| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Z-tests (known variance) |
| Type | One-tail z-test (lower tail) |
| Difficulty | Standard +0.3 This is a straightforward application of standard hypothesis testing procedures with clear setup: one-tail z-test with given values, followed by a routine confidence interval calculation. Part (a) requires simple reasoning about sampling. While it involves multiple parts and careful attention to tail direction and significance levels, all techniques are standard S2 content with no novel problem-solving required. |
| Spec | 5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Batteries unusable after testing OR population too big or too costly or too time consuming to use the whole population | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu = 150\), \(H_1: \mu < 150\) | B1 | Or population mean = 150; not just 'mean' = 150 |
| \(\dfrac{147 - 150}{\sqrt{225 \div \sqrt{120}}}\) | M1 | Allow with continuity correction; need \(\sqrt{120}\) |
| \(-2.191\) | A1 | Condone \(-2.19\) |
| \(-2.191 < -2.054\) [or \(-2.055\)] | M1 | OE. For valid comparison with 2.054 or 2.055; or \(0.0143\) (or \(0.0142\)) \(< 0.02\); for two tail test allow comp \(-2.326\) OE if \(H_1: \mu \neq 150\) (can score B0M1A1M1A0 max 3/5) |
| [Reject \(H_0\)] There is evidence that the (mean) life of type \(B\) is less than type \(A\) (or less than 150) | A1 FT | In context, not definite with no contradictions; accept critical value method \(147.19\) M1A1; \(147 < 147.19\) M1 conclusion A1; or \(150 > 149.81\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(147 \pm z \times \dfrac{15}{\sqrt{120}}\) | M1 | Expression of correct form must be a \(z\) value |
| \(z = 1.881\) [or 1.882] | B1 | |
| 144 to 150 (3 s.f.) | A1 | Must be an interval; incorrect \(z\) value can only score M1B0A0 |
## Question 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Batteries unusable after testing OR population too big or too costly or too time consuming to use the whole population | B1 | |
---
## Question 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu = 150$, $H_1: \mu < 150$ | B1 | Or population mean = 150; not just 'mean' = 150 |
| $\dfrac{147 - 150}{\sqrt{225 \div \sqrt{120}}}$ | M1 | Allow with continuity correction; need $\sqrt{120}$ |
| $-2.191$ | A1 | Condone $-2.19$ |
| $-2.191 < -2.054$ [or $-2.055$] | M1 | OE. For valid comparison with 2.054 or 2.055; or $0.0143$ (or $0.0142$) $< 0.02$; for two tail test allow comp $-2.326$ OE if $H_1: \mu \neq 150$ (can score B0M1A1M1A0 max 3/5) |
| [Reject $H_0$] There is evidence that the (mean) life of type $B$ is less than type $A$ (or less than 150) | A1 FT | In context, not definite with no contradictions; accept critical value method $147.19$ M1A1; $147 < 147.19$ M1 conclusion A1; or $150 > 149.81$ |
---
## Question 3(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $147 \pm z \times \dfrac{15}{\sqrt{120}}$ | M1 | Expression of correct form must be a $z$ value |
| $z = 1.881$ [or 1.882] | B1 | |
| 144 to 150 (3 s.f.) | A1 | Must be an interval; incorrect $z$ value can only score M1B0A0 |
---3 Batteries of type $A$ are known to have a mean life of 150 hours. It is required to test whether a new type of battery, type $B$, has a shorter mean life than type $A$ batteries.
\begin{enumerate}[label=(\alph*)]
\item Give a reason for using a sample rather than the whole population in carrying out this test.\\
A random sample of 120 type $B$ batteries are tested and it is found that their mean life is 147 hours, and an unbiased estimate of the population variance is 225 hours $^ { 2 }$.
\item Test, at the $2 \%$ significance level, whether type $B$ batteries have a shorter mean life than type $A$ batteries.
\item Calculate a $94 \%$ confidence interval for the population mean life of type $B$ batteries.
\end{enumerate}
\hfill \mbox{\textit{CAIE S2 2022 Q3 [9]}}