Further Mathematics · OCR H245 · Option S · Topic 5.06

Chi-squared tests

Contingency tables, fitting a theoretical distribution, goodness of fit test.

  • 3 sub-topics
  • 2 in Year 12, 1 in Year 13
  • 81 practice questions
  • 3 animated lessons
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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. 5.06aContingency tablesY12
  2. 5.06b-cFitting a theoretical distributionY12
  3. 5.06dGoodness of fit testY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 81 questions in this topic, free, no account needed.

Chi-squared tests for association

Easy

A small expected frequency · Year 12 · 1 mark

  1. In a chi-squared test, one expected frequency is less than 5. What should be done?
    (1)

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  1. The chi-squared distribution is a good approximation only when the expected frequencies are large enough.BB: Combine that class with a neighbouring one, and reduce the degrees of freedom to match

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Chi-squared tests for association

Medium

Degrees of freedom after combining · Year 12 · 1 mark

  1. A contingency table has 2 rows and 4 columns. Two of the columns are combined because of small expected frequencies. How many degrees of freedom does the test now have?
    (1)

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  1. The table now has 2 rows and 3 columns: (2−1)(3−1)(2 - 1)(3 - 1).CC: 2

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Chi-squared tests

Hard

Chi-squared goodness of fit: Poisson · Year 12 · 8 marks

The numbers of emails received in 100 randomly chosen five-minute intervals are: 0 emails: 46; 1: 34; 2: 15; 3: 4; 4 or more: 1. Treat '4 or more' as 4.
  1. (a)
    Show that the mean number of emails per interval is 0.80.8 to 2 decimal places.
    (1)
  2. (b)
    Test, at the 5% level, whether a Poisson distribution is a suitable model.
    (7)
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Part (a)1 mark

  1. ∑xf100=0.800\frac{\sum xf}{100} = 0.800

Part (b)7 marks

  1. H₁: it is not.H0:a Poisson distribution is a suitable modelH_0: \text{a Poisson distribution is a suitable model}
  2. N×P(X=x)N \times \mathrm{P}(X = x) with λ=0.8\lambda = 0.8; the last class is P(X≥4)\mathrm{P}(X \ge 4).EE=44.93,35.95,14.38,3.83,0.91{}= 44.93, 35.95, 14.38, 3.83, 0.91
  3. Classes with E<5E < 5 are combined: 0, 1, 2+.
  4. X2X^2=∑(O−E)2E{}= \sum\frac{(O - E)^2}{E}=0.171{}= 0.171
  5. Subtract 1 for the total and 1 for estimating λ\lambda.ν\nu=3−2{}= 3 - 2=1, cv{}= 1, \ \text{cv}=3.841{}= 3.841
  6. Do not reject H₀: a Poisson distribution is a suitable model.

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