| Exam Board | Edexcel |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2022 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Chi-squared goodness of fit: Poisson |
| Difficulty | Standard +0.8 This is a standard chi-squared goodness of fit test for a Poisson distribution, requiring calculation of mean, expected frequencies using Poisson probabilities, and hypothesis testing. While it involves multiple steps and careful handling of the ≥6 category, it follows a well-established procedure taught in S3 with no novel insights required. The difficulty is elevated slightly above average due to the computational demands and need for precise application of the chi-squared test framework. |
| Spec | 5.02j Poisson formula: P(X=x) = e^(-lambda)*lambda^x/x!5.02k Calculate Poisson probabilities5.06b Fit prescribed distribution: chi-squared test |
| Number of emails per hour | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Frequencies | 1 | 10 | 23 | 15 | 19 | 9 | 3 |
| Number of emails per hour | Expected Frequencies |
| 0 | \(r\) |
| 1 | 11.949 |
| 2 | 17.923 |
| 3 | 17.923 |
| 4 | 13.443 |
| 5 | \(s\) |
| \(\geqslant 6\) | \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{0\times1+1\times10+2\times23+3\times15+4\times19+5\times9+6\times3}{80} = 3\) | B1 | For a correct method to show that the mean is 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(r = e^{-3} \times 80 = 3.983\); \(s = \frac{e^{-3}\times 3^5}{5!}\times 80 = 8.066\) | M1 A1 | Use of \(\frac{e^{-\lambda}\times\lambda^r}{r!}\times 80\); \(r=3.983\) and \(s=8.066\) |
| \(t = 80-(r+11.949+17.923+17.923+13.443+s) = 6.713\) | M1 A1 | A correct method to ensure expected totals = 80; \(t=6.713\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0\): Poisson distribution is a reasonable/suitable/sensible model; \(H_1\): Poisson distribution is not a reasonable/suitable/sensible model | B1 | Both hypotheses correct. Must mention Poisson at least once |
| Combined observed and expected frequencies table with \(\leq1\), 2, 3, 4, 5, \(\geq6\) rows | M1 | Combining 0 emails and 1 email. Must have both observed and expected frequencies |
| \(\frac{(O-E)^2}{E}\) values: 1.5267, 1.4381, 0.4767, 2.2971, 0.1083, 2.0544; Total = 7.901 | M1 A1 | At least 2 correct expressions/values (to awrt 2dp); awrt 7.90. Accept 7.9 if no incorrect working |
| \(X^2 = \sum\frac{(O-E)^2}{E} = 7.901...\) awrt 7.90 | — | — |
| \(\nu = 6-1-1 = 4\) | B1 | \(\nu=4\), can be implied by correct critical value of 7.779 |
| \(c^2_4(0.10) = 7.779 \Rightarrow\) CR: \(X^2 \geq 7.779\) | B1 | 7.779 |
| Since \(X^2 = 7.90\) lies in CR: Sufficient evidence to say that Poisson is not a reasonable model | A1 | A correct conclusion based on their \(X^2\) value and their \(\chi^2\) critical value |
## Question 6:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{0\times1+1\times10+2\times23+3\times15+4\times19+5\times9+6\times3}{80} = 3$ | B1 | For a correct method to show that the mean is 3 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $r = e^{-3} \times 80 = 3.983$; $s = \frac{e^{-3}\times 3^5}{5!}\times 80 = 8.066$ | M1 A1 | Use of $\frac{e^{-\lambda}\times\lambda^r}{r!}\times 80$; $r=3.983$ and $s=8.066$ |
| $t = 80-(r+11.949+17.923+17.923+13.443+s) = 6.713$ | M1 A1 | A correct method to ensure expected totals = 80; $t=6.713$ |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0$: Poisson distribution is a reasonable/suitable/sensible model; $H_1$: Poisson distribution is not a reasonable/suitable/sensible model | B1 | Both hypotheses correct. Must mention Poisson at least once |
| Combined observed and expected frequencies table with $\leq1$, 2, 3, 4, 5, $\geq6$ rows | M1 | Combining 0 emails and 1 email. Must have both observed and expected frequencies |
| $\frac{(O-E)^2}{E}$ values: 1.5267, 1.4381, 0.4767, 2.2971, 0.1083, 2.0544; Total = 7.901 | M1 A1 | At least 2 correct expressions/values (to awrt 2dp); awrt 7.90. Accept 7.9 if no incorrect working |
| $X^2 = \sum\frac{(O-E)^2}{E} = 7.901...$ awrt 7.90 | — | — |
| $\nu = 6-1-1 = 4$ | B1 | $\nu=4$, can be implied by correct critical value of 7.779 |
| $c^2_4(0.10) = 7.779 \Rightarrow$ CR: $X^2 \geq 7.779$ | B1 | 7.779 |
| Since $X^2 = 7.90$ lies in CR: Sufficient evidence to say that Poisson is not a reasonable model | A1 | A correct conclusion based on their $X^2$ value and their $\chi^2$ critical value |\begin{enumerate}
\item The number of emails per hour received by a helpdesk were recorded. The results for a random sample of 80 one-hour periods are shown in the table.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | }
\hline
Number of emails per hour & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
Frequencies & 1 & 10 & 23 & 15 & 19 & 9 & 3 \\
\hline
\end{tabular}
\end{center}
(a) Show that the mean number of emails per hour in the sample is 3
The manager believes that the number of emails per hour received could be modelled by a Poisson distribution.
The following table shows some of the expected frequencies.
\begin{center}
\begin{tabular}{|l|l|}
\hline
Number of emails per hour & Expected Frequencies \\
\hline
0 & $r$ \\
\hline
1 & 11.949 \\
\hline
2 & 17.923 \\
\hline
3 & 17.923 \\
\hline
4 & 13.443 \\
\hline
5 & $s$ \\
\hline
$\geqslant 6$ & $t$ \\
\hline
\end{tabular}
\end{center}
(b) Find the values of $r , s$ and $t$, giving your answers to 3 decimal places.\\
(c) Using a 10\% significance level, test whether or not a Poisson model is reasonable. You should clearly state your hypotheses, test statistic and the critical value used.
\hfill \mbox{\textit{Edexcel S3 2022 Q6 [12]}}