| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | Recover parameters from given CI |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question requiring understanding of both normal and chi-squared confidence intervals. Part (a) requires working backwards from a confidence interval for the mean to find the variance estimate (non-routine manipulation), and part (b) requires applying chi-squared distribution tables for variance confidence intervals. While the techniques are standard for FS2, the reverse-engineering in part (a) and the need to handle two different distributions elevates this above typical A-level questions. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Answer | Marks | Guidance |
|---|---|---|
| 95% CI for \(\mu\) uses \(t\) value of 2.064 | B1 | Realising that the \(t\)-distribution must be used as a model and finding the correct value awrt 2.06 |
| \(\frac{\hat{\sigma}}{\sqrt{25}} \times\) "2.064" \(= \frac{1}{2}(2.232-1.128)\) or \(\frac{1}{2}(2.232+1.128)+\) "2.064"\(\times\) \(\frac{\hat{\sigma}}{\sqrt{25}} = 2.232\) (oe) | M1 | Using the correct formula with a \(t\)-value, \(\frac{\hat{\sigma}}{\sqrt{25}} \times\) "\(t\) value" \(= \frac{1}{2}(2.232-1.128)\) or \(\frac{1}{2}(2.232+1.128)+\) "\(t\) value"\(\times\) \(\frac{\hat{\sigma}}{\sqrt{25}} = 2.232\) or \(\frac{1}{2}(2.232+1.128)-\) "\(t\) value"\(\times\) \(\frac{\hat{\sigma}}{\sqrt{25}} = 1.128\) |
| Rearranging one of these formula accurately to find a value of \(\hat{\sigma}\) | M1 | |
| \(\hat{\sigma} = \frac{2.76}{\text{"2.064"}}\) or \(1.3372...\) | M1 | |
| \(\hat{\sigma}^2 = 1.788...[= 1.79\) (3sf)] \(*\) | A1*cso | A correct solution only using awrt 1.79 |
| Answer | Marks | Guidance |
|---|---|---|
| \(12.401_< \frac{24 \times 1.79}{\sigma^2} < 39.364\) | B1 | awrt 12.4 or 39.4 May be implied by a correct confidence interval |
| M1 | \(\frac{24 \times 1.79}{\sigma^2}\). May be implied by a correct confidence interval | |
| \(\mathbf{1.09} < \sigma^2 < \mathbf{3.46}\) | A1 | awrt 1.09 and awrt 3.46 |
**(a)**
| 95% CI for $\mu$ uses $t$ value of **2.064** | B1 | Realising that the $t$-distribution must be used as a model and finding the correct value awrt 2.06 |
| $\frac{\hat{\sigma}}{\sqrt{25}} \times$ "2.064" $= \frac{1}{2}(2.232-1.128)$ **or** $\frac{1}{2}(2.232+1.128)+$ "2.064"$\times$ $\frac{\hat{\sigma}}{\sqrt{25}} = 2.232$ (oe) | M1 | Using the correct formula with a $t$-value, $\frac{\hat{\sigma}}{\sqrt{25}} \times$ "$t$ value" $= \frac{1}{2}(2.232-1.128)$ or $\frac{1}{2}(2.232+1.128)+$ "$t$ value"$\times$ $\frac{\hat{\sigma}}{\sqrt{25}} = 2.232$ **or** $\frac{1}{2}(2.232+1.128)-$ "$t$ value"$\times$ $\frac{\hat{\sigma}}{\sqrt{25}} = 1.128$ |
| Rearranging one of these formula accurately to find a value of $\hat{\sigma}$ | M1 | |
| $\hat{\sigma} = \frac{2.76}{\text{"2.064"}}$ or $1.3372...$ | M1 | |
| $\hat{\sigma}^2 = 1.788...[= 1.79$ (3sf)] $*$ | A1*cso | A correct solution only using awrt 1.79 |
**(b)**
| $12.401_< \frac{24 \times 1.79}{\sigma^2} < 39.364$ | B1 | awrt 12.4 or 39.4 May be implied by a correct confidence interval |
| | M1 | $\frac{24 \times 1.79}{\sigma^2}$. May be implied by a correct confidence interval |
| $\mathbf{1.09} < \sigma^2 < \mathbf{3.46}$ | A1 | awrt 1.09 and awrt 3.46 |
---\begin{enumerate}
\item A nutritionist studied the levels of cholesterol, $X \mathrm { mg } / \mathrm { cm } ^ { 3 }$, of male students at a large college. She assumed that $X$ was distributed $\mathrm { N } \left( \mu , \sigma ^ { 2 } \right)$ and examined a random sample of 25 male students. Using this sample she obtained unbiased estimates of $\mu$ and $\sigma ^ { 2 }$ as $\hat { \mu }$ and $\hat { \sigma } ^ { 2 }$
\end{enumerate}
A $95 \%$ confidence interval for $\mu$ was found to be $( 1.128,2.232 )$\\
(a) Show that $\hat { \sigma } ^ { 2 } = 1.79$ (correct to 3 significant figures)\\
(b) Obtain a $95 \%$ confidence interval for $\sigma ^ { 2 }$
\hfill \mbox{\textit{Edexcel FS2 Q3 [7]}}