| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | F-test two variances hypothesis |
| Difficulty | Challenging +1.2 This is a Further Maths Statistics question requiring chi-squared confidence intervals and F-test for variance equality. While these are advanced topics beyond A-level core, the question follows standard procedures with clear data and straightforward application of formulas. The multi-step nature and need for critical value lookup elevate it above average, but it remains a textbook application without requiring novel insight. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Sample size | \(\sum x\) | \(\sum x ^ { 2 }\) |
| Machine \(A\) | 9 | 2268 | 571700 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([\bar{x} = \frac{2268}{9} = 252]\), \(s_A^2 = \frac{571700 - 9 \times 252^2}{8} [= 20.5]\) | M1 | Correct expression for \(s_A^2\) |
| \(\chi^2_{8,0.025} = 17.535\), \(\chi^2_{8,0.975} = 2.180\) | B1 | For selecting \(\chi^2\) with \(\nu = 8\). May be indicated in notation or either of 17.535, 2.180 |
| \(\frac{8 \times 20.5}{17.535} < \sigma_A^2 < \frac{8 \times 20.5}{2.180}\) | M1 | Setting up 95% CI with their sample variance, chi-squared values, and 8 |
| \(9.3527... < \sigma_A^2 < 75.2293...\) | A1 | Correct CI with awrt 9.35 and awrt 75.2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \sigma_A^2 = \sigma_B^2\), \(H_1: \sigma_A^2 \neq \sigma_B^2\) | B1 | Both hypotheses correct using \(\sigma\) or \(\sigma^2\) |
| \(\frac{s_A^2}{s_B^2} = \frac{\text{"20.5"}}{12.7} = 1.61...\) | M1 | Using the \(F\)-distribution as the model e.g. \(\frac{s_A^2}{s_B^2}\) |
| \(F_{8,10}(0.05) = 3.07\) | B1 | awrt 3.07 |
| There is insufficient evidence to suggest that the variances are different. | A1 | Drawing a correct inference following through their CV and value. Allow \(\sigma_B^2 = \sigma_A^2\). Allow standard deviation instead of variance. NB: Allow candidates to use \(\frac{s_B^2}{s_A^2}\) with \(\frac{1}{3.35}\) = awrt 0.299 for final M1B1A1 in (b) |
# Question 3:
## Part 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[\bar{x} = \frac{2268}{9} = 252]$, $s_A^2 = \frac{571700 - 9 \times 252^2}{8} [= 20.5]$ | M1 | Correct expression for $s_A^2$ |
| $\chi^2_{8,0.025} = 17.535$, $\chi^2_{8,0.975} = 2.180$ | B1 | For selecting $\chi^2$ with $\nu = 8$. May be indicated in notation or either of 17.535, 2.180 |
| $\frac{8 \times 20.5}{17.535} < \sigma_A^2 < \frac{8 \times 20.5}{2.180}$ | M1 | Setting up 95% CI with their sample variance, chi-squared values, and 8 |
| $9.3527... < \sigma_A^2 < 75.2293...$ | A1 | Correct CI with awrt **9.35** and awrt **75.2** |
## Part 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \sigma_A^2 = \sigma_B^2$, $H_1: \sigma_A^2 \neq \sigma_B^2$ | B1 | Both hypotheses correct using $\sigma$ or $\sigma^2$ |
| $\frac{s_A^2}{s_B^2} = \frac{\text{"20.5"}}{12.7} = 1.61...$ | M1 | Using the $F$-distribution as the model e.g. $\frac{s_A^2}{s_B^2}$ |
| $F_{8,10}(0.05) = 3.07$ | B1 | awrt **3.07** |
| There is insufficient evidence to suggest that the variances are different. | A1 | Drawing a correct inference following through their CV and value. Allow $\sigma_B^2 = \sigma_A^2$. Allow standard deviation instead of variance. NB: Allow candidates to use $\frac{s_B^2}{s_A^2}$ with $\frac{1}{3.35}$ = awrt 0.299 for final M1B1A1 in (b) |
---\begin{enumerate}
\item Two machines, $A$ and $B$, are used to fill bottles of water. The amount of water dispensed by each machine is normally distributed.
\end{enumerate}
Samples are taken from each machine and the amount of water, $x \mathrm { ml }$, dispensed in each bottle is recorded. The table shows the summary statistics for Machine $A$.
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Sample size & $\sum x$ & $\sum x ^ { 2 }$ \\
\hline
Machine $A$ & 9 & 2268 & 571700 \\
\hline
\end{tabular}
\end{center}
(a) Find a 95\% confidence interval for the variance of the amount of water dispensed in each bottle by Machine $A$.
For Machine $B$, a random sample of 11 bottles is taken. The sample variance of the amount of water dispensed in bottles is $12.7 \mathrm { ml } ^ { 2 }$\\
(b) Test, at the $10 \%$ level of significance, whether there is evidence that the variances of the amounts of water dispensed in bottles by the two machines are different. You should state the hypotheses and the critical value used.
\hfill \mbox{\textit{Edexcel FS2 2023 Q3 [8]}}