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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 χ 2  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 , f x f = 0 × 4 + 1 × 18 + 2 × 19 + + 5 × 8 80 = 200 80 = 2.5         A1

Theoretical frequencies are

f ( 0 ) = 8.0 e 2.5 = 6.5668         (M1)(A1)

f ( 1 ) = 2.5 1 × 6.5668 = 16.4170         A1

f ( 2 ) = 2.5 2 × 16.4170 = 20.5212

f ( 3 ) = 2.5 3 × 20.5212 = 17.1010

f ( 4 ) = 2.5 4 × 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

χ 2 = ( 4 6.5668 ) 2 6.5668 + ( 18 16.4170 ) 2 16.4170 + ( 19 20.5212 ) 2 20.5212 + ( 20 17.1010 ) 2 17.1010 + ( 11 10.6882 ) 2 10.6882 + ( 8 8.7058 ) 2 8.7058

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

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

       χ 2 ( 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

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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
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Topic 4—Statistics and probability

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