| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics A AS (Further Statistics A AS) |
| Year | 2023 |
| Session | June |
| Marks | 15 |
| 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 straightforward chi-squared goodness of fit test with routine calculations. Part (a) involves basic probability and standard deviation checks for a discrete uniform distribution. Part (b) requires filling in missing spreadsheet values using the standard formula (O-E)²/E and conducting a standard hypothesis test. All steps are mechanical applications of learned procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 5.02e Discrete uniform distribution5.06b Fit prescribed distribution: chi-squared test |
| \multirow[b]{2}{*}{1} | A | B | C | D |
| Score | Observed frequency | Expected frequency | Chi-squared contribution | |
| 2 | 1 | 14 | 10 | 1.6 |
| 3 | 2 | 4 | 10 | 3.6 |
| 4 | 3 | 10 | 10 | 0 |
| 5 | 4 | 15 | 10 | |
| 6 | 5 | 6 | 10 | 1.6 |
| 7 | 6 | 11 | 10 | 0.1 |
| 8 | 7 | 7 | 10 | 0.9 |
| 9 | 8 | 10 | 0.9 |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (a) | (i) |
| Answer | Marks | Guidance |
|---|---|---|
| 8 | B1 | |
| [1] | 1.1 | |
| 6 | (a) | (ii) |
| Answer | Marks |
|---|---|
| values. | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.1a |
| Answer | Marks |
|---|---|
| 1.1 | BC |
| Answer | Marks | Guidance |
|---|---|---|
| 6 | (b) | (i) |
| Answer | Marks | Guidance |
|---|---|---|
| 𝟖 | B1 | |
| [1] | 2.2a | |
| 6 | (b) | (ii) |
| Answer | Marks |
|---|---|
| = 2.5 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | SCB1 for 2.5 if no method shown |
Question 6:
6 | (a) | (i) | 3
= 0.375
8 | B1
[1] | 1.1
6 | (a) | (ii) | E(X) = 4.5
Var(X) = 1 (82−1) (⇒ SD(X) = 2.29..)
12
Need P(−0.1 < X < 9.1)
= 𝟏, because this interval includes all the possible
values. | B1
B1
M1
A1
[4] | 1.1a
1.1
3.1b
1.1 | BC
BC
M1 for “4.5” + 2× “2.29” or “4.5” − 2× “2.29”
AG For A1 needs to be fully correct. The given answer
must be stated.
6 | (b) | (i) | 𝟏
Because P(score = n) × 80 = ×𝟖𝟎 = 10
𝟖 | B1
[1] | 2.2a
6 | (b) | (ii) | B9 = 13
(15−10)2
D5 contribution =
10
= 2.5 | B1
M1
A1
[3] | 1.1
3.4
1.1 | SCB1 for 2.5 if no method shown
[6]
PMT
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6 An eight-sided dice has its faces numbered $1,2 , \ldots , 8$.
\begin{enumerate}[label=(\alph*)]
\item In this part of the question you should assume that the dice is fair.
\begin{enumerate}[label=(\roman*)]
\item State the probability that, when the dice is rolled once, the score is at least 6 .
\item Show that the probability that the score is within 2 standard deviations of its mean is 1 .
\end{enumerate}\item A student thinks that the dice may be biased. To investigate this, the student decides to roll the dice 80 times and then carry out a $\chi ^ { 2 }$ goodness of fit test of a uniform distribution. The spreadsheet below shows the data for the test, where some of the values have been deliberately omitted.
\begin{center}
\begin{tabular}{|l|l|l|l|l|}
\hline
\multirow[b]{2}{*}{1} & A & B & C & D \\
\hline
& Score & Observed frequency & Expected frequency & Chi-squared contribution \\
\hline
2 & 1 & 14 & 10 & 1.6 \\
\hline
3 & 2 & 4 & 10 & 3.6 \\
\hline
4 & 3 & 10 & 10 & 0 \\
\hline
5 & 4 & 15 & 10 & \\
\hline
6 & 5 & 6 & 10 & 1.6 \\
\hline
7 & 6 & 11 & 10 & 0.1 \\
\hline
8 & 7 & 7 & 10 & 0.9 \\
\hline
9 & 8 & & 10 & 0.9 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\roman*)]
\item Explain why all of the expected frequencies are equal to 10 .
\item Determine the missing values in each of the following cells.
\begin{itemize}
\end{enumerate}\item B9
\item D5\\
(iii) In this question you must show detailed reasoning.
\end{itemize}
Carry out the $\chi ^ { 2 }$ test at the $5 \%$ significance level.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Further Statistics A AS 2023 Q6 [15]}}