| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2024 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Wilcoxon tests |
| Type | Paired t-test |
| Difficulty | Standard +0.3 This is a straightforward paired t-test question with standard parts: identifying why pairing is needed, stating assumptions, calculating a confidence interval (routine formula application with 8 data points), and interpreting it for a hypothesis test. All components are textbook exercises requiring no novel insight, though it's slightly easier than average due to clear structure and small dataset. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Set of twin lambs | A | \(B\) | C | D | \(E\) | \(F\) | \(G\) | \(H\) | |
| \multirow{2}{*}{Weight gain (kg)} | With food supplement | 4.1 | 5.3 | 6.0 | 3.6 | 5.9 | 4.2 | 7.1 | 6.4 |
| No food supplement | 5.0 | 4.8 | 5.2 | 3.4 | 5.1 | 3.9 | 7.0 | 6.5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| The samples are not independent | B1 | The idea that samples are not independent. Condone other irrelevant comments provided they do not contradict this. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| They should consider the birth weight, gender, or whether or not the lambs are premature | B1, B1 | For one suitable comment on twins being identical relating to selecting the sample. Condone start weight. Do not accept age/diets of the lambs. For a second suitable comment. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Need the assumption that the underlying distribution of the difference between the weight gains must be normally distributed | B1 | Need the emboldened words |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Differences: \(-0.9, \ 0.5, \ 0.8, \ 0.2, \ 0.8, \ 0.3, \ 0.1, \ -0.1\) | M1 | Attempting differences (at least 4 correct) implied by awrt 0.304 or 0.551 but not 0.2125 |
| \(\bar{w} = 0.2125, \quad s^2 = 0.3041\ldots \ (s = 0.551\ldots)\) | M1 | Attempt to find \(\bar{w}\) and \(s\) or \(s^2\) for their differences implied by 0.2125 and either awrt 0.304 or 0.551 |
| Confidence interval: \(\text{"0.2125"} \pm t \times \sqrt{\dfrac{\text{"0.3041"}}{8}}\) | M1 | For using the correct formula \(\bar{w} \pm t \times \sqrt{\dfrac{s^2}{8}}\) where \( |
| \(\text{"0.2125"} \pm 2.998 \times \sqrt{\dfrac{\text{"0.3041"}}{8}}\) | A1ft | For \(\bar{w} \pm\) awrt \(2.998 \times \sqrt{\dfrac{s^2}{8}}\), all values need to be substituted in |
| \(= (-0.37202\ldots, \ 0.79702\ldots)\) | A1 | Dependent on all previous method marks (awrt \(-0.372\), awrt \(0.797\)). SC: If CI for two independent samples allow 3rd M for difference of means and pooled variance, A1ft using correct formula with awrt 2.624 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_w = 0.2 \quad H_1: \mu_w > 0.2\) | B1 | For both hypotheses correct in terms of \(\mu\) or \(\mu_w\). Condone 200 |
| \(200\text{g} = 0.2\text{ kg}\) is in the interval | M1 | For changing 200g to 0.2 kg (ignore units) and comparing to their CI |
| There is no evidence that \(\mu_w\) is greater than 0.2 | A1ft | Independent of hypotheses. Drawing a correct inference following through on their CI provided 0.2 is within their CI, with no contradictory statements. Accept "insufficient evidence to support the (researcher's) belief" |
## Question 6:
**Part (a):**
| Answer | Mark | Guidance |
|--------|------|----------|
| The samples are not independent | B1 | The idea that samples are not independent. Condone other irrelevant comments provided they do not contradict this. |
**Part (b):**
| Answer | Mark | Guidance |
|--------|------|----------|
