| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2021 |
| Session | November |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Assess model suitability before testing |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test for a Poisson distribution with given parameter. Part (a) requires calculating mean and variance (routine), parts (b-c) involve filling in spreadsheet cells using given formulas and explaining the combining of categories rule (standard procedure), and part (d) is a straightforward hypothesis test with 4 degrees of freedom. All steps are textbook procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02l Poisson conditions: for modelling5.06b Fit prescribed distribution: chi-squared test |
| Number of muons | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| Frequency | 34 | 65 | 55 | 24 | 14 | 6 | 2 | 0 |
| A | B | C | D | E | |
| 1 | Number of muons | Observed frequency | Poisson probability | Expected frequency | Chi-squared contribution |
| 2 | 0 | 34 | 0.1827 | 36.5367 | 0.1761 |
| 3 | 1 | 65 | |||
| 4 | 2 | 55 | 0.2640 | 52.7955 | 0.0920 |
| 5 | 3 | 24 | 0.1496 | 29.9175 | 1.1704 |
| 6 | 4 | 14 | 0.1299 | ||
| 7 | \(\geqslant 5\) | 8 | 0.0296 | 5.9230 | 0.7284 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | Mean = 1.725 |
| Answer | Marks |
|---|---|
| does support the suitability of a Poisson model | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2.2b | Condone 1.759 (using divisor n) | Or |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | Cell C3 = 0.3106 |
| Answer | Marks |
|---|---|
| = 0.1342 | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.4 |
| Answer | Marks | Guidance |
|---|---|---|
| 1.1 | 200 × their C3 (62.12 if use 0.3106) | Do not allow 0.311 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (c) | Because otherwise some expected frequencies would |
| be less than 5 so too small for the test to be valid | E1 | |
| [1] | 3.5b | For ‘less than 5 so invalid’ |
| 6 | (d) | H : Poisson model is a good fit |
| Answer | Marks |
|---|---|
| Po(1.7) model is not a good fit. | B1 |
| Answer | Marks |
|---|---|
| [6] | 2.5 |
| Answer | Marks |
|---|---|
| 2.2b | FT Their value of E3 |
| Answer | Marks |
|---|---|
| Conclusion in context | Allow M1 (not A1) for |
Question 6:
6 | (a) | Mean = 1.725
Variance = 1.768
The variance is reasonably close to the mean so this
does support the suitability of a Poisson model | B1
B1
E1
[3] | 1.1
1.1
2.2b | Condone 1.759 (using divisor n) | Or
345
200
Dep on mean and
variance correct
6 | (b) | Cell C3 = 0.3106
Cell D3 = 62.1124
(65−62.1224)2
Cell E3 =
62.1224
= 0.1342 | B1
B1FT
M1FT
A1
[4] | 3.4
2.2a
1.1a
1.1 | 200 × their C3 (62.12 if use 0.3106) | Do not allow 0.311
Allow 62.2 from 0.311
Must show working to
get M1
Allow 0.126 from 62.2
6 | (c) | Because otherwise some expected frequencies would
be less than 5 so too small for the test to be valid | E1
[1] | 3.5b | For ‘less than 5 so invalid’
6 | (d) | H : Poisson model is a good fit
0
H : Poisson model is not a good fit
1
X 2 = 2.43
Refer to χ2
5
Critical value at 5% level = 11.07
2.43 < 11.07 so result is not significant
There is insufficient evidence to suggest that the
Po(1.7) model is not a good fit. | B1
B1FT
B1
B1
M1
A1
[6] | 2.5
1.1
3.4
1.1
1.1
2.2b | FT Their value of E3
For degrees of freedom = 5 soi
For comparison with critical value
Conclusion in context | Allow M1 (not A1) for
comparison with any
chi squared critical
value eg 1.145 or
5.9916 Cosmic rays passing through the upper atmosphere cause muons, and other types of particle, to be formed. Muons can be detected when they reach the surface of the earth. It is known that the mean number of muons reaching a particular detector is 1.7 per second. The numbers of muons reaching this detector in 200 randomly selected periods of 1 second are shown in Fig. 6.1.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Number of muons & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Frequency & 34 & 65 & 55 & 24 & 14 & 6 & 2 & 0 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Use the values of the sample mean and sample variance to discuss the suitability of a Poisson distribution as a model.
The screenshot in Fig. 6.2 shows part of a spreadsheet to assess the goodness of fit of the distribution Po(1.7).
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
& A & B & C & D & E \\
\hline
1 & Number of muons & Observed frequency & Poisson probability & Expected frequency & Chi-squared contribution \\
\hline
2 & 0 & 34 & 0.1827 & 36.5367 & 0.1761 \\
\hline
3 & 1 & 65 & & & \\
\hline
4 & 2 & 55 & 0.2640 & 52.7955 & 0.0920 \\
\hline
5 & 3 & 24 & 0.1496 & 29.9175 & 1.1704 \\
\hline
6 & 4 & 14 & & & 0.1299 \\
\hline
7 & $\geqslant 5$ & 8 & 0.0296 & 5.9230 & 0.7284 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{table}
\item Calculate the missing values in each of the following cells.
\begin{itemize}
\item C3
\item D3
\item E3
\item Explain why the numbers for 5, 6 and at least 7 muons have been combined into the single category of at least 5 muons, as shown in Fig. 6.2.
\item In this question you must show detailed reasoning.
\end{itemize}
Carry out the test at the 5\% significance level.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2021 Q6 [14]}}