Further Mathematics · OCR H245 · Option S · Topic 5.05

Hypothesis tests and confidence intervals

The distribution of the sample mean and the central limit theorem, unbiased estimates of population mean and variance, using the normal distribution in hypothesis tests, confidence intervals.

  • 4 sub-topics
  • Taught in Year 13
  • 77 practice questions
  • 4 animated lessons

What the specification says

Every sub-topic OCR 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. 5.05aThe distribution of the sample mean and the central limit theoremY13
  2. 5.05bUnbiased estimates of population mean and varianceY13
  3. 5.05cUsing the normal distribution in hypothesis testsY13
  4. 5.05dConfidence intervalsY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 77 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

Total1

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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The central limit theorem

Not needed for Year 12Medium

The central limit theorem · Year 13 · 5 marks

A random sample of 80 observations is taken from a continuous uniform distribution on [2, 8].
  1. (a)
    Use the Central Limit Theorem to find the approximate distribution of the sample mean.
    (3)
  2. (b)
    Hence find the probability that the sample mean is greater than 5.3.
    (2)
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Part (a)3 marks

  1. μ=5, σ2=3\mu = 5, \ \sigma^2 = 3
  2. For large n, X̄ ~ N(μ, σ²/n) approximately.Xˉ\bar{X}≈N(5,380){}\approx \mathrm{N}\left(5, \frac{3}{80}\right)

Part (b)2 marks

  1. Standardise.P(Xˉ>5.3)P(\bar{X} > 5.3)
  2. =0.0607= 0.0607

Total5

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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Estimation, confidence intervals and tests using a normal distribution

Not needed for Year 12Hard

Estimators and bias · Year 13 · 6 marks

X1,X2,X3X_1, X_2, X_3 is a random sample from a population with mean μ\mu and variance σ2\sigma^2. Two estimators of μ\mu are T1T_1=13(X1+X2+X3){}= \tfrac{1}{3}(X_1 + X_2 + X_3) and T2T_2=18X1{}= \frac{1}{8}X_1+38X2{}+ \frac{3}{8}X_2+kX3{}+ kX_3.
  1. (a)
    Find the value of kk for which T2T_2 is unbiased.
    (2)
  2. (b)
    Find the variances of T1T_1 and T2T_2, and state which is the better estimator.
    (4)

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Part (a)2 marks

  1. E(T2)\mathrm{E}(T_2)=(18+38+k)μ{}= (\frac{1}{8} + \frac{3}{8} + k)\mu=μ{}= \mu
  2. k=12k = \frac{1}{2}

Part (b)4 marks

  1. Var⁡(T1)\operatorname{Var}(T_1)=13σ2{}= \tfrac{1}{3}\sigma^2
  2. Var⁡(T2)\operatorname{Var}(T_2)=(182+382+122)σ2{}= \left(\frac{1}{8}^2 + \frac{3}{8}^2 + \frac{1}{2}^2\right)\sigma^2=1332σ2{}= \frac{13}{32}\sigma^2
  3. T1T_1: both are unbiased and it has the smaller variance.

Total6

  • MMethod mark: for a correct method, even with a slip.
  • AAccuracy mark: for a correct answer from a correct method.
  • BIndependent mark: for a correct fact or result on its own, with no method needed.
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