| They should consider the birth weight, gender, or whether or not the lambs are premature | B1, B1 | For one suitable comment on twins being identical relating to selecting the sample. Condone start weight. Do not accept age/diets of the lambs. For a second suitable comment. |
**Part (c):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Need the assumption that the underlying distribution of the **difference** between the **weight gains** must be **normally distributed** | B1 | Need the emboldened words |
**Part (d):**
| Answer | Mark | Guidance |
|--------|------|----------|
| Differences: $-0.9, \ 0.5, \ 0.8, \ 0.2, \ 0.8, \ 0.3, \ 0.1, \ -0.1$ | M1 | Attempting differences (at least 4 correct) implied by awrt 0.304 or 0.551 but not 0.2125 |
| $\bar{w} = 0.2125, \quad s^2 = 0.3041\ldots \ (s = 0.551\ldots)$ | M1 | Attempt to find $\bar{w}$ **and** $s$ or $s^2$ for their differences implied by 0.2125 **and** either awrt 0.304 or 0.551 |
| Confidence interval: $\text{"0.2125"} \pm t \times \sqrt{\dfrac{\text{"0.3041"}}{8}}$ | M1 | For using the correct formula $\bar{w} \pm t \times \sqrt{\dfrac{s^2}{8}}$ where $|t| > 2$, all values need to be substituted in |
| $\text{"0.2125"} \pm 2.998 \times \sqrt{\dfrac{\text{"0.3041"}}{8}}$ | A1ft | For $\bar{w} \pm$ awrt $2.998 \times \sqrt{\dfrac{s^2}{8}}$, all values need to be substituted in |
| $= (-0.37202\ldots, \ 0.79702\ldots)$ | A1 | Dependent on all previous method marks (awrt $-0.372$, awrt $0.797$). SC: If CI for two independent samples allow 3rd M for difference of means and pooled variance, A1ft using correct formula with awrt 2.624 |
**Part (e):**
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_w = 0.2 \quad H_1: \mu_w > 0.2$ | B1 | For both hypotheses correct in terms of $\mu$ or $\mu_w$. Condone 200 |
| $200\text{g} = 0.2\text{ kg}$ is in the interval | M1 | For changing 200g to 0.2 kg (ignore units) and comparing to their CI |
| There is no evidence that $\mu_w$ is **greater** than 0.2 | A1ft | Independent of hypotheses. Drawing a correct inference following through on their CI provided 0.2 is within their CI, with no contradictory statements. Accept "insufficient evidence to support the (researcher's) belief" |
---\begin{enumerate}
\item A researcher set up a trial to assess the effect that a food supplement has on the increase in weight of Herdwick lambs. The researcher randomly selected 8 sets of twin lambs. One of each set of twins was given the food supplement and the other had no food supplement. The gain in weight, in kg, of each lamb over the period of the trial was recorded.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|l|l|l|l|l|l|}
\hline
& Set of twin lambs & A & $B$ & C & D & $E$ & $F$ & $G$ & $H$ \\
\hline
\multirow{2}{*}{Weight gain (kg)} & With food supplement & 4.1 & 5.3 & 6.0 & 3.6 & 5.9 & 4.2 & 7.1 & 6.4 \\
\hline
& No food supplement & 5.0 & 4.8 & 5.2 & 3.4 & 5.1 & 3.9 & 7.0 & 6.5 \\
\hline
\end{tabular}
\end{center}
(a) State why a two sample $t$-test is not suitable for use with these data.\\
(b) Suggest 2 other factors about the lambs that the researcher may need to control when selecting the sample.\\
(c) State one assumption, in context, that needs to be made for a paired $t$-test to be valid.
For a pair of twin lambs, the random variable $W$ represents the weight gain of the lamb given the food supplement minus the weight gain of the lamb not given the food supplement.\\
(d) Using the data in the table, calculate a $98 \%$ confidence interval for the mean of $W$ Show your working clearly.
The researcher believes that the mean of $W$ is greater than 200 g\\
(e) Stating your hypotheses clearly, use your confidence interval to explain whether or not there is evidence to support the researcher's belief.
\hfill \mbox{\textit{Edexcel FS2 2024 Q6 [12]}}