| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Marks | 16 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | F-test and chi-squared for variance |
| Type | F-test then t-test sequential |
| Difficulty | Standard +0.8 This S4 question requires conducting an F-test for variance equality, a two-sample t-test, stating assumptions, and constructing a pooled variance confidence interval. While the individual techniques are standard for Further Maths Statistics, the multi-part structure, need to carefully handle degrees of freedom, and the pooled variance CI calculation make this moderately challenging, requiring solid understanding rather than just routine application. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| New method | Conventional method | |
| Mean \(( \bar { x } )\) | 82.3 | 78.2 |
| Standard deviation \(( s )\) | 3.5 | 5.7 |
| Number of students \(( n )\) | 10 | 9 |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \sigma_C^2 = \sigma_N^2\), \(H_1: \sigma_C^2 > \sigma_N^2\) | B1 | |
| \(\dfrac{s_C^2}{s_N^2} = \dfrac{5.7^2}{3.5^2} = 2.652\ldots\); \(\quad F_{8,9}\) (5%) critical value \(= 3.23\) | M1; B1 | |
| Not significant so do not reject \(H_0\) — insufficient evidence that variance using conventional method is greater | A1 \(\checkmark\) | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(H_0: \mu_N = \mu_C\), \(H_1: \mu_N > \mu_C\) | B1 | |
| \(s^2 = \dfrac{8 \times 5.7^2 + 9 \times 3.5^2}{17} = \dfrac{370.17}{17} = 21.774\ldots\) | M1 | |
| Test statistic \(t = \dfrac{82.3 - 78.2}{\sqrt{21.774\ldots\left(\frac{1}{9} + \frac{1}{10}\right)}} = 1.9122\ldots\) awrt 1.91 | M1 A1 | |
| \(t_{17}\) (5%) 1-tail critical value \(= 1.740\) | B1 | |
| Significant — reject \(H_0\). There is evidence that new style leads to an increase in mean | A1 \(\checkmark\) | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Assumed population of marks obtained were normally distributed | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(7.564 < \dfrac{17s^2}{\sigma^2} < 30.191\) | B1 M1 B1 | |
| \(\sigma^2 > \dfrac{17 \times 21.774\ldots}{30.191} = 12.3\) (1 d.p.) | A1 | |
| \(\sigma^2 < \dfrac{17 \times 21.774\ldots}{7.564} = 48.9\) (1 d.p.) | A1 | (5 marks) (16 marks total) |
# Question 5:
## Part (a)(i)
$H_0: \sigma_C^2 = \sigma_N^2$, $H_1: \sigma_C^2 > \sigma_N^2$ | B1 |
$\dfrac{s_C^2}{s_N^2} = \dfrac{5.7^2}{3.5^2} = 2.652\ldots$; $\quad F_{8,9}$ (5%) critical value $= 3.23$ | M1; B1 |
Not significant so do not reject $H_0$ — insufficient evidence that variance using conventional method is greater | A1 $\checkmark$ | (4 marks)
## Part (a)(ii)
$H_0: \mu_N = \mu_C$, $H_1: \mu_N > \mu_C$ | B1 |
$s^2 = \dfrac{8 \times 5.7^2 + 9 \times 3.5^2}{17} = \dfrac{370.17}{17} = 21.774\ldots$ | M1 |
Test statistic $t = \dfrac{82.3 - 78.2}{\sqrt{21.774\ldots\left(\frac{1}{9} + \frac{1}{10}\right)}} = 1.9122\ldots$ awrt 1.91 | M1 A1 |
$t_{17}$ (5%) 1-tail critical value $= 1.740$ | B1 |
Significant — reject $H_0$. There is evidence that new style leads to an increase in mean | A1 $\checkmark$ | (6 marks)
## Part (b)
Assumed population of marks obtained were normally distributed | B1 | (1 mark)
## Part (c)
Unbiased estimate of common variance is $s^2$ in (ii)
$7.564 < \dfrac{17s^2}{\sigma^2} < 30.191$ | B1 M1 B1 |
$\sigma^2 > \dfrac{17 \times 21.774\ldots}{30.191} = 12.3$ (1 d.p.) | A1 |
$\sigma^2 < \dfrac{17 \times 21.774\ldots}{7.564} = 48.9$ (1 d.p.) | A1 | (5 marks) **(16 marks total)**
---5. An educational researcher is testing the effectiveness of a new method of teaching a topic in mathematics. A random sample of 10 children were taught by the new method and a second random sample of 9 children, of similar age and ability, were taught by the conventional method. At the end of the teaching, the same test was given to both groups of children.
The marks obtained by the two groups are summarised in the table below.
\begin{center}
\begin{tabular}{ | l | c | c | }
\hline
& New method & Conventional method \\
\hline
Mean $( \bar { x } )$ & 82.3 & 78.2 \\
\hline
Standard deviation $( s )$ & 3.5 & 5.7 \\
\hline
Number of students $( n )$ & 10 & 9 \\
\hline
\end{tabular}
\end{center}
\begin{enumerate}[label=(\alph*)]
\item Stating your hypotheses clearly and using a $5 \%$ level of significance, investigate whether or not
\begin{enumerate}[label=(\roman*)]
\item the variance of the marks of children taught by the conventional method is greater than that of children taught by the new method,
\item the mean score of children taught by the conventional method is lower than the mean score of those taught by the new method.\\[0pt]
[In each case you should give full details of the calculation of the test statistics.]
\end{enumerate}\item State any assumptions you made in order to carry out these tests.
\item Find a 95\% confidence interval for the common variance of the marks of the two groups.
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 Q5 [16]}}