| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2021 |
| Session | June |
| Marks | 15 |
| 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 F-test for variance comparison, chi-squared confidence interval calculation, and pooled t-test. While it involves multiple statistical techniques and careful hypothesis testing, the procedures are standard for FS2 level with straightforward calculations from summary statistics. The multi-part structure and need to apply several different tests elevates it above average difficulty, but it remains a textbook application of learned techniques without requiring novel insight. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| \(\sum w\) | \(\sum w ^ { 2 }\) | |||
| Ringlet | 8 | 410 | 21032 | ||
| Meadow Brown | 6 | 294 | 14426 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(H_0: \sigma_r^2 = \sigma_{mb}^2 \quad H_1: \sigma_r^2 \neq \sigma_{mb}^2\) | B1 | For both hypotheses in terms of \(\sigma\) |
| \(s_r^2 = \frac{1}{7}\!\left(21032 - 8\times\left(\frac{410}{8}\right)^2\right) = 2.7857...\) | M1, A1 | Correct method for \(s_r^2\) or \(s_{mb}^2\); awrt 2.79 (allow \(\frac{39}{14}\)) |
| \(s_{mb}^2 = \frac{1}{5}\!\left(14426 - 6\times\left(\frac{294}{6}\right)^2\right) = 4\) | A1 | 4 cao |
| \(\frac{s_{mb}^2}{s_r^2} = 1.4358...\) | M1 | Using correct model for test statistic with correct ratio |
| CV \(F_{5,7} = 7.46\) | B1 | Correct CV |
| \(1.4358... < 7.46\) so insufficient evidence to suggest the variances of the wingspans are different | A1ft | Drawing correct inference in context using CV and test statistic (dep on both M marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\chi^2_{5,\alpha} = \frac{5\times{'}4{'}}{1.194}\) or \(\chi^2_{5,\alpha} = \frac{5\times{'}4{'}}{48.54}\) | M1 | Either correct attempt at \(\chi^2_{5,\alpha}\) with \(\nu=5\) |
| \(\chi^2_{5,\alpha} = 16.75 \rightarrow \alpha = 0.005\) or \(\chi^2_{5,\alpha} = 0.412 \rightarrow \alpha = 0.995\) | M1 | Using tables to find appropriate probability |
| \(k = 99\) | A1 | 99 cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(s_p^2 = \frac{7\times{'}2.7857...{'} + 5\times{'}4{'}}{8+6-2} = 3.29...\) | M1 | Correct expression for \(s_p^2\) |
| \(t_{12} = 2.179\) | B1 | Correct 95% \(t\)-value |
| \(\left(\frac{410}{8} - \frac{294}{6}\right) \pm {'}2.179{'} \times \sqrt{{'3.29{'}}}\sqrt{\frac{1}{8}+\frac{1}{6}}\) | M1 | CI in the correct form |
| (awrt 0.115, awrt 4.39) | A1, A1 | awrt 0.115; awrt 4.39 |
# Question 6:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $H_0: \sigma_r^2 = \sigma_{mb}^2 \quad H_1: \sigma_r^2 \neq \sigma_{mb}^2$ | B1 | For both hypotheses in terms of $\sigma$ |
| $s_r^2 = \frac{1}{7}\!\left(21032 - 8\times\left(\frac{410}{8}\right)^2\right) = 2.7857...$ | M1, A1 | Correct method for $s_r^2$ or $s_{mb}^2$; awrt 2.79 (allow $\frac{39}{14}$) |
| $s_{mb}^2 = \frac{1}{5}\!\left(14426 - 6\times\left(\frac{294}{6}\right)^2\right) = 4$ | A1 | 4 cao |
| $\frac{s_{mb}^2}{s_r^2} = 1.4358...$ | M1 | Using correct model for test statistic with correct ratio |
| CV $F_{5,7} = 7.46$ | B1 | Correct CV |
| $1.4358... < 7.46$ so insufficient evidence to suggest the variances of the wingspans are different | A1ft | Drawing correct inference in context using CV and test statistic (dep on both M marks) |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\chi^2_{5,\alpha} = \frac{5\times{'}4{'}}{1.194}$ or $\chi^2_{5,\alpha} = \frac{5\times{'}4{'}}{48.54}$ | M1 | Either correct attempt at $\chi^2_{5,\alpha}$ with $\nu=5$ |
| $\chi^2_{5,\alpha} = 16.75 \rightarrow \alpha = 0.005$ or $\chi^2_{5,\alpha} = 0.412 \rightarrow \alpha = 0.995$ | M1 | Using tables to find appropriate probability |
| $k = 99$ | A1 | 99 cao |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| $s_p^2 = \frac{7\times{'}2.7857...{'} + 5\times{'}4{'}}{8+6-2} = 3.29...$ | M1 | Correct expression for $s_p^2$ |
| $t_{12} = 2.179$ | B1 | Correct 95% $t$-value |
| $\left(\frac{410}{8} - \frac{294}{6}\right) \pm {'}2.179{'} \times \sqrt{{'3.29{'}}}\sqrt{\frac{1}{8}+\frac{1}{6}}$ | M1 | CI in the correct form |
| (awrt 0.115, awrt 4.39) | A1, A1 | awrt 0.115; awrt 4.39 |
---\begin{enumerate}
\item Elsa is collecting information on the wingspan of two different species of butterfly, Ringlet and Meadow Brown. She takes a random sample of each type of butterfly. The wingspans, $w \mathrm {~cm}$, are summarised in the table below. The wingspans of Ringlet and Meadow Brown butterflies each follow normal distributions.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | }
\hline
& \begin{tabular}{ c }
Number of \\
butterflies \\
\end{tabular} & $\sum w$ & $\sum w ^ { 2 }$ \\
\hline
Ringlet & 8 & 410 & 21032 \\
\hline
Meadow Brown & 6 & 294 & 14426 \\
\hline
\end{tabular}
\end{center}
(a) Test, at the $2 \%$ level of significance, whether or not there is evidence that the variance of the wingspans of Ringlet butterflies is different from the variance of the wingspans of Meadow Brown butterflies. You should state your hypotheses clearly.
The $k \%$ confidence interval for the variance of the wingspans of Meadow Brown butterflies is (1.194, 48.54)\\
(b) Find the value of $k$\\
(c) Calculate a $95 \%$ confidence interval for the difference between the mean wingspan of the Ringlet butterfly and the mean wingspan of the Meadow Brown butterfly.
\hfill \mbox{\textit{Edexcel FS2 2021 Q6 [15]}}