| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Poisson distribution |
| Type | Goodness-of-fit test for Poisson |
| Difficulty | Standard +0.3 This is a standard goodness-of-fit test for a Poisson distribution with the parameter given (not estimated). Part (a) requires simple subtraction, (b) is routine hypothesis statement, (c) tests understanding of degrees of freedom (cells minus 1, no parameter estimation), and (d) is straightforward chi-squared table lookup and comparison. All parts follow textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.06b Fit prescribed distribution: chi-squared test |
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | ||
| Number of squares | 30 | 42 | 35 | 26 | 11 | 6 | 0 |
| 0 | 1 | 2 | 3 | 4 | 5 | More than 5 | ||
| Number of squares | 26.07 | 45.62 | 39.91 | 23.28 | 10.19 | 3.57 | \(r\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(1.36\) or \(1.37\) | B1 | Accept 1.36 or 1.37 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): Po(1.75) is a suitable model; \(H_1\): Po(1.75) is not a suitable model | B1 | Must have Po(1.75) or Poisson with mean 1.75 attached to \(H_0\) and \(H_1\) the right way round |
| Answer | Marks | Guidance |
|---|---|---|
| Cells combined for expected frequencies \(< 5\), combine last 3 cells | B1 | Must mention combine the 3 cells when frequencies \(< 5\) |
| Subtract 1 since totals agree | B1 | Must say/show 1 is subtracted and totals agree or total frequency must be 150 or only need 4 pieces of data. NB: B0 for "only 1 constraint" on its own |
| Answer | Marks | Guidance |
|---|---|---|
| \(\chi^2_4 = 9.488\) | B1 | awrt 9.49 |
| Researcher's belief is supported / evidence that Po(1.75) is a good model for number of orchids per square metre | B1ft | ft their critical value only; if hypotheses wrong way round or absent award B0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(P(\text{exactly 6 orchids}) = \) awrt \(0.00353\) | B1 | May be implied by awrt 0.706 for mean |
| \(X \sim B(200, \text{"0.00353"})\), mean \(= 200 \times \text{"0.00353"} = \) awrt \(0.706\) | M1 | Selecting model \(B(200, \text{"their } p\text{(exactly 6 orchids)"})\) and using \(np\) \((0 < p < 1)\) |
| \(Y \sim Po(\text{"0.706"})\), \(1 - P(Y=0) = 1 - e^{-\text{"0.706"}}\) | M1 | Using Po(their \(np\)) and using \(1-P(Y=0)\) or \(1-P(Y \leq 0)\) or \(1-e^{-\text{"0.706"}}\) |
| \(= 0.506^*\) | A1* | Only award if previous 3 marks awarded and 0.506 stated |
# Question 1:
## Part (a)
| $1.36$ or $1.37$ | B1 | Accept 1.36 or 1.37 |
## Part (b)
| $H_0$: Po(1.75) is a suitable model; $H_1$: Po(1.75) is not a suitable model | B1 | Must have Po(1.75) or Poisson with mean 1.75 attached to $H_0$ and $H_1$ the right way round |
## Part (c)
| Cells combined for expected frequencies $< 5$, combine last 3 cells | B1 | Must mention combine the 3 cells when frequencies $< 5$ |
| Subtract 1 since totals agree | B1 | Must say/show 1 is subtracted and totals agree or total frequency must be 150 or only need 4 pieces of data. NB: B0 for "only 1 constraint" on its own |
## Part (d)
| $\chi^2_4 = 9.488$ | B1 | awrt 9.49 |
| Researcher's belief is supported / evidence that Po(1.75) is a good model for number of orchids per square metre | B1ft | ft their critical value only; if hypotheses wrong way round or absent award B0 |
## Part (e)
| $P(\text{exactly 6 orchids}) = $ awrt $0.00353$ | B1 | May be implied by awrt 0.706 for mean |
| $X \sim B(200, \text{"0.00353"})$, mean $= 200 \times \text{"0.00353"} = $ awrt $0.706$ | M1 | Selecting model $B(200, \text{"their } p\text{(exactly 6 orchids)"})$ and using $np$ $(0 < p < 1)$ |
| $Y \sim Po(\text{"0.706"})$, $1 - P(Y=0) = 1 - e^{-\text{"0.706"}}$ | M1 | Using Po(their $np$) and using $1-P(Y=0)$ or $1-P(Y \leq 0)$ or $1-e^{-\text{"0.706"}}$ |
| $= 0.506^*$ | A1* | Only award if previous 3 marks awarded and 0.506 stated |
---\begin{enumerate}
\item A researcher is investigating the distribution of orchids in a field. He believes that the Poisson distribution with a mean of 1.75 may be a good model for the number of orchids in each square metre. He randomly selects 150 non-overlapping areas, each of one square metre, and counts the number of orchids present in each square.
\end{enumerate}
The results are recorded in the table below.
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | }
\hline
\begin{tabular}{ l }
Number of orchids in \\
each square metre \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Number of squares & 30 & 42 & 35 & 26 & 11 & 6 & 0 \\
\hline
\end{tabular}
\end{center}
He calculates the expected frequencies as follows
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | }
\hline
\begin{tabular}{ l }
Number of orchids in \\
each square metre \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 & More than 5 \\
\hline
Number of squares & 26.07 & 45.62 & 39.91 & 23.28 & 10.19 & 3.57 & $r$ \\
\hline
\end{tabular}
\end{center}
(a) Find the value of $r$ giving your answer to 2 decimal places.
The researcher will test, at the $5 \%$ level of significance, whether or not the data can be modelled by a Poisson distribution with mean 1.75\\
(b) State clearly the hypotheses required to test whether or not this Poisson distribution is a suitable model for these data.
The test statistic for this test is 2.0 and the number of degrees of freedom to be used is 4\\
(c) Explain fully why there are 4 degrees of freedom.\\
(d) Stating your critical value clearly, determine whether or not these data support the researcher's belief.
The researcher works in another field where the number of orchids in each square metre is known to have a Poisson distribution with mean 1.5
He randomly selects 200 non-overlapping areas, each of one square metre, in this second field, and counts the number of orchids present in each square.\\
(e) Using a Poisson approximation, show that the probability that he finds at least one square with exactly 6 orchids in it is 0.506 to 3 decimal places.
\hfill \mbox{\textit{Edexcel FS1 AS 2018 Q1 [10]}}