| Exam Board | OCR |
|---|---|
| Module | Further Statistics AS (Further Statistics AS) |
| Year | 2024 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Chi-squared test of independence |
| Type | Expected frequencies partially provided |
| Difficulty | Standard +0.3 This is a standard chi-squared test of independence with straightforward application of formulas. Part (a) requires stating standard hypotheses, (b) involves calculating one expected frequency and contribution (routine), (c) requires summing contributions and comparing to critical value, and (d) asks for the largest contribution. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 5.06a Chi-squared: contingency tables |
| Grade | Walk or cycle | Private motorised transport | Public transport |
| A | 9 | 13 | 6 |
| B | 16 | 43 | 41 |
| C | 11 | 8 | 13 |
| Grade | Walk or cycle | Private motorised transport | Public transport |
| A | 1.157 | 0.289 | 1.929 |
| B | 1.878 | 0.225 | 0.327 |
| C | 2.006 | 1.800 | 0.083 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0\): there is no association between grade and method of transport; \(H_1\): there is association | B1 [1] | Needs context. Allow "relationship" but *not* "correlation"; allow "grade and method are independent/dependent"; *not* "insufficient evidence that there is association…". Can be recovered from later part |
| Answer | Marks | Guidance |
|---|---|---|
| Expected frequency is \(64 \times 100/160\ (= 40)\) | M1 | Clear method for expected frequency, allow without working only if at least two other expected frequencies shown correctly |
| \((43 - 40)^2/40\) | M1 | Clear indication of \((O-E)^2/E\) used |
| \(= 0.225\) AG | A1 [3] | Correctly obtain AG 0.225, needs both previous marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Sigma x^2 = 9.69\) | B1 | Awrt 9.69 |
| \(> 9.488\) | B1 | Explicit comparison with 9.488 |
| Reject \(H_0\). There is significant evidence of association between grade and method of transport | B1ft [3] | Contextualised, not over-assertive. FT on their TS but *not* on wrong 9.488; *not* FT from reversed hypotheses. "Reject \(H_0\)" can be equivalent to "Accept \(H_1\)", or omitted. Allow "evidence that grade and method are dependent", but *not* "significant evidence that grade is dependent upon method of transport". *Not* "insufficient evidence of no association", *not* "sufficient evidence that there may be association" |
| Answer | Marks | Guidance |
|---|---|---|
| Grade C/walk-or-cycle, as biggest contribution to TS | B1 [1] | Correct combination identified, and correct justification needed, but don't need to state 2.006. *Not* with wrong reason, e.g. "as this is closest to critical value" or "biggest difference between O and E" |
# Question 2:
## Part (a)
$H_0$: there is no association between grade and method of transport; $H_1$: there is association | B1 [1] | Needs context. Allow "relationship" but *not* "correlation"; allow "grade and method are independent/dependent"; *not* "insufficient evidence that there is association…". Can be recovered from later part
## Part (b)
Expected frequency is $64 \times 100/160\ (= 40)$ | M1 | Clear method for expected frequency, allow without working only if at least two other expected frequencies shown correctly
$(43 - 40)^2/40$ | M1 | Clear indication of $(O-E)^2/E$ used
$= 0.225$ **AG** | A1 [3] | Correctly obtain AG 0.225, needs both previous marks
## Part (c)
$\Sigma x^2 = 9.69$ | B1 | Awrt 9.69
$> 9.488$ | B1 | Explicit comparison with 9.488
Reject $H_0$. There is significant evidence of association between grade and method of transport | B1ft [3] | Contextualised, not over-assertive. FT on their TS but *not* on wrong 9.488; *not* FT from reversed hypotheses. "Reject $H_0$" can be equivalent to "Accept $H_1$", or omitted. Allow "evidence that grade and method are dependent", but *not* "significant evidence that grade is dependent upon method of transport". *Not* "insufficient evidence of no association", *not* "sufficient evidence that there may be association"
## Part (d)
Grade C/walk-or-cycle, as biggest contribution to TS | B1 [1] | Correct combination identified, and correct justification needed, but don't need to state 2.006. *Not* with wrong reason, e.g. "as this is closest to critical value" or "biggest difference between O and E"
---2 For a random sample of 160 employees of a large company, the principal method of transport for getting to work, arranged according to grade of employee, is shown in the table.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
Grade & Walk or cycle & Private motorised transport & Public transport \\
\hline
A & 9 & 13 & 6 \\
\hline
B & 16 & 43 & 41 \\
\hline
C & 11 & 8 & 13 \\
\hline
\end{tabular}
\end{center}
A test is carried out at the $5 \%$ significance level of whether there is association between grade of employee and method of transport.
\begin{enumerate}[label=(\alph*)]
\item State appropriate hypotheses for the test.
The contributions to the test statistic are shown in the following table, correct to 3 decimal places.
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
Grade & Walk or cycle & Private motorised transport & Public transport \\
\hline
A & 1.157 & 0.289 & 1.929 \\
\hline
B & 1.878 & 0.225 & 0.327 \\
\hline
C & 2.006 & 1.800 & 0.083 \\
\hline
\end{tabular}
\end{center}
\item Show how the value 0.225 is obtained.
\item Complete the test, stating the conclusion.
\item Which combination of grade of employee and method of transport most strongly suggests association? Justify your answer.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics AS 2024 Q2 [8]}}