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9231 P41 - Nov 2025 - Q2 - 6 marks
6602
The manager of a car park claims that the number of cars entering the car park follows a Poisson distribution with mean \(2.8\). The numbers of cars entering the car park are recorded on a working day during successive 5-minute periods. The following table contains the observed frequencies, together with most of the expected frequencies and their contributions to the \(\chi^{2}\)-test statistic.
Number of cars
\(0\)
\(1\)
\(2\)
\(3\)
\(4\)
\(5\)
\(\geq 6\)
Observed frequency
\(2\)
\(15\)
\(31\)
\(29\)
\(13\)
\(3\)
\(7\)
Expected frequency
\(6.081\)
\(17.03\)
\(23.84\)
\(p\)
\(15.57\)
\(8.721\)
\(6.511\)
\(\chi^{2}\)-test statistic
\(2.739\)
\(0.241\)
\(2.152\)
\(q\)
\(0.425\)
\(3.753\)
\(0.037\)
(a) Find the value of \(p\) and the value of \(q\).
(b) Carry out a goodness of fit test at the \(5\%\) significance level to investigate the manager's claim.
Solution
Answer:
(a) \(p=22.248\), so \(p\approx 22.25\). Also \(q\approx 2.05\).
(b) The test statistic is \(\chi^2\approx 11.396\). With \(6\) degrees of freedom, the 5% critical value is \(12.59\). Since \(11.396\) is less than \(12.59\), do not reject the Poisson model. There is insufficient evidence to suggest that the manager’s claim is false.
(a) The claimed model is \(X\sim \mathrm{Po}(2.8)\). The expected frequency for 3 cars is found using \(P(X=3)\), multiplied by the total number of observed 5-minute periods, which is \(100\).
(b) Let \(H_0\) be that \(\mathrm{Po}(2.8)\) is a suitable model for the number of cars entering the car park in a 5-minute period. Let \(H_1\) be that \(\mathrm{Po}(2.8)\) is not a suitable model.
Using the contributions in the table, including the value of \(q\) found in part (a),
There are \(7\) classes and the mean \(2.8\) has been specified in the claim, not estimated from the data. Therefore the number of degrees of freedom is \(7-1=6\).
At the 5% significance level, the critical value for \(\chi^2_6\) is \(12.59\). Since \(11.396\) is less than \(12.59\), the result is not significant at the 5% level.
Therefore we do not reject \(H_0\). There is insufficient evidence to suggest that the Poisson distribution with mean \(2.8\) is not a good fit, so there is insufficient evidence to suggest that the manager’s claim is false.
