| Exam Board | Edexcel |
|---|---|
| Module | FS2 (Further Statistics 2) |
| Year | 2019 |
| 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 | Standard +0.8 This is a Further Maths FS2 question requiring an F-test for equality of variances. Students must calculate two sample variances from summary statistics, set up appropriate hypotheses, perform a two-tailed F-test, and state the normality assumption. While methodical, it requires knowledge of a specialized topic beyond standard A-level and careful handling of the two-tailed critical region. |
| Spec | 2.01a Population and sample: terminology2.02g Calculate mean and standard deviation5.05c Hypothesis test: normal distribution for population mean |
| Number of plants | \(\sum x\) | \(\sum x ^ { 2 }\) | |
| \(A\) | 25 | 194.7 | 1637.37 |
| \(B\) | 26 | 227.5 | 2031.19 |
| 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 \(\sigma^2\) |
| \(s_A^2 = \frac{1}{24}\left(1637.37 - 25\times\left(\frac{194.7}{25}\right)^2\right) = 5.0436\) | M1, A1 | Using correct method for either \(s_A^2\) or \(s_B^2\); awrt 5.04 |
| \(s_B^2 = \frac{1}{25}\left(2031.19 - 26\times\left(\frac{227.5}{26}\right)^2\right) = 1.6226\) | A1 | awrt 1.62 |
| \(\frac{s_A^2}{s_B^2} = 3.108...\) | M1 | Using F-distribution; allow \(\frac{s_B^2}{s_A^2}[=0.321...]\) |
| Critical value upper tail \(F_{24,25} = 1.96\) | B1 | awrt 1.96 or 0.506; must match their method |
| There is evidence that the two variances are different | A1ft | Correct inference following through their CV; allow \(\sigma_B^2 \neq \sigma_A^2\); do not allow \(\sigma_B^2 = \sigma_A^2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The yields are normally distributed | B1 | Recalling that the variable yield needs to be normally distributed |
# Question 3(a):
| 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 $\sigma^2$ |
| $s_A^2 = \frac{1}{24}\left(1637.37 - 25\times\left(\frac{194.7}{25}\right)^2\right) = 5.0436$ | M1, A1 | Using correct method for either $s_A^2$ or $s_B^2$; awrt 5.04 |
| $s_B^2 = \frac{1}{25}\left(2031.19 - 26\times\left(\frac{227.5}{26}\right)^2\right) = 1.6226$ | A1 | awrt 1.62 |
| $\frac{s_A^2}{s_B^2} = 3.108...$ | M1 | Using F-distribution; allow $\frac{s_B^2}{s_A^2}[=0.321...]$ |
| Critical value upper tail $F_{24,25} = 1.96$ | B1 | awrt 1.96 or 0.506; must match their method |
| There is evidence that the two **variances** are different | A1ft | Correct inference following through their CV; allow $\sigma_B^2 \neq \sigma_A^2$; do not allow $\sigma_B^2 = \sigma_A^2$ |
# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The **yields** are normally distributed | B1 | Recalling that the variable **yield** needs to be normally distributed |
---3 Yin grows two varieties of potato, plant $A$ and plant $B$. A random sample of each variety of potato is taken and the yield, $x \mathrm {~kg}$, produced by each plant is measured. The following statistics are obtained from the data.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& Number of plants & $\sum x$ & $\sum x ^ { 2 }$ \\
\hline
$A$ & 25 & 194.7 & 1637.37 \\
\hline
$B$ & 26 & 227.5 & 2031.19 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly, test, at the $10 \%$ significance level, whether or not the variances of the yields of the two varieties of potato are the same.
\item State an assumption you have made in order to carry out the test in part (a).
\end{enumerate}
\hfill \mbox{\textit{Edexcel FS2 2019 Q3 [8]}}