EXM.1.AHL.TZ0.58

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

The number of telephone calls received by a helpline over 80 one-minute periods are summarized in the table below.

Find the exact value of the mean of this distribution.

[2]

Test, at the 5% level of significance, whether or not the data can be modelled by a Poisson distribution.

[12]

Markscheme / solution

Mean $\lambda = \frac{{\left( {9 \times 0 + 12 \times 1 + 22 \times 2 + 10 \times 3 + 11 \times 4 + 8 \times 5 + 8 \times 6} \right)}}{{80}}$ (M1)

$= 2.725 = \left( {\frac{{109}}{{40}}} \right)$ A1

Note: Do not accept 2.73.

[2 marks]

H₀: the data can be modelled by a Poisson distribution A1

H₁: the data cannot be modelled by a Poisson distribution A1

Note: Award A2 for one error, A1 for two errors, A0 for three or more errors.

Combining last two columns (M1)

Note: Allow FT from not combining the last two columns and / or getting 2.98 for the last expected frequency.

EITHER

${\chi ^2} = \frac{{{9^2}}}{{5.244}} + \frac{{{{12}^2}}}{{14.289}} + \frac{{{{22}^2}}}{{19.469}} + \frac{{{{10}^2}}}{{17.684}} + \frac{{{{11}^2}}}{{12.047}} + \frac{{{{16}^2}}}{{11.267}} - 80$ (M1)(A1)

= 8.804 (accept 8.8) A1

$v = 6 - 2 = 4$ , $\chi _{5{\text{% }}}^2 = 9.488$ A1A1

Hence 8.804 is not significant since 8.804 < 9.488 and we accept H₀ R1

p-value = 0.0662 (accept 0.066) which is not significant since A5

0.0662 > 0.05 and we accept H₀ R1 N0

[12 marks]

Examiners’ report

[N/A]