| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Single sample t-test |
| Difficulty | Standard +0.3 This is a straightforward application of standard one-sample t-test and chi-squared confidence interval procedures from S4. Part (a) requires calculating sample mean and standard deviation from summary statistics, then performing a routine one-tailed t-test. Part (b) applies the standard chi-squared formula for variance confidence intervals. Both parts are textbook exercises with no novel insight required, though slightly above average difficulty due to being Further Maths content and requiring careful calculation with summary statistics. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu = 135\), \(H_1: \mu < 135\) | B1 | Both hypotheses |
| \(\bar{x} = 131\), \(s^2 = 10\) | B1 B1 | B1 for 131; B1 for 10 or awrt 3.16 |
| \(t = \dfrac{131 - 135}{\sqrt{10/5}} = -2.828\ldots\) | M1A1 | M1: allow \(\pm \dfrac{\text{"their }131\text{"}-135}{\sqrt{\text{"their }10\text{"}/5}}\); A1 awrt \(-2.83\) or \(-2\sqrt{2}\) |
| Critical value \(t_4(10\%) = -1.533\) | B1 | \(\pm 1.533\); sign must match \(t\)-value or be \(\pm\) |
| Sufficient evidence that the mean length of wing is less than 135 mm | A1ft | ft \(t\)-value if awarded 1st and 4th B marks; words 'mean length' and '135' must be included |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{4 \times 10}{9.488} < \sigma^2 < \dfrac{4 \times 10}{0.711}\) | M1 | M1: \(\dfrac{4 \times \text{"their }10\text{"}}{\chi^2 \text{ value}}\) |
| B1B1 | B1 awrt 9.49; B1 awrt 0.711 | |
| \((4.22,\ 56.3)\) | A1 | awrt 4.22/4.21 and awrt 56.3 |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu = 135$, $H_1: \mu < 135$ | B1 | Both hypotheses |
| $\bar{x} = 131$, $s^2 = 10$ | B1 B1 | B1 for 131; B1 for 10 or awrt 3.16 |
| $t = \dfrac{131 - 135}{\sqrt{10/5}} = -2.828\ldots$ | M1A1 | M1: allow $\pm \dfrac{\text{"their }131\text{"}-135}{\sqrt{\text{"their }10\text{"}/5}}$; A1 awrt $-2.83$ or $-2\sqrt{2}$ |
| Critical value $t_4(10\%) = -1.533$ | B1 | $\pm 1.533$; sign must match $t$-value or be $\pm$ |
| Sufficient evidence that the **mean length** of wing is less than **135** mm | A1ft | ft $t$-value if awarded 1st and 4th B marks; words 'mean length' and '135' must be included |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{4 \times 10}{9.488} < \sigma^2 < \dfrac{4 \times 10}{0.711}$ | M1 | M1: $\dfrac{4 \times \text{"their }10\text{"}}{\chi^2 \text{ value}}$ |
| | B1B1 | B1 awrt 9.49; B1 awrt 0.711 |
| $(4.22,\ 56.3)$ | A1 | awrt 4.22/4.21 and awrt 56.3 |
---3. The lengths, $X \mathrm {~mm}$, of the wings of adult blackbirds follow a normal distribution. A random sample of 5 adult blackbirds is taken and the lengths of the wings are measured. The results are summarised below
$$\sum x = 655 \text { and } \sum x ^ { 2 } = 85845$$
\begin{enumerate}[label=(\alph*)]
\item Test, at the $10 \%$ level of significance, whether or not the mean length of an adult blackbird's wing is less than 135 mm . State your hypotheses clearly.
\item Find the $90 \%$ confidence interval for the variance of the lengths of adult blackbirds' wings. Show your working clearly.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2017 Q3 [11]}}