| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2011 |
| Session | January |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard two-sample t-test with pooled variance, requiring routine calculations of sample statistics, pooled variance, and test statistic. While it involves multiple steps and careful arithmetic, it follows a well-established procedure taught directly in S3 with no novel problem-solving or conceptual insight required—making it slightly easier than average for an A-level statistics question. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When independent samples are drawn from populations having a common variance | B1 1 | For common variance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Lung capacities should have normal distributions with a common variance | B1 1 | Normal distributions required, in context |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H_0: \mu_1=\mu_2,\quad H_1: \mu_1 > \mu_2\) | B1 | Or equivalent |
| \(s_1^2 = \frac{1}{19}(90.43 - 42.4^2/20)\) | M1 | For 1 correct \(s^2\) |
| \(s_2^2 = \frac{1}{21}(82.93 - 42.5^2/22)\) | ||
| \(\bar{x}_1 = 2.12\qquad \bar{x}_2 = 1.93(2)\) | B1 | For both |
| PEV, \(s^2 = (19s_1^2 + 21s_2^2)/(20+22-2)\) | M1 | |
| \(= 0.03424(3)\) | A1 | |
| Test statistic \(= \dfrac{2.12-1.932}{\sqrt{s^2(20^{-1}+22^{-1})}}\) | M1A1 | ft \(s^2\) |
| \(= 3.29(15)\) | A1\(\sqrt{}\) | Accept answer rounding to 3.3 |
| \(CV = 2.423\) | B1 | If \(z\) used: B0M0A0 |
| \(\text{TS} > \text{CV}\) | M1 | Compare with CV |
| Sufficient evidence at 1% SL that mean lung capacity greater for children whose parents do not smoke | A1 11 | Or equivalent, in context |
| SR1: For 2-tail test lose 1st B1 and last 3. Max 8/11 | ||
| SR2: If \(s^2 = s_1^2/20 + s_2^2/22\), B1M1A0A0M1A0A1(3.32) B1M1A1 Max 8/11 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = 2.704\) | B1 | |
| \(0.1882\ldots \pm ts(20^{-1}+22^{-1})^{1/2}\) | M1 | Accept 0.19 |
| \((0.0336,\ 0.3423)\) | A1 3 [16] | \((0.033\text{-}0.036,\ 0.342\text{-}0.346)\) |
## Question 8:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| When independent samples are drawn from populations having a common variance | B1 **1** | For common variance |
### Part (ii)(a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Lung capacities should have normal distributions with a common variance | B1 **1** | Normal distributions required, in context |
### Part (ii)(b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H_0: \mu_1=\mu_2,\quad H_1: \mu_1 > \mu_2$ | B1 | Or equivalent |
| $s_1^2 = \frac{1}{19}(90.43 - 42.4^2/20)$ | M1 | For 1 correct $s^2$ |
| $s_2^2 = \frac{1}{21}(82.93 - 42.5^2/22)$ | | |
| $\bar{x}_1 = 2.12\qquad \bar{x}_2 = 1.93(2)$ | B1 | For both |
| PEV, $s^2 = (19s_1^2 + 21s_2^2)/(20+22-2)$ | M1 | |
| $= 0.03424(3)$ | A1 | |
| Test statistic $= \dfrac{2.12-1.932}{\sqrt{s^2(20^{-1}+22^{-1})}}$ | M1A1 | ft $s^2$ |
| $= 3.29(15)$ | A1$\sqrt{}$ | Accept answer rounding to 3.3 |
| $CV = 2.423$ | B1 | If $z$ used: B0M0A0 |
| $\text{TS} > \text{CV}$ | M1 | Compare with CV |
| Sufficient evidence at 1% SL that mean lung capacity greater for children whose parents do not smoke | A1 **11** | Or equivalent, in context |
| SR1: For 2-tail test lose 1st B1 and last 3. Max 8/11 | | |
| SR2: If $s^2 = s_1^2/20 + s_2^2/22$, B1M1A0A0M1A0A1(3.32) B1M1A1 Max 8/11 | | |
### Part (ii)(c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = 2.704$ | B1 | |
| $0.1882\ldots \pm ts(20^{-1}+22^{-1})^{1/2}$ | M1 | Accept 0.19 |
| $(0.0336,\ 0.3423)$ | A1 **3** [16] | $(0.033\text{-}0.036,\ 0.342\text{-}0.346)$ |8 (i) State circumstances under which it would be necessary to calculate a pooled estimate of variance when carrying out a two-sample hypothesis test.\\
(ii) An investigation into whether passive smoking affects lung capacity considered a random sample of 20 children whose parents did not smoke and a random sample of 22 children whose parents did smoke. None of the children themselves smoked. The lung capacity, in litres, of each child was measured and the results are summarised as follows.
For the children whose parents did not smoke: $n _ { 1 } = 20 , \Sigma x _ { 1 } = 42.4$ and $\Sigma x _ { 1 } ^ { 2 } = 90.43$.\\
For the children whose parents did smoke: $\quad n _ { 2 } = 22 , \Sigma x _ { 2 } = 42.5$ and $\Sigma x _ { 2 } ^ { 2 } = 82.93$.\\
The means of the two populations are denoted by $\mu _ { 1 }$ and $\mu _ { 2 }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item State conditions for which a $t$-test would be appropriate for testing whether $\mu _ { 1 }$ exceeds $\mu _ { 2 }$.
\item Assuming the conditions are valid, carry out the test at the $1 \%$ significance level and comment on the result.
\item Calculate a 99\% confidence interval for $\mu _ { 1 } - \mu _ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2011 Q8 [16]}}