Further Mathematics · AQA 7367 · Option S · Topic SH

Confidence intervals

Confidence intervals for a mean, with known variance, large samples and the t-distribution.

  • 4 sub-topics
  • 3 in Year 12, 1 in Year 13
  • 61 practice questions
  • 4 animated lessons
Practice questions (61) Lessons Worksheets All AQA Further Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. SH1Confidence interval for a normal mean with known varianceY12
  2. SH2Large-sample confidence interval with unknown varianceY12
  3. SH3Inferences from confidence intervalsY12
  4. SH4Confidence intervals using the t-distributionY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 61 questions in this topic, free, no account needed.

Estimation, confidence intervals and tests using a normal distribution

Not needed for Year 12Easy

Why large samples help · Year 13 · 1 mark

  1. Which result lets the sample mean be treated as approximately Normal when the sample is large, whatever the distribution of the population?
    (1)

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  1. For large nn, Xˉ\bar{X} is approximately N ⁣(μ,σ2n)N\!\left(\mu, \dfrac{\sigma^2}{n}\right).A: The Central Limit Theorem

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Confidence intervals

Medium

Confidence intervals · Year 12 · 5 marks

A measurement is normally distributed with standard deviation 4. A 99% confidence interval for the mean is to have width at most 2.
  1. (a)
    Find the smallest sample size needed.
    (3)
  2. (b)
    A sample of this size has mean 61.7. Find the 99% confidence interval.
    (2)

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Part (a)3 marks

  1. The width is twice zσ/nz \sigma /\sqrt{n}.2×2.5758×4n≤22 \times 2.5758 \times \frac{4}{\sqrt{n}} \leq 2
  2. n≥106.16n \geq 106.16
  3. Round up.n=107n = 107

Part (b)2 marks

  1. 61.7±2.5758×410761.7 \pm 2.5758 \times \frac{4}{\sqrt{107}}
  2. (60.704,62.696)(60.704, 62.696)

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Confidence intervals

Not needed for Year 12Hard

Confidence intervals using the t-distribution · Year 13 · 6 marks

The masses, in grams, of a random sample of 6 biscuits are: 22.2, 23.6, 23.4, 23.5, 24.6, 20.3. Assume the masses are normally distributed.
  1. (a)
    Find a 95% confidence interval for the mean mass.
    (5)
  2. (b)
    The manufacturer claims the mean mass is 20.7 g. Comment on this claim.
    (1)
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Part (a)5 marks

  1. ss is the unbiased estimate.xˉ=22.933, s=1.4989\bar{x} = 22.933, \ s = 1.4989
  2. t5=2.571t_{5} = 2.571
  3. 22.933±2.571×1.4989622.933 \pm 2.571 \times \frac{1.4989}{\sqrt{6}}
  4. (21.36,24.51)(21.36, 24.51)

Part (b)1 mark

  1. 20.7 is outside the interval, so the claim is not supported.

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