| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | Find critical region for F-test |
| Difficulty | Standard +0.8 This is a Further Maths Statistics question requiring F-test application with critical value lookup, reverse calculation to find a range of sample variances, and chi-squared confidence interval construction. While the individual techniques are standard for FS2, the combination of F-distribution and chi-squared distribution applications, plus the non-routine part (b) requiring algebraic manipulation to find acceptable values of s²_B, elevates this above typical A-level questions. The 2% significance level and two-tailed test add computational complexity. |
| Spec | 5.05a Sample mean distribution: central limit theorem5.05c Hypothesis test: normal distribution for population mean |
| \cline { 2 - 3 } \multicolumn{1}{c|}{} | number of samples | \(S _ { A } ^ { 2 }\) |
| Site \(A\) | 13 | 6.39 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| The concentration of air pollutant (for each site) follows a normal distribution. | B1 | Correct modelling assumption; samples are independent is B0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(F_{12,8(0.01)} = 5.67\) or \(F_{8,12(0.01)} = 4.50\) | B1 | Setting up either \(F\) statistic |
| \(\frac{6.39}{s_B^2} > 5.67'\), \(\quad \frac{s_B^2}{6.39} > 4.50'\) | M1M1 | M1: either correct inequality (condone equation); M1: understanding both correct inequalities needed |
| \((0 <)\ s_B^2 < 1.12698...,\quad s_B^2 > 28.755\) | A1 | Correct range; awrt 1.13 and awrt 28.8; note \(s_B^2 < 1.12698...\) and \(s_B^2 > 28.755\) is A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\chi^2_{12,0.005} = 28.3\) or \(\chi^2_{12,0.995} = 3.074\) | M1 | Realising \(\chi^2\) distribution must be used; allow \(\chi^2_{12,0.01} = 26.217\) or \(\chi^2_{12,0.99} = 3.571\) for this mark |
| \(\frac{12 \times 6.39}{28.3} < \sigma_A^2 < \frac{12 \times 6.39}{3.074}\) | M1 | Setting up 99% CI |
| \(2.7095... < \sigma_A^2 < 24.94469...\) | A1 | Correct CI with awrt 2.71 and awrt 24.9 |
# Question 5:
## Part 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| The concentration of air pollutant (for each site) follows a normal distribution. | B1 | Correct modelling assumption; samples are independent is B0 |
## Part 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $F_{12,8(0.01)} = 5.67$ or $F_{8,12(0.01)} = 4.50$ | B1 | Setting up either $F$ statistic |
| $\frac{6.39}{s_B^2} > 5.67'$, $\quad \frac{s_B^2}{6.39} > 4.50'$ | M1M1 | M1: either correct inequality (condone equation); M1: understanding both correct inequalities needed |
| $(0 <)\ s_B^2 < 1.12698...,\quad s_B^2 > 28.755$ | A1 | Correct range; awrt 1.13 and awrt 28.8; note $s_B^2 < 1.12698...$ and $s_B^2 > 28.755$ is A0 |
## Part 5(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\chi^2_{12,0.005} = 28.3$ or $\chi^2_{12,0.995} = 3.074$ | M1 | Realising $\chi^2$ distribution must be used; allow $\chi^2_{12,0.01} = 26.217$ or $\chi^2_{12,0.99} = 3.571$ for this mark |
| $\frac{12 \times 6.39}{28.3} < \sigma_A^2 < \frac{12 \times 6.39}{3.074}$ | M1 | Setting up 99% CI |
| $2.7095... < \sigma_A^2 < 24.94469...$ | A1 | Correct CI with awrt 2.71 and awrt 24.9 |
---\begin{enumerate}
\item The concentration of an air pollutant is measured in micrograms $/ \mathrm { m } ^ { 3 }$
\end{enumerate}
Samples of air were taken at two different sites and the concentration of this particular air pollutant was recorded.
For Site $A$ the summary statistics are shown below.
\begin{center}
\begin{tabular}{ | l | c | c | }
\cline { 2 - 3 }
\multicolumn{1}{c|}{} & number of samples & $S _ { A } ^ { 2 }$ \\
\hline
Site $A$ & 13 & 6.39 \\
\hline
\end{tabular}
\end{center}
For Site $B$ there were 9 samples of air taken.\\
A test of the hypothesis $\mathrm { H } _ { 0 } : \sigma _ { A } ^ { 2 } = \sigma _ { B } ^ { 2 }$ against the hypothesis $\mathrm { H } _ { 1 } : \sigma _ { A } ^ { 2 } \neq \sigma _ { B } ^ { 2 }$ is carried out using a $2 \%$ level of significance.\\
(a) State a necessary assumption required to carry out the test.
Given that the assumption in part (a) holds,\\
(b) find the set of values of $s _ { B } ^ { 2 }$ that would lead to the null hypothesis being rejected,\\
(c) find a 99\% confidence interval for the variance of the concentration of the air pollutant at Site A.
\hfill \mbox{\textit{Edexcel FS2 2022 Q5 [8]}}