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How do you use sample evidence to decide whether a claim about a population proportion, mean or correlation is supported?

Null and alternative hypotheses, one- and two-tailed tests, significance levels and critical regions, hypothesis tests for a binomial proportion, for a Normal mean, and for a correlation coefficient.

A focused answer to the OCR A-Level Mathematics A hypothesis testing content, covering null and alternative hypotheses, one- and two-tailed tests, significance levels and critical regions, tests for a binomial proportion, tests for the mean of a Normal distribution, and tests for a product moment correlation coefficient.

Generated by Claude Opus 4.812 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

OCR wants you to set up null and alternative hypotheses, choose a one- or two-tailed test, work at a stated significance level, find and use a critical region, carry out a hypothesis test for a binomial proportion, test the mean of a Normal distribution using the distribution of the sample mean, and test a product moment correlation coefficient against critical values, always concluding in context.

The answer

Hypotheses and significance

A hypothesis test weighs sample evidence against a claim. The null hypothesis H0H_0 is the default (the parameter equals a stated value); the alternative H1H_1 is what you test for (greater, less, or not equal). The significance level (often 5%5\% or 1%1\%) is the probability of rejecting H0H_0 when it is actually true, that is the risk you accept of a false alarm.

One- and two-tailed tests

If H1H_1 has a direction (p>0.2p > 0.2 or μ<500\mu < 500), the test is one-tailed and the whole significance level sits in one tail. If H1H_1 is "not equal", the test is two-tailed and the level is split between the two tails (so 2.5%2.5\% in each tail for a 5%5\% test).

The critical region

The critical region is the set of outcomes so extreme that you reject H0H_0. You either compare the probability of the observed result (or more extreme) with the significance level, or find the critical region first and check whether the observation falls in it.

The method, step by step

State the hypotheses and the model, identify the tail(s) and significance level, find the relevant probability or critical region, compare, and state the conclusion in the context of the question. The contextual conclusion is always worth a mark.

Examples in context

A binomial proportion test

A correlation test

For bivariate data you can test whether there is genuine correlation in the population. The hypotheses are H0:ρ=0H_0: \rho = 0 (no correlation) against a one- or two-tailed alternative, and you compare the sample correlation coefficient rr with a critical value from tables for the sample size and significance level.

Try this

Q1. For a two-tailed test at the 5%5\% level, what probability sits in each tail? [1 mark]

  • Cue. 2.5%2.5\% in each tail.

Q2. A sample of 2525 from N(μ,9)N(\mu, 9) has mean xˉ\bar{x}. State the standard deviation of Xˉ\bar{X}. [2 marks]

  • Cue. σn=325=35=0.6\dfrac{\sigma}{\sqrt{n}} = \dfrac{3}{\sqrt{25}} = \dfrac{3}{5} = 0.6.

Exam-style practice questions

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

OCR 20196 marksA seed supplier claims at least 80%80\% of seeds germinate. In a sample of 2020 seeds, 1313 germinate. Test at the 5%5\% significance level whether the germination rate is less than 80%80\%.
Show worked answer →

State hypotheses (B1): let pp be the germination probability and XX the number that germinate, XB(20,0.8)X \sim B(20, 0.8). H0:p=0.8H_0: p = 0.8, H1:p<0.8H_1: p < 0.8 (one-tailed).

This is a lower-tail test at 5%5\% (M1). Find P(X13)P(X \le 13) under H0H_0 (M1).

From the cumulative binomial, P(X13)0.0867P(X \le 13) \approx 0.0867 (A1).

Compare with 5%5\%: 0.0867>0.050.0867 > 0.05, so the result is not in the critical region; do not reject H0H_0 (M1).

There is insufficient evidence at the 5%5\% level that the germination rate is below 80%80\% (A1).

Markers reward the hypotheses with the model, the correct lower-tail probability, the comparison, and a contextual conclusion.

OCR 20216 marksA machine fills bottles with volume modelled by N(μ,4)N(\mu, 4) ml. The target mean is 500500 ml. A sample of 1616 bottles has mean 498.5498.5 ml. Test at the 5%5\% level whether the mean volume has fallen below 500500 ml.
Show worked answer →

State hypotheses (B1): H0:μ=500H_0: \mu = 500, H1:μ<500H_1: \mu < 500 (one-tailed).

The sample mean is distributed as XˉN ⁣(500,416)=N(500,0.25)\bar{X} \sim N\!\left(500, \dfrac{4}{16}\right) = N(500, 0.25), so its standard deviation is 0.50.5 (M1).

Standardise the observed mean (M1): Z=498.55000.5=3.0Z = \dfrac{498.5 - 500}{0.5} = -3.0 (A1).

The critical zz for a 5%5\% lower-tail test is 1.645-1.645 (M1). Since 3.0<1.645-3.0 < -1.645, reject H0H_0 (A1).

There is evidence at the 5%5\% level that the mean volume has fallen below 500500 ml.

Markers reward the hypotheses, the distribution of the sample mean with variance σ2n\frac{\sigma^2}{n}, standardising, and comparing with the critical value.

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