| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Two-sample t-test (unknown variances) |
| Difficulty | Challenging +1.2 This is a standard two-sample hypothesis testing question requiring an F-test for variances followed by a two-sample t-test. While it involves multiple steps and careful calculation of sample variances from summary statistics, the procedures are routine for S4 students with no novel problem-solving required. The difficulty is elevated slightly above average due to the computational demands and the need to correctly interpret the directional claim (μ_girls - μ_boys > 5), but remains a textbook-style question. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Number of children | Sample mean \(\bar { x }\) | \(\sum x ^ { 2 }\) | |
| Boys | 9 | 22.8 | 4693.60 |
| Girls | 6 | 29.5 | 5236.12 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \sigma_B^2 = \sigma_G^2\), \(H_1: \sigma_B^2 \neq \sigma_G^2\) | B1 | Both hypotheses. Must use \(\sigma\) or \(\sigma^2\), make clear which is \(H_0\) and \(H_1\). Do not allow in words |
| \(s_B^2 = \frac{1}{8}(4693.6 - 9 \times 22.8^2) = 1.88\) | M1 | Correct method for either \(s_B^2\) or \(s_G^2\) |
| \(s_G^2 = \frac{1}{5}(5236.12 - 6 \times 29.5^2) = 2.924\) awrt 2.92 | A1 | Both \(s_B^2\) and \(s_G^2\) correct to 3sf; allow sd's. A1 awrt 1.56 or 0.643 |
| \(\frac{s_G^2}{s_B^2} = 1.555\ldots\ [0.643]\) | M1 A1 | M1 allow use of \(s_B\) and \(s_G\) instead of \(s_B^2\) or \(s_G^2\) |
| Critical value \(F_{5,8} = 3.69\ [0.271]\) | B1 | Correct CV for their \(F\) or correct comparison if using \(p\) |
| Not significant, variances are the same | A1 cso (7) | All previous marks must be awarded. Variances are the same or are not different |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_G = \mu_B + 5\), \(H_1: \mu_G > \mu_B + 5\) | B1 | Both hypotheses using \(\mu\). Do not allow \(\geq\) instead of \(>\). May use different letters but must be defined |
| Pooled estimate: \(s_p^2 = \frac{8 \times 1.88 + 5 \times 2.924}{13} = 2.2815\ldots\) or \(s_p = 1.51046\ldots\) | M1 | Only allow use of \(s_B\) and \(s_G\) instead of \(s_B^2\) or \(s_G^2\) if seen in part (a). Condone missing 5 |
| \(t = \pm\left(\dfrac{29.5 - 22.8 - 5}{s_p\sqrt{\frac{1}{9}+\frac{1}{6}}}\right) = \pm\ \text{awrt}\ 2.14\) or \(p = 0.0262\) | M1 M1 A1 | M1 correct formula with their \(s_p\); M1 correct formula with their \(s_p\) (must have been attempted) |
| Critical value \(t_{13}(1\%) = \pm 2.650\) or \(0.0262 > 0.01\) | B1 | Correct CV but must match \(t\)-value or correct comparison if using \(p\) |
| Insufficient evidence to support Headteacher's claim or: the time taken for girls is not more than 5 seconds greater than for boys | A1 cso (7) | Correct statement with either the word Headteacher/Teacher/Head or time and not more than 5 oe; do not allow contradicting statements |
## Question 1:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \sigma_B^2 = \sigma_G^2$, $H_1: \sigma_B^2 \neq \sigma_G^2$ | B1 | Both hypotheses. Must use $\sigma$ or $\sigma^2$, make clear which is $H_0$ and $H_1$. Do not allow in words |
| $s_B^2 = \frac{1}{8}(4693.6 - 9 \times 22.8^2) = 1.88$ | M1 | Correct method for either $s_B^2$ or $s_G^2$ |
| $s_G^2 = \frac{1}{5}(5236.12 - 6 \times 29.5^2) = 2.924$ awrt 2.92 | A1 | Both $s_B^2$ and $s_G^2$ correct to 3sf; allow sd's. A1 awrt 1.56 or 0.643 |
| $\frac{s_G^2}{s_B^2} = 1.555\ldots\ [0.643]$ | M1 A1 | M1 allow use of $s_B$ and $s_G$ instead of $s_B^2$ or $s_G^2$ |
| Critical value $F_{5,8} = 3.69\ [0.271]$ | B1 | Correct CV for their $F$ or correct comparison if using $p$ |
| Not significant, variances are the same | A1 cso (7) | All previous marks must be awarded. Variances are the same or are not different |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_G = \mu_B + 5$, $H_1: \mu_G > \mu_B + 5$ | B1 | Both hypotheses using $\mu$. Do not allow $\geq$ instead of $>$. May use different letters but must be defined |
| Pooled estimate: $s_p^2 = \frac{8 \times 1.88 + 5 \times 2.924}{13} = 2.2815\ldots$ or $s_p = 1.51046\ldots$ | M1 | Only allow use of $s_B$ and $s_G$ instead of $s_B^2$ or $s_G^2$ if seen in part (a). Condone missing 5 |
| $t = \pm\left(\dfrac{29.5 - 22.8 - 5}{s_p\sqrt{\frac{1}{9}+\frac{1}{6}}}\right) = \pm\ \text{awrt}\ 2.14$ or $p = 0.0262$ | M1 M1 A1 | M1 correct formula with their $s_p$; M1 correct formula with their $s_p$ (must have been attempted) |
| Critical value $t_{13}(1\%) = \pm 2.650$ or $0.0262 > 0.01$ | B1 | Correct CV but must match $t$-value or correct comparison if using $p$ |
| Insufficient evidence to support **Headteacher**'s claim or: the **time** taken for girls is **not more than 5** seconds greater than for boys | A1 cso (7) | Correct statement with either the word Headteacher/Teacher/Head or time and not more than 5 oe; do not allow contradicting statements |
**Total: 14 marks**\begin{enumerate}
\item The times taken by children to run 150 m are normally distributed. The times taken, $x$ seconds, by a random sample of 9 boys and an independent random sample of 6 girls are recorded. The following statistics are obtained.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\hline
& Number of children & Sample mean $\bar { x }$ & $\sum x ^ { 2 }$ \\
\hline
Boys & 9 & 22.8 & 4693.60 \\
\hline
Girls & 6 & 29.5 & 5236.12 \\
\hline
\end{tabular}
\end{center}
(a) Test, at the $10 \%$ level of significance, whether or not the variances of the two distributions are equal. State your hypotheses clearly.
The Headteacher claims that the mean time taken for the girls is more than 5 seconds greater than the mean time taken for the boys.\\
(b) Stating your hypotheses clearly, test the Headteacher's claim. Use a $1 \%$ level of significance and show your working clearly.
\hfill \mbox{\textit{Edexcel S4 2017 Q1 [14]}}