| Exam Board | Edexcel |
|---|---|
| Module | FS1 AS (Further Statistics 1 AS) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Poisson |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test for a Poisson distribution with straightforward parts: stating hypotheses, calculating expected frequencies from given Poisson parameter, computing test statistics, and determining degrees of freedom. All steps are routine FS1 procedures with no novel problem-solving required, making it slightly easier than average A-level difficulty. |
| Spec | 5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.02l Poisson conditions: for modelling5.02m Poisson: mean = variance = lambda5.06b Fit prescribed distribution: chi-squared test5.06c Fit other distributions: discrete and continuous |
| 0 | 1 | 2 | 3 | 4 | 5 | ||
| Frequency | 12 | 11 | 19 | 14 | 3 | 1 |
| Number of car breakdowns | 0 | 1 | 2 | 3 | 4 | \(\geqslant 5\) |
| Observed frequency (O) | 12 | 11 | 19 | 14 | 3 | 1 |
| Expected frequency ( \(\mathbf { E } _ { \mathbf { i } }\) ) | 9.92 | 9.64 | 4.34 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(H_0\): No. of car breakdowns per month follows a Poisson distribution; \(H_1\): No. of car breakdowns per month does not follow a Poisson distribution | B1 | Correct hypotheses mentioning "breakdowns" and "Poisson"; \(Po(1.8)\) o.e. is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| A Poisson distribution will assign some probability to (all) values greater than 5 (o.e.) | B1 | For a suitable reason mentioning the Poisson distribution taking values beyond 5, e.g. sample space for Poisson being \([0, \infty)\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Need \(\hat{\lambda}\): \(\hat{\lambda} = \frac{0\times12 + 1\times11 + 2\times19 + 3\times14 + 4\times3 + 5\times1}{(12+11+19+14+3+1)}\) or 1.8 | M1 | For an expression for the mean with at least 3 correct products and correct denominator; M0 if found \(\lambda\) by working backwards from \(P(X=0) \Rightarrow 60e^{-\lambda} = 9.92\) |
| [Under \(H_0\) \(X \sim Po(1.8)\)] \(E_1 = 60 \times P(X=1) = 17.85(227\ldots)\) | M1 | For a correct method for finding \(E_1\) or \(E_2\) (implied by one correct value) |
| \(E_2 = 60 \times P(X=2) = 16.06(705\ldots)\) | A1 | For awrt 17.85 and awrt 16.07 |
| \(E_{\geq 5} = 60 - \sum_{0}^{4} E_i = 2.18(43996\ldots)\) | B1ft | For awrt 2.18, or using \(60 - (\text{sum of their } E_i)\) or \(60 \times (1 - P(X \leq 4))\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\frac{(11 - 17.85\ldots)^2}{17.85\ldots} = 2.6287\ldots\) or \(\frac{(14 - 9.64)^2}{9.64} = 1.97195\ldots\) | M1 | For either correct expression (ft their \(E_1\)) |
| awrt 2.63 and awrt 1.97 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Need to combine last two columns since \(E_i < 5\) | B1 | For explaining need to pool columns since \(E_i < 5\); allow candidates describing combining last three columns as long as they say \(E_i < 5\) |
| Degrees of freedom therefore \(5 - 2\) since mean for Poisson estimated from \(O_i\) | B1 | For mentioning mean/rate/parameter/\(\lambda\) estimated from \(O_i\) and 2 constraints |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\chi_3^2(5\%) = 7.815\) | B1 | For correct cv of 7.815 (or better) |
| (Not significant) insufficient evidence to reject Anja's belief | B1 | For correct conclusion in context mentioning "breakdowns" and "Poisson"; allow equivalent words for belief such as "theory"; must be consistent with their cv; allow \(Po(1.8)\) in conclusion instead of Poisson |
