CAIE Further Paper 4 2022 November — Question 6 10 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2022
SessionNovember
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNon-parametric tests
TypeWilcoxon rank-sum test
DifficultyStandard +0.3 This is a straightforward application of the Wilcoxon rank-sum test with clear data and standard procedure. Students must rank combined data, sum ranks, compare to critical values, and explain why a paired t-test is inappropriate. While it requires careful calculation, it's a routine textbook exercise with no novel insight needed, making it slightly easier than average for Further Maths statistics.
Spec5.07d Paired vs two-sample: selection5.07e Test medians

6 The manager of a technology company \(A\) claims that his employees earn more per year than the employees at technology company \(B\). The amounts earned per year, in hundreds of dollars, by a random sample of 12 employees from company \(A\) and an independent random sample of 12 employees from company \(B\) are shown below.
Company \(A\)461482374512415452502427398545612359
Company \(B\)454506491384361443401472414342355437
  1. Carry out a Wilcoxon rank-sum test at the \(5 \%\) significance level to test whether the manager's claim is supported by the data.
  2. Explain whether a paired sample \(t\)-test would be appropriate to test the manager's claim if earnings are normally distributed.
    If you use the following page to complete the answer to any question, the question number must be clearly shown.

Question 6(a):
AnswerMarks Guidance
AnswerMark Guidance
Rankings table with values: 359→3, 374→5, 398→7, 415→10, 427→11, 452→14, 461→16, 482→18, 502→20, 512→22, 545→23, 612→24 and 342→1, 355→2, 361→4, 384→6, 401→8, 414→9, 437→12, 443→13, 454→15, 472→17, 491→19, 506→21M1 Attempt at rankings (allow up to 4 errors). Wrong test: max B1 for hypotheses, B1B1 for correct mean and variance
Test statistic \(= 127\)A1 Clearly identified
\(H_0\): population medians are equal; \(H_1\): population median for \(X\) is greater than population median for \(Y\)B1
Mean \(= \frac{1}{2} \times 12 \times 25 = 150\)B1
Variance \(= \frac{1}{12} \times 12 \times 12 \times 25 = 300\)B1
\(\frac{127.5 - 150}{\sqrt{300}}\)M1 Allow incorrect or no continuity correction
\(-1.299\)A1
Compare with \(-1.645\): \(-1.299 > -1.645\), or \(0.097 > 0.05\), Accept \(H_0\)M1 Valid comparison with 1.645 or 0.05 and reach correct conclusion
Insufficient evidence to support manager's claimA1 Correct conclusion in context, following correct work, level of uncertainty in language. 'Prove' is A0
Total9
Question 6(b):
AnswerMarks Guidance
AnswerMark Guidance
Not appropriate/no, not the same peopleB1 OE. No reason needed, e.g. individuals in the samples cannot be paired up
Total1
## Question 6(a):

| Answer | Mark | Guidance |
|--------|------|----------|
| Rankings table with values: 359→3, 374→5, 398→7, 415→10, 427→11, 452→14, 461→16, 482→18, 502→20, 512→22, 545→23, 612→24 and 342→1, 355→2, 361→4, 384→6, 401→8, 414→9, 437→12, 443→13, 454→15, 472→17, 491→19, 506→21 | M1 | Attempt at rankings (allow up to 4 errors). Wrong test: max B1 for hypotheses, B1B1 for correct mean and variance |
| Test statistic $= 127$ | A1 | Clearly identified |
| $H_0$: population medians are equal; $H_1$: population median for $X$ is greater than population median for $Y$ | B1 | |
| Mean $= \frac{1}{2} \times 12 \times 25 = 150$ | B1 | |
| Variance $= \frac{1}{12} \times 12 \times 12 \times 25 = 300$ | B1 | |
| $\frac{127.5 - 150}{\sqrt{300}}$ | M1 | Allow incorrect or no continuity correction |
| $-1.299$ | A1 | |
| Compare with $-1.645$: $-1.299 > -1.645$, or $0.097 > 0.05$, Accept $H_0$ | M1 | Valid comparison with 1.645 or 0.05 and reach correct conclusion |
| Insufficient evidence to support manager's claim | A1 | Correct conclusion in context, following correct work, level of uncertainty in language. 'Prove' is A0 |
| **Total** | **9** | |

---

## Question 6(b):

| Answer | Mark | Guidance |
|--------|------|----------|
| Not appropriate/no, not the same people | B1 | OE. No reason needed, e.g. individuals in the samples cannot be paired up |
| **Total** | **1** | |
6 The manager of a technology company $A$ claims that his employees earn more per year than the employees at technology company $B$. The amounts earned per year, in hundreds of dollars, by a random sample of 12 employees from company $A$ and an independent random sample of 12 employees from company $B$ are shown below.

\begin{center}
\begin{tabular}{ l l l l l l l l l l l l l }
Company $A$ & 461 & 482 & 374 & 512 & 415 & 452 & 502 & 427 & 398 & 545 & 612 & 359 \\
Company $B$ & 454 & 506 & 491 & 384 & 361 & 443 & 401 & 472 & 414 & 342 & 355 & 437 \\
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Carry out a Wilcoxon rank-sum test at the $5 \%$ significance level to test whether the manager's claim is supported by the data.
\item Explain whether a paired sample $t$-test would be appropriate to test the manager's claim if earnings are normally distributed.\\

If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}

\hfill \mbox{\textit{CAIE Further Paper 4 2022 Q6 [10]}}