| Exam Board | CAIE |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2015 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Central limit theorem |
| Type | Sampling method explanation |
| Difficulty | Moderate -0.8 This is a straightforward hypothesis testing question with standard bookwork explanations. Part (i) requires a simple practical reason (destructive testing), part (ii) is a routine one-tailed z-test with given parameters requiring no problem-solving, and part (iii) asks for recall of when CLT applies (large sample size). All parts are direct application of standard S2 content with no novel insight required. |
| Spec | 2.01a Population and sample: terminology5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Cables broken / or not all cables can be accessed / or too many cables / or too time consuming | B1 [1] | e.g. previous days' stocks may have gone |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0\): Pop mean brk str \((\mu) = 5\); \(H_1\): Pop mean brk str \((\mu) < 5\) | B1 | Not just "mean" |
| \((\pm)\frac{4.95-5}{\frac{0.15}{\sqrt{60}}}\) | M1 | Allow 60 instead of \(\sqrt{60}\) |
| \((= \pm 2.582)\) | A1 | |
| comp \(\pm 2.326\); There is evidence that mean breaking strength is less than it should be / Or reject \(H_0\) (\(H_0\) correctly defined) | B1ft [4] | Ft their \(-2.582\); (No ft 2 tailed test); Correct comparison shown, no errors seen; Accept area comparison 0.0049 with 0.01. [CR method \((x-5)/(0.15/\sqrt{60}) = -2.326\) M1 A1 leading to \(x=4.955\) compared to 4.95 and correct conclusion B1ft; OR \(((x-4.95)/0.15/\sqrt{60})\) leading to 4.995 M1 A1 compared to 5 and correct conclusion B1ft] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Population not necessarily normal, so yes | B1, B1dep [2] | SR B1 For "it" is not necessarily normal (no mention of population) AND Yes |
## Question 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Cables broken / or not all cables can be accessed / or too many cables / or too time consuming | B1 [1] | e.g. previous days' stocks may have gone |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0$: Pop mean brk str $(\mu) = 5$; $H_1$: Pop mean brk str $(\mu) < 5$ | B1 | Not just "mean" |
| $(\pm)\frac{4.95-5}{\frac{0.15}{\sqrt{60}}}$ | M1 | Allow 60 instead of $\sqrt{60}$ |
| $(= \pm 2.582)$ | A1 | |
| comp $\pm 2.326$; There is evidence that mean breaking strength is less than it should be / Or reject $H_0$ ($H_0$ correctly defined) | B1ft [4] | Ft their $-2.582$; (No ft 2 tailed test); Correct comparison shown, no errors seen; Accept area comparison 0.0049 with 0.01. [CR method $(x-5)/(0.15/\sqrt{60}) = -2.326$ M1 A1 leading to $x=4.955$ compared to 4.95 and correct conclusion B1ft; OR $((x-4.95)/0.15/\sqrt{60})$ leading to 4.995 M1 A1 compared to 5 and correct conclusion B1ft] |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Population not necessarily normal, so yes | B1, B1dep [2] | SR B1 For "it" is not necessarily normal (no mention of population) AND Yes |
---5 The mean breaking strength of cables made at a certain factory is supposed to be 5 tonnes. The quality control department wishes to test whether the mean breaking strength of cables made by a particular machine is actually less than it should be. They take a random sample of 60 cables. For each cable they find the breaking strength by gradually increasing the tension in the cable and noting the tension when the cable breaks.\\
(i) Give a reason why it is necessary to take a sample rather then testing all the cables produced by the machine.\\
(ii) The mean breaking strength of the 60 cables in the sample is found to be 4.95 tonnes. Given that the population standard deviation of breaking strengths is 0.15 tonnes, test at the $1 \%$ significance level whether the population mean breaking strength is less than it should be.\\
(iii) Explain whether it was necessary to use the Central Limit theorem in the solution to part (ii).
\hfill \mbox{\textit{CAIE S2 2015 Q5 [7]}}