# Question 4:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $H_0$: No. of car breakdowns per month follows a Poisson distribution; $H_1$: No. of car breakdowns per month does not follow a Poisson distribution | B1 | Correct hypotheses mentioning "breakdowns" and "Poisson"; $Po(1.8)$ o.e. is B0 |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| A Poisson distribution will assign some probability to (all) values greater than 5 (o.e.) | B1 | For a suitable reason mentioning the Poisson distribution taking values beyond 5, e.g. sample space for Poisson being $[0, \infty)$ |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Need $\hat{\lambda}$: $\hat{\lambda} = \frac{0\times12 + 1\times11 + 2\times19 + 3\times14 + 4\times3 + 5\times1}{(12+11+19+14+3+1)}$ or 1.8 | M1 | For an expression for the mean with at least 3 correct products and correct denominator; M0 if found $\lambda$ by working backwards from $P(X=0) \Rightarrow 60e^{-\lambda} = 9.92$ |
| [Under $H_0$ $X \sim Po(1.8)$] $E_1 = 60 \times P(X=1) = 17.85(227\ldots)$ | M1 | For a correct method for finding $E_1$ or $E_2$ (implied by one correct value) |
| $E_2 = 60 \times P(X=2) = 16.06(705\ldots)$ | A1 | For awrt 17.85 and awrt 16.07 |
| $E_{\geq 5} = 60 - \sum_{0}^{4} E_i = 2.18(43996\ldots)$ | B1ft | For awrt 2.18, or using $60 - (\text{sum of their } E_i)$ or $60 \times (1 - P(X \leq 4))$ |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\frac{(11 - 17.85\ldots)^2}{17.85\ldots} = 2.6287\ldots$ or $\frac{(14 - 9.64)^2}{9.64} = 1.97195\ldots$ | M1 | For either correct expression (ft their $E_1$) |
| awrt 2.63 and awrt 1.97 | A1 | |
## Part (e):
| Working | Mark | Guidance |
|---------|------|----------|
| Need to combine last two columns since $E_i < 5$ | B1 | For explaining need to pool columns since $E_i < 5$; allow candidates describing combining last three columns as long as they say $E_i < 5$ |
| Degrees of freedom therefore $5 - 2$ since mean for Poisson estimated from $O_i$ | B1 | For mentioning mean/rate/parameter/$\lambda$ estimated from $O_i$ and 2 constraints |
## Part (f):
| Working | Mark | Guidance |
|---------|------|----------|
| $\chi_3^2(5\%) = 7.815$ | B1 | For correct cv of 7.815 (or better) |
| (Not significant) insufficient evidence to reject Anja's belief | B1 | For correct conclusion in context mentioning "breakdowns" and "Poisson"; allow equivalent words for belief such as "theory"; must be consistent with their cv; allow $Po(1.8)$ in conclusion instead of Poisson |
\begin{enumerate}
\item Table 1 below shows the number of car breakdowns in the Snoreap district in each of 60 months.
\end{enumerate}
\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | }
\hline
\begin{tabular}{ c }
Number of car \\
breakdowns \\
\end{tabular} & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
Frequency & 12 & 11 & 19 & 14 & 3 & 1 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 1}
\end{center}
\end{table}
Anja believes that the number of car breakdowns per month in Snoreap can be modelled by a Poisson distribution. Table 2 below shows the results of some of her calculations.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|}
\hline
Number of car breakdowns & 0 & 1 & 2 & 3 & 4 & $\geqslant 5$ \\
\hline
Observed frequency (O) & 12 & 11 & 19 & 14 & 3 & 1 \\
\hline
Expected frequency ( $\mathbf { E } _ { \mathbf { i } }$ ) & 9.92 & & & 9.64 & 4.34 & \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 2}
\end{center}
\end{table}
(a) State suitable hypotheses for a test to investigate Anja's belief.\\
(b) Explain why Anja has changed the label of the final column to $\geqslant 5$\\
(c) Showing your working clearly, complete Table 2\\
(d) Find the value of $\frac { \left( O _ { i } - E _ { i } \right) ^ { 2 } } { E _ { i } }$ when the number of car breakdowns is\\
(i) 1\\
(ii) 3\\
(e) Explain why Anja used 3 degrees of freedom for her test.
The test statistic for Anja's test is 6.54 to 2 decimal places.\\
(f) Stating the critical value and using a $5 \%$ level of significance, complete Anja's test.
\hfill \mbox{\textit{Edexcel FS1 AS 2023 Q4 [12]}}