EXM.1.AHL.TZ0.57

pestleMathematicsAIHLPaper 1EXM· sl-4-11-expected-observed-hypotheses-chi-squared-gof-t-testsource ↗

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 ${\chi ^2}$ goodness of fit test at the 5% significance level to decide if the above data can be modelled by a Poisson distribution.

Markscheme / solution

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

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

$\sum {f = 80{\text{,}}\,\,} \frac{{\sum {fx} }}{{\sum f }} = \frac{{0 \times 4 + 1 \times 18 + 2 \times 19 + \ldots + 5 \times 8}}{{80}} = \frac{{200}}{{80}} = 2.5$ A1

Theoretical frequencies are

$f\left( 0 \right) = 8.0{{\text{e}}^{ - 2.5}} = 6.5668$ (M1)(A1)

$f\left( 1 \right) = \frac{{2.5}}{1} \times 6.5668 = 16.4170$ A1

$f\left( 2 \right) = \frac{{2.5}}{2} \times 16.4170 = 20.5212$

$f\left( 3 \right) = \frac{{2.5}}{3} \times 20.5212 = 17.1010$

$f\left( 4 \right) = \frac{{2.5}}{4} \times 17.1010 = 10.6882$ A1

Note: Award A1 for $f\left( 2 \right)$ , $f\left( 4 \right)$ , $f\left( 4 \right)$ .

$f$ (5 or more) $= 80 - \left( {6.5668 + 16.4170 + 20.5212 + 17.1010 + 10.6882} \right)$ A1

$= 8.7058$

${\chi ^2} = \frac{{{{\left( {4 - 6.5668} \right)}^2}}}{{6.5668}} + \frac{{{{\left( {18 - 16.4170} \right)}^2}}}{{16.4170}} + \frac{{{{\left( {19 - 20.5212} \right)}^2}}}{{20.5212}} + \frac{{{{\left( {20 - 17.1010} \right)}^2}}}{{17.1010}} + \frac{{{{\left( {11 - 10.6882} \right)}^2}}}{{10.6882}} + \frac{{{{\left( {8 - 8.7058} \right)}^2}}}{{8.7058}}$

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

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

${\chi ^2}\left( 4 \right) = 9.488$ 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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