| Exam Board | AQA |
|---|---|
| Module | S2 (Statistics 2) |
| Year | 2016 |
| Session | June |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | One-sample z-test known variance |
| Difficulty | Standard +0.3 This is a standard S2 hypothesis testing question covering z-tests, t-tests, and confidence intervals with straightforward calculations. Part (a) is routine one-tailed z-test, part (b) is standard t-distribution confidence interval, and part (c) requires simple comparison. All techniques are core S2 material with no novel problem-solving required, making it slightly easier than average A-level difficulty. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Island | \(\sum x\) | \(n\) |
| A | 1384.5 | 78 |
| B | 116.9 | 7 |
| C | 394.6 | 20 |
| Answer | Marks | Guidance |
|---|---|---|
| Part | Answer/Working | Marks |
| (a) | \(H_0: \mu = 18.2\) (or \(\mu \geq 18.2\)) | B1 |
| \(H_1: \mu < 18.2\) | ||
| \(\bar{x} = 1384.5 \div 78 = 17.75\) | B1 | CAO |
| test stat = \(\frac{'17.75' - 18.2}{(1.8 \div \sqrt{78})}\) | M1 | Condone 18.2 – '17.75' for M1 |
| \(= -2.208\) | A1 | AWRT –2.21. Must be negative |
| \(z_{crit} = \pm 2.0537\) | B1 | AWRT ±2.05 |
| \(-2.208 < -2.0537\) or \(2.208 > 2.0537\) in critical region, reject \(H_0\) or accept \(H_1\) | A1 dep | Comparison stated or diagram; Dep on previous A1 and B1 |
| There is significant evidence at the 2% level of significance to support Gerald's belief. | E1 dep | Context conclusion. Dep on A1 dep; Definitive conclusions (Eg Gerald is correct) score E0. If not referring to belief, must use "mean" or "average" |
| Total: 7 | ||
| Alternative Calculation of critical region boundary value or using confidence interval | ||
| \(H_0\) and \(H_1\) as above | (B1) (B1) (B1) | CAO |
| \(\bar{x} = 1384.5 \div 78 = 17.75\) | (B1) | AWRT ±2.05; Must be subtracting/adding as appropriate |
| \(z_{crit} = \pm 2.0537\) | (M1) | |
| \(18.2 - 2.0537 \times \frac{1.8}{\sqrt{78}}\) \[\left\vert 17.75 - 2.0537 \times \frac{1.8}{\sqrt{78}}\right\vert\] = 17.78; 17.75 < 17.78 so reject \(H_0\) \[\left\vert -18.17\right.\] so 18.2 > 18.17 so reject \(H_0\) | (A1) | AWRT |
| Comparison stated or diagram. Dep on previous A1; Dep on A1 dep | (A1 dep) | |
| Context conclusion as above | (E1 dep) | Dep on A1 dep; Dep on A1 dep |
| (b)(i) | \(\bar{x} = 16.7, s = 1.94(3)\) | B1 |
| \(t_5 = 2.447\) | B1 | AWFW 2.44 to 2.45 |
| \(16.7 \pm 2.447 \times '1.94'/ \sqrt{7}\) | M1 | Use of √7 & rest of formula correct |
| \(= 14.9, 18.5\) or \(16.7 \pm 1.8\) | m1 A1 | Their 1.94 & rest of formula correct; AWRT 14.9, 18.5; Correct answer seen, no working scores all 5 marks. |
| Total: 5 | ||
| (ii) | 18.2 (or the mean for mainland lizards) lies within this confidence interval so no evidence to support Gerald's belief. | A1FT |
| Total: 1 | ||
| (c) | \(\bar{x} = 19.73\) which is > 18.2 so cannot provide evidence to support Gerald's belief. | B1 E1 dep |
| Total: 2 | ||
| TOTAL FOR Q6: 16 |
| Part | Answer/Working | Marks | Guidance |
|------|---|---|---|
| (a) | $H_0: \mu = 18.2$ (or $\mu \geq 18.2$) | B1 | For both, $\mu$ or "population mean" |
| | $H_1: \mu < 18.2$ | | |
| | $\bar{x} = 1384.5 \div 78 = 17.75$ | B1 | CAO |
| | test stat = $\frac{'17.75' - 18.2}{(1.8 \div \sqrt{78})}$ | M1 | Condone 18.2 – '17.75' for M1 |
| | $= -2.208$ | A1 | AWRT –2.21. Must be **negative** |
| | $z_{crit} = \pm 2.0537$ | B1 | AWRT ±2.05 |
| | $-2.208 < -2.0537$ or $2.208 > 2.0537$ in critical region, reject $H_0$ or accept $H_1$ | A1 dep | Comparison stated or diagram; Dep on previous A1 and B1 |
