10,p).
    Explain why the power of the test is greater when p=0.9p=0.9 than when p=0.7p=0.7.2 marks
  2. The number of faults in a 5 m length of cable is modelled by Y∼Po(λ)Y\sim\mathrm{Po}(\lambda). To test H0:λ=5\mathrm{H}_0:\lambda=5 against H1:λ>5\mathrm{H}_1:\lambda>5, a single 5 m length is examined and H0\mathrm{H}_0 is rejected if Y≥10Y\ge10.
    Find the probability that the test makes a Type I error.2 marks
  3. The mass of sugar in a bag is normally distributed with standard deviation 44 g. A test of H0:μ=500\mathrm{H}_0:\mu=500 against H1:μ<500\mathrm{H}_1:\mu<500 uses the mean xˉ\bar x of a random sample of 16 bags. H0\mathrm{H}_0 is rejected if xˉ<498\bar x<498.
    The true mean mass of sugar in the bags is 499.5499.5 g. Find the probability of a Type II error.3 marks
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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).