Edexcel A-level Maths coverage

Statistical hypothesis testing

Section S5
3 spec leafs

Notes and three levels of exam-style practice for each registered specification leaf in this section.

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S5.1

Apply the language of hypothesis testing via a binomial model: null/alternative hypothesis, significance level, test statistic, 1- and 2-tail tests, critical value/region, acceptance region, p-value; extend to correlation coefficients.

  • The null hypothesis H0H_0 gives the reference parameter value; the alternative H1H_1 states the direction or difference supported by the claim being tested.
  • Use a one-tailed test for a pre-specified directional alternative and a two-tailed test for any change; choose this before observing the data.
  • The critical region contains outcomes sufficiently unlikely under H0H_0; the acceptance region is its complement, and the p-value is the probability under H0H_0 of an outcome at least as extreme as observed.
  • For a correlation test, use H0:ρ=0H_0:\rho=0 and compare the sample product-moment correlation coefficient with the supplied critical value; significance does not establish causation.

Tier 1 · Easy

2 marks
ORIGINAL

A company claims that the probability pp of a customer choosing its premium plan has increased from 0.400.40. State suitable hypotheses and identify the number of tails.

Tier 2 · Standard

4 marks
ORIGINAL

Under H0H_0, XB(20,0.2)X\sim\operatorname{B}(20,0.2). For an upper-tailed test at the 5%5\% level, P(X7)=0.0867P(X\geq7)=0.0867 and P(X8)=0.0321P(X\geq8)=0.0321. State the critical region, the critical value and the acceptance region.

Tier 3 · Hard

6 marks
ORIGINAL

For a sample of 1818 paired observations, a two-tailed 5%5\% correlation test has critical values 0.468-0.468 and 0.4680.468. The sample product-moment correlation coefficient is r=0.520r=-0.520, with p-value 0.0260.026. State the hypotheses, carry out the test and interpret the p-value without claiming causation.

S5.2

Conduct a hypothesis test for the proportion in the binomial distribution and interpret results in context; a sample makes an inference about the population; the significance level is the probability of incorrectly rejecting H0.

  • Model the number of sample successes by XB(n,p0)X\sim\operatorname{B}(n,p_0) under H0:p=p0H_0:p=p_0, provided the binomial assumptions are defensible.
  • Calculate the probability, under H0H_0, of the observed result or one more extreme in the direction specified by H1H_1; double an appropriate tail for a symmetric two-tailed binomial test.
  • Reject H0H_0 when the p-value is at most the significance level; otherwise say there is insufficient evidence to reject H0H_0, not that H0H_0 has been proved.
  • The significance level is the probability of incorrectly rejecting H0H_0 when it is true; discreteness often makes the actual probability of the critical region smaller than the nominal level.

Tier 1 · Easy

4 marks
ORIGINAL

A coin is tested with H0:p=0.5H_0:p=0.5 against H1:p>0.5H_1:p>0.5. It lands heads 1010 times in 1212 tosses. Given P(X10)=0.0193P(X\geq10)=0.0193 for XB(12,0.5)X\sim\operatorname{B}(12,0.5), conduct the test at the 5%5\% level.

Tier 2 · Standard

5 marks
ORIGINAL

A process is tested using H0:p=0.5H_0:p=0.5 against H1:p0.5H_1:p\neq0.5. In 2020 independent trials there are 55 successes. Given P(X5)=0.02069P(X\leq5)=0.02069 under XB(20,0.5)X\sim\operatorname{B}(20,0.5), conduct a two-tailed test at the 5%5\% level.

Tier 3 · Hard

6 marks
ORIGINAL

A one-tailed test uses H0:p=0.2H_0:p=0.2 against H1:p>0.2H_1:p>0.2 with a sample of 1515. Under H0H_0, P(X6)=0.0611P(X\geq6)=0.0611 and P(X7)=0.0181P(X\geq7)=0.0181. Find the 5%5\% critical region. If 66 successes are observed, state the conclusion and explain the actual probability of a Type I error.

S5.3

Conduct a statistical hypothesis test for the mean of a Normal distribution with known, given or assumed variance and interpret the results in context.

  • For a Normal population with known or assumed standard deviation σ\sigma, the sample mean satisfies XN(μ,σ2/n)\overline{X}\sim\operatorname{N}(\mu,\sigma^2/n).
  • Under H0:μ=μ0H_0:\mu=\mu_0, standardise the observed mean using Z=xμ0σ/nZ=\frac{\overline{x}-\mu_0}{\sigma/\sqrt n}.
  • Use the tail or tails specified by H1H_1, compare the p-value with the significance level, and give the conclusion in the language of the population mean.
  • The method relies on a random, independent sample from a Normal population and uses a known, given or assumed variance rather than estimating it within this specified test.

Tier 1 · Easy

5 marks
ORIGINAL

A Normal population has known standard deviation 1212. Test H0:μ=100H_0:\mu=100 against H1:μ>100H_1:\mu>100 using a random sample of 3636 with mean 104104, at the 5%5\% level.

Tier 2 · Standard

6 marks
ORIGINAL

A Normal population has known standard deviation 55. For a sample of 2525, test H0:μ=50H_0:\mu=50 against H1:μ50H_1:\mu\neq50 at the 1%1\% level. The critical standard Normal values are ±2.576\pm2.576. Find the critical values of the sample mean and decide what to conclude if x=47.3\overline{x}=47.3.

Tier 3 · Hard

7 marks
ORIGINAL

A Normal population has known standard deviation 66. To test H0:μ=80H_0:\mu=80 against H1:μ>80H_1:\mu>80 at the 5%5\% level, find the smallest sample size nn for which an observed mean of 8282 would lead to rejection. Use the critical value 1.6451.645, then find the p-value for this minimum nn.