| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hypothesis test of a Poisson distribution |
| Type | Comment on test validity or assumptions |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness-of-fit test for a Poisson distribution with routine calculations. Part (a) is basic mean calculation, part (b) uses the Poisson formula directly, and part (c) follows a standard hypothesis testing procedure. While it requires multiple steps and understanding of the chi-squared test, it's a textbook application with no novel insight required, making it slightly easier than average. |
| Spec | 2.02g Calculate mean and standard deviation2.05a Hypothesis testing language: null, alternative, p-value, significance5.02i Poisson distribution: random events model5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.06b Fit prescribed distribution: chi-squared test |
| Number of calls, \(\boldsymbol { x }\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| Frequency | 3 | 13 | 14 | 15 | 10 | 8 | 8 | 6 | 3 |
| \(\boldsymbol { x }\) | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| ||
| 2.42 | 8.46 | 14.80 | \(r\) | 15.10 | 10.57 | 6.17 | 3.08 | \(s\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\hat{\lambda} = \dfrac{0(3)+1(13)+2(14)+3(15)+4(10)+5(8)+6(8)+7(6)+8(3)}{80} = \dfrac{280}{80} = 3.5\) | B1cso | At least 2 non-zero products shown and divide by 80 to achieve 3.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = 80 \times \dfrac{e^{-3.5}(3.5)^3}{3!} = 17.26283752...\) or \(r = 80 \times (0.5366 - 0.3208) = 17.264\) | M1 | Correct method for pooling classes at both ends (ft their \(s\) value) |
| \(s = 80 - (2.42 + 8.46 + 14.80 + r + 15.10 + 10.57 + 6.17 + 3.08) = 2.14\) or \(s = 80 \times (1 - 0.9733) = 2.136\) | ||
| \(r = 17.26\) (2dp), \(s = 2.14\) (2dp) | A1, A1 | At least one of \(r =\) awrt 17.26 or \(s =\) awrt 2.14; Both awrt 17.26 and awrt 2.14 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): Poisson distribution is a suitable model. \(H_1\): Poisson distribution is not a suitable model. | B1 | Must have both hypotheses mentioning Poisson. Inclusion of 3.5 for \(\lambda\) is in 1st B0 |
| Combining rows 0,1 → \(O_i = 16\), \(E_i = 10.88\); rows 7, \(\geq 8\) → \(O_i = 9\), \(E_i = 5.22\) | M1 | Correct method of pooling classes at both ends |
| At least 3 correct expressions/values to awrt/truncated 2 d.p. in \(\frac{(O_i-E_i)^2}{E_i}\) or \(\frac{O_i^2}{E_i}\) columns | M1 | |
| \(X^2 = \sum\dfrac{(O-E)^2}{E}\) or \(\sum\dfrac{O^2}{E} - 80 =\) awrt 8.38 | A1 | awrt 8.38 or awrt 8.39 |
| \(\nu = 7 - 1 - 1 = 5\) | B1ft | For their evaluated \(n-1-1\), i.e. realising they subtract 2 from \(n\) |
| \(\chi^2_5(0.05) = 11.070 \Rightarrow\) CR: \(X^2 \geq 11.070\) | B1ft | Correct ft for \(\chi^2_k(0.05)\) where \(k = n-1-1\) from their \(n\). (May see 9.488, 12.592, 14.067) |
| Not in CR / not significant / Do not reject \(H_0\) | ||
| Poisson distribution is a suitable model | A1 | Dep. on at least 1 M1 for correct conclusion accepting \(H_0\). Contradictory statements score A0 |
# Question 4:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\hat{\lambda} = \dfrac{0(3)+1(13)+2(14)+3(15)+4(10)+5(8)+6(8)+7(6)+8(3)}{80} = \dfrac{280}{80} = 3.5$ | B1cso | At least 2 non-zero products shown and divide by 80 to achieve 3.5 |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = 80 \times \dfrac{e^{-3.5}(3.5)^3}{3!} = 17.26283752...$ or $r = 80 \times (0.5366 - 0.3208) = 17.264$ | M1 | Correct method for pooling classes at both ends (ft their $s$ value) |
| $s = 80 - (2.42 + 8.46 + 14.80 + r + 15.10 + 10.57 + 6.17 + 3.08) = 2.14$ or $s = 80 \times (1 - 0.9733) = 2.136$ | | |
| $r = 17.26$ (2dp), $s = 2.14$ (2dp) | A1, A1 | At least one of $r =$ awrt 17.26 or $s =$ awrt 2.14; Both awrt 17.26 and awrt 2.14 |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: Poisson distribution is a suitable model. $H_1$: Poisson distribution is not a suitable model. | B1 | Must have both hypotheses mentioning Poisson. Inclusion of 3.5 for $\lambda$ is in 1st B0 |
| Combining rows 0,1 → $O_i = 16$, $E_i = 10.88$; rows 7, $\geq 8$ → $O_i = 9$, $E_i = 5.22$ | M1 | Correct method of pooling classes at both ends |
| At least 3 correct expressions/values to awrt/truncated 2 d.p. in $\frac{(O_i-E_i)^2}{E_i}$ or $\frac{O_i^2}{E_i}$ columns | M1 | |
| $X^2 = \sum\dfrac{(O-E)^2}{E}$ or $\sum\dfrac{O^2}{E} - 80 =$ awrt 8.38 | A1 | awrt **8.38** or awrt **8.39** |
| $\nu = 7 - 1 - 1 = 5$ | B1ft | For their evaluated $n-1-1$, i.e. realising they subtract 2 from $n$ |
| $\chi^2_5(0.05) = 11.070 \Rightarrow$ CR: $X^2 \geq 11.070$ | B1ft | Correct ft for $\chi^2_k(0.05)$ where $k = n-1-1$ from their $n$. (May see 9.488, 12.592, 14.067) |
| Not in CR / not significant / Do not reject $H_0$ | | |
| Poisson distribution is a suitable model | A1 | Dep. on at least 1 M1 for correct conclusion accepting $H_0$. Contradictory statements score A0 |
---4. The number of emergency plumbing calls received per day by a local council was recorded over a period of 80 days. The results are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | c | }
\hline
Number of calls, $\boldsymbol { x }$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\
\hline
Frequency & 3 & 13 & 14 & 15 & 10 & 8 & 8 & 6 & 3 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Show that the mean number of emergency plumbing calls received per day is 3.5
A council officer suggests that a Poisson distribution can be used to model the number of emergency plumbing calls received per day. He uses the mean from the sample above and calculates the expected frequencies shown in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | }
\hline
$\boldsymbol { x }$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & \begin{tabular}{ c }
8 or \\
more \\
\end{tabular} \\
\hline
\begin{tabular}{ c }
Expected \\
frequency \\
\end{tabular} & 2.42 & 8.46 & 14.80 & $r$ & 15.10 & 10.57 & 6.17 & 3.08 & $s$ \\
\hline
\end{tabular}
\end{center}
\item Calculate the value of $r$ and the value of $s$, giving your answers correct to 2 decimal places.
\item Test, at the $5 \%$ level of significance, whether or not the Poisson distribution is a suitable model for the number of emergency plumbing calls received per day. State your hypotheses clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S3 2017 Q4 [11]}}