| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics Major (Further Statistics Major) |
| Year | 2020 |
| Session | November |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared goodness of fit |
| Type | Spreadsheet-based chi-squared test |
| Difficulty | Standard +0.3 This is a standard chi-squared goodness of fit test with binomial distribution, requiring routine calculations (estimating p from data, computing expected frequencies, and performing the test). The spreadsheet provides most values, making it more straightforward than typical chi-squared questions. The only slightly elevated aspect is combining categories, but this is a standard technique taught explicitly in Further Statistics. |
| Spec | 5.02l Poisson conditions: for modelling5.06b Fit prescribed distribution: chi-squared test |
| Number of rotten peaches | 0 | 1 | 2 | 3 | 4 | 5 | 6 | \(\geqslant 7\) |
| Frequency | 39 | 39 | 33 | 19 | 8 | 8 | 4 | 0 |
| - | A | B | C | D | E |
| 1 | Number of rotten peaches | Observed frequency | Binomial probability | Expected frequency | Chi-squared contribution |
| 2 | 0 | 39 | |||
| 3 | 1 | 39 | 1.4229 | ||
| 4 | 2 | 33 | 0.2941 | 44.1167 | 2.8012 |
| 5 | 3 | 19 | 0.1629 | 24.4383 | 1.2102 |
| 6 | \(\geqslant 4\) | 20 | 0.0769 | 11.5311 | 6.2199 |
| 7 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (a) | p = 1.72/10 |
| = 0.172 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.3 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (b) | Cell C2 = 0.1515 |
| Answer | Marks |
|---|---|
| = 11.6668 | B1 |
| Answer | Marks |
|---|---|
| [4] | 3.4 |
| Answer | Marks |
|---|---|
| 1.1 | Allow 22.725 from 0.1515 |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (c) | Because otherwise at least one expected frequency |
| Answer | Marks | Guidance |
|---|---|---|
| valid. | E1 | |
| [1] | 3.5b | For ‘less than 5 so invalid’ |
| 9 | (d) | H : the binomial model is appropriate |
| Answer | Marks |
|---|---|
| binomial model is a good fit. | B1 |
| Answer | Marks |
|---|---|
| [6] | 2.4 |
| Answer | Marks |
|---|---|
| 2.2b | For both. Ignore any reference to |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (e) | The contribution of 11.67 suggests that far more trays |
| Answer | Marks |
|---|---|
| appropriate. | E1 |
| Answer | Marks |
|---|---|
| [3] | 3.5a |
| Answer | Marks |
|---|---|
| 3.5a | Allow answers suggesting less than |
Question 9:
9 | (a) | p = 1.72/10
= 0.172 | M1
A1
[2] | 3.3
1.1
9 | (b) | Cell C2 = 0.1515
Cell D2 = 22.7193
(22.7193−39)2
Cell E2 =
22.7193
= 11.6668 | B1
B1
M1
A1
[4] | 3.4
1.1
1.1a
1.1 | Allow 22.725 from 0.1515
For correct form
Allow 11.6557 from 22.725
9 | (c) | Because otherwise at least one expected frequency
would be less than 5 so too small for the test to be
valid. | E1
[1] | 3.5b | For ‘less than 5 so invalid’
9 | (d) | H : the binomial model is appropriate
0
H : the binomial model is not appropriate
1
X 2 = 23.32
Refer to χ 2
3
Critical value at 1% level = 11.34
23.32 > 11.34 Result is significant
There is insufficient evidence to suggest that the
binomial model is a good fit. | B1
B1
B1
B1
M1
A1
[6] | 2.4
1.1
3.4
1.1
1.1
2.2b | For both. Ignore any reference to
value of binomial parameter.
For degrees of freedom = 3 soi
For comparison with critical value
Conclusion in context
9 | (e) | The contribution of 11.67 suggests that far more trays
have no rotten peaches than would be expected if a
binomial model were appropriate.
The contribution of 6.22 suggests that far more trays
have four or more rotten peaches than would be
expected if a binomial model were appropriate.
The other three smaller contributions suggest that the
numbers of trays with 1, 2 or 3 rotten peaches are very
roughly as expected if a binomial model were
appropriate. | E1
E1
E1
[3] | 3.5a
3.5a
3.5a | Allow answers suggesting less than
expected for 2 trays (or similar valid
answers).9 A supermarket sells trays of peaches. Each tray contains 10 peaches. Often some of the peaches in a tray are rotten. The numbers of rotten peaches in a random sample of 150 trays are shown in Table 9.1.
\begin{table}[h]
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Number of rotten peaches & 0 & 1 & 2 & 3 & 4 & 5 & 6 & $\geqslant 7$ \\
\hline
Frequency & 39 & 39 & 33 & 19 & 8 & 8 & 4 & 0 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Table 9.1}
\end{center}
\end{table}
A manager at the supermarket thinks that the number of rotten peaches in a tray may be modelled by a binomial distribution.
\begin{enumerate}[label=(\alph*)]
\item Use these data to estimate the value of the parameter $p$ for the binomial model $\mathrm { B } ( 10 , p )$.
The manager decides to carry out a goodness of fit test to investigate further. The screenshot in Fig. 9.2 shows part of a spreadsheet to assess the goodness of fit of the distribution $\mathrm { B } ( 10 , p )$, using the value of $p$ estimated from the data.
\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
- & A & B & C & D & E \\
\hline
1 & Number of rotten peaches & Observed frequency & Binomial probability & Expected frequency & Chi-squared contribution \\
\hline
2 & 0 & 39 & & & \\
\hline
3 & 1 & 39 & & & 1.4229 \\
\hline
4 & 2 & 33 & 0.2941 & 44.1167 & 2.8012 \\
\hline
5 & 3 & 19 & 0.1629 & 24.4383 & 1.2102 \\
\hline
6 & $\geqslant 4$ & 20 & 0.0769 & 11.5311 & 6.2199 \\
\hline
7 & & & & & \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 9.2}
\end{center}
\end{table}
\item Calculate the missing values in each of the following cells.
\begin{itemize}
\item C2
\item D2
\item E2
\item Explain why the numbers for 4, 5, 6 and at least 7 rotten peaches have been combined into the single category of at least 4 rotten peaches, as shown in the spreadsheet.
\item Carry out the test at the $1 \%$ significance level.
\item Using the values of the contributions, comment on the results of the test.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics Major 2020 Q9 [16]}}