| | There is significant evidence at the 2% level of significance to support Gerald's belief. | E1 dep | **Context** conclusion. Dep on A1 dep; Definitive conclusions (Eg Gerald is correct) score E0. If not referring to belief, must use "mean" or "average" |
| | **Total: 7** | | |
| | **Alternative Calculation of critical region boundary value or using confidence interval** | | |
| | $H_0$ and $H_1$ as above | (B1) (B1) (B1) | CAO |
| | $\bar{x} = 1384.5 \div 78 = 17.75$ | (B1) | AWRT ±2.05; Must be subtracting/adding as appropriate |
| | $z_{crit} = \pm 2.0537$ | (M1) | |
| | $18.2 - 2.0537 \times \frac{1.8}{\sqrt{78}}$ $$\left\vert 17.75 - 2.0537 \times \frac{1.8}{\sqrt{78}}\right\vert$$ = 17.78; 17.75 < 17.78 so reject $H_0$ $$\left\vert -18.17\right.$$ so 18.2 > 18.17 so reject $H_0$ | (A1) | AWRT |
| | Comparison stated or diagram. Dep on previous A1; Dep on A1 dep | (A1 dep) | |
| | Context conclusion as above | (E1 dep) | Dep on A1 dep; Dep on A1 dep |
| (b)(i) | $\bar{x} = 16.7, s = 1.94(3)$ | B1 | For both, CAO & AWRT 1.94. PI |
| | $t_5 = 2.447$ | B1 | AWFW 2.44 to 2.45 |
| | $16.7 \pm 2.447 \times '1.94'/ \sqrt{7}$ | M1 | Use of √7 & rest of formula correct |
| | $= 14.9, 18.5$ or $16.7 \pm 1.8$ | m1 A1 | Their 1.94 & rest of formula correct; AWRT 14.9, 18.5; Correct answer seen, no working scores all 5 marks. |
| | **Total: 5** | | |
| (ii) | 18.2 (or the mean for mainland lizards) lies within this confidence interval so no evidence to support Gerald's belief. | A1FT | FT provided both M marks earned and CI includes 18.2; Must be a clear statement of this. Dep on A1 FT |
| | **Total: 1** | | |
| (c) | $\bar{x} = 19.73$ which is > 18.2 so cannot provide evidence to support Gerald's belief. | B1 E1 dep | Calculation and this comparison; Accept 19.73 > mean on mainland; Comment in context. Dep on B1 |
| | **Total: 2** | | |
| | **TOTAL FOR Q6: 16** | | |
---Gerald is a scientist who studies sand lizards. He believes that sand lizards on islands are, on average, shorter than those on the mainland. The population of sand lizards on the mainland has a mean length of 18.2 cm and a standard deviation of 1.8 cm.
Gerald visited three islands, A, B and C, and measured the length, $X$ centimetres, of each of a sample of $n$ sand lizards on each island. The samples may be regarded as random. The data are shown in the table.
\begin{tabular}{|c|c|c|}
\hline
Island & $\sum x$ & $n$ \\
\hline
A & 1384.5 & 78 \\
\hline
B & 116.9 & 7 \\
\hline
C & 394.6 & 20 \\
\hline
\end{tabular}
\begin{enumerate}[label=(\alph*)]
\item Carry out a hypothesis test to investigate whether the data from Island A provide support for Gerald's belief at the 2% significance level. Assume that the standard deviation of the lengths of sand lizards on Island A is 1.8 cm. [7 marks]
\item For Island B, it is also given that
$$\sum(x - \bar{x})^2 = 22.64$$
\begin{enumerate}[label=(\roman*)]
\item Construct a 95% confidence interval for $\mu_B$, where $\mu_B$ centimetres is the mean length of sand lizards on Island B. Assume that the lengths of sand lizards on Island B are normally distributed with unknown standard deviation.
\item Comment on whether your confidence interval provides support for Gerald's belief.
\end{enumerate}
[7 marks]
\item Comment on whether the data from Island C provide support for Gerald's belief. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA S2 2016 Q6 [16]}}