| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2024 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | Two-tail z-test |
| Difficulty | Moderate -0.3 This is a straightforward application of a z-test with known variance following a standard template: identify test type, calculate sample mean, compute z-statistic, compare to critical value, and conclude. All steps are routine with no conceptual challenges beyond basic hypothesis testing procedure. The only minor complexity is the two-tailed test requiring ±2.576 critical values, but this is standard S2 content. |
| Spec | 2.05e Hypothesis test for normal mean: known variance5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a) | Two-tailed because looking for difference | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 5(b) | H : μ = 12.7 H : μ ≠ 12.7 | |
| 0 1 | B1 | No ft from part (a). |
| Answer | Marks |
|---|---|
| 50 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept – 2.336 or – 2.337 | A1 | or 0.00989 or 0.0099. |
| Answer | Marks | Guidance |
|---|---|---|
| or `0.0097` > 0.005 | M1 | Accept 2.574 to 2.579. |
| Answer | Marks | Guidance |
|---|---|---|
| not 12.7 | A1 FT | OE ft their z |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(a) ---
5(a) | Two-tailed because looking for difference | B1
1
Question | Answer | Marks | Guidance
--- 5(b) ---
5(b) | H : μ = 12.7 H : μ ≠ 12.7
0 1 | B1 | No ft from part (a).
597.1−12.7
50
2.3
50 | M1
= −2.330
Accept – 2.336 or – 2.337 | A1 | or 0.00989 or 0.0099.
or 0.0097 if area comparison used.
‘−2.330’ > −2.576 or ‘2.330’ < 2.576
or ‘0.00989’ > 0.005
or `0.0097` > 0.005 | M1 | Accept 2.574 to 2.579.
Or use of CV.
12.7- 2.576 x ( 2.3 / sqrt 50 ) = 11.862 M1A1.
11.942 > 11.862 M1A1.
[Not reject H ] There is insufficient evidence to suggest that µ is
0
not 12.7 | A1 FT | OE ft their z
calc.
In context, not definite, e.g. not ‘µ =12.7’.
No contradictions.
SC use of 1 tailed test can score B0M1A1M1 for comparison with
0.01 A0 max 3/5.
5
Question | Answer | Marks | GuidanceThe lengths, in centimetres, of worms of a certain kind are normally distributed with mean $\mu$ and standard deviation $2.3$. An article in a magazine states that the value of $\mu$ is $12.7$. A scientist wishes to test whether this value is correct. He measures the lengths, $x$ cm, of a random sample of $50$ worms of this kind and finds that $\sum x = 597.1$. He plans to carry out a test, at the $1\%$ significance level, of whether the true value of $\mu$ is different from $12.7$.
\begin{enumerate}[label=(\alph*)]
\item State, with a reason, whether he should use a one-tailed or a two-tailed test. [1]
\item Carry out the test. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2024 Q5 [6]}}