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Date May Example question Marks available 11 Reference code EXM.1.AHL.TZ0.57
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Test Question number 57 Adapted from N/A

Question

The number of cars passing a certain point in a road was recorded during 80 equal time intervals and summarized in the table below.

Carry out a χ2 goodness of fit test at the 5% significance level to decide if the above data can be modelled by a Poisson distribution.

Markscheme

H0 : The data can be modeled by a Poisson distribution.

H1 : The data cannot be modeled by a Poisson distribution.

f=80,fxf=0×4+1×18+2×19++5×880=20080=2.5        A1

Theoretical frequencies are

f(0)=8.0e2.5=6.5668        (M1)(A1)

f(1)=2.51×6.5668=16.4170        A1

f(2)=2.52×16.4170=20.5212

f(3)=2.53×20.5212=17.1010

f(4)=2.54×17.1010=10.6882        A1

Note:    Award A1 for f(2), f(4), f(4).

f(5 or more) =80(6.5668+16.4170+20.5212+17.1010+10.6882)        A1

          =8.7058

χ2=(46.5668)26.5668+(1816.4170)216.4170+(1920.5212)220.5212+(2017.1010)217.1010+(1110.6882)210.6882+(88.7058)28.7058

      =1.83  (accept 1.82)        (M1)(A1)

      v=4  (six frequencies and two restrictions)        (A1)

      χ2(4)=9.488 at the 5% level.        A1

       Since 1.83 < 9.488 we accept H0 and conclude that the distribution can be modeled by a Poisson distribution.        R1    N0

[11 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.11—Expected, observed, hypotheses, chi squared, gof, t-test
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Topic 4—Statistics and probability » AHL 4.12—Data collection, reliability and validity tests
Topic 4—Statistics and probability » AHL 4.17—Poisson distribution
Topic 4—Statistics and probability

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