| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2016 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.8 This S4 question requires conducting both an F-test for variance equality and a two-sample t-test with pooled variance, interpreting the connection between them. While the calculations are standard for Further Maths Statistics, the multi-part structure, need to correctly apply pooled variance based on part (a), and conceptual explanation in part (c) elevate this above routine exercises. The small sample sizes and specific hypothesis (difference = 5) add modest complexity. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| \cline { 2 - 4 } \multicolumn{1}{c|}{} | Sample size | Sample mean |
| ||
| \(X\) | 9 | 14.8 | 6.76 | ||
| \(Y\) | 6 | 7.2 | 5.42 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \sigma^2_X = \sigma^2_Y \quad H_1: \sigma^2_X \neq \sigma^2_Y\) | B1 | Both hypotheses |
| \(F_{8,5} = \frac{6.76^2}{5.42^2} = 1.556\) | M1 A1 | M1 allow use of 6.76 and 5.42 instead of \(6.76^2\) and \(5.42^2\); A1 awrt 1.56 |
| \(F_{8,5}\) is 4.82 | B1 | Allow p value 0.650 instead of critical value |
| There is evidence that the variances are the same | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \mu_X = \mu_Y + 5 \quad H_1: \mu_X > \mu_Y + 5\) | B1 | Both hypotheses |
| \(s_p^2 = \frac{8\times 6.76^2 + 5\times 5.42^2}{13} = 39.42...\) or \(s_p = 6.278...\) | M1 A1 | M1 allow use of 6.76 and 5.42; A1 awrt 39.4 or 6.28 |
| \((t_{13}=)(\pm)\frac{14.8-7.2-5}{s_p\sqrt{\frac{1}{9}+\frac{1}{6}}} = (\pm)0.78578...\) | M1 M1d A1 | M1 use of correct formula with their \(S_p\) (condone missing 5); M1d use of correct formula; awrt 0.786 |
| Critical value \(t_{13}(2.5\%) = 1.771\) | B1 | |
| There is no evidence to Reject \(H_0\). There is evidence that the fire brigade in \(X\) does not take more than 5 minutes longer than those in \(Y\). | A1cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Test in part (b) requires the variances to be equal. The test in part (a) showed that the variances could be assumed to be equal. | B1 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \sigma^2_X = \sigma^2_Y \quad H_1: \sigma^2_X \neq \sigma^2_Y$ | B1 | Both hypotheses |
| $F_{8,5} = \frac{6.76^2}{5.42^2} = 1.556$ | M1 A1 | M1 allow use of 6.76 and 5.42 instead of $6.76^2$ and $5.42^2$; A1 awrt 1.56 |
| $F_{8,5}$ is 4.82 | B1 | Allow p value 0.650 instead of critical value |
| There is evidence that the variances are the same | A1 | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \mu_X = \mu_Y + 5 \quad H_1: \mu_X > \mu_Y + 5$ | B1 | Both hypotheses |
| $s_p^2 = \frac{8\times 6.76^2 + 5\times 5.42^2}{13} = 39.42...$ or $s_p = 6.278...$ | M1 A1 | M1 allow use of 6.76 and 5.42; A1 awrt 39.4 or 6.28 |
| $(t_{13}=)(\pm)\frac{14.8-7.2-5}{s_p\sqrt{\frac{1}{9}+\frac{1}{6}}} = (\pm)0.78578...$ | M1 M1d A1 | M1 use of correct formula with their $S_p$ (condone missing 5); M1d use of correct formula; awrt 0.786 |
| Critical value $t_{13}(2.5\%) = 1.771$ | B1 | |
| There is no evidence to Reject $H_0$. There is evidence that the fire brigade in $X$ does not take more than 5 minutes longer than those in $Y$. | A1cso | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Test in part (b) requires the variances to be equal. The test in part (a) showed that the variances could be assumed to be equal. | B1 | |
---5. Fire brigades in cities $X$ and $Y$ are in similar locations. The response times, in minutes, during a particular month, for randomly selected calls are summarised in the table below.
\begin{center}
\begin{tabular}{ | c | c | c | c | }
\cline { 2 - 4 }
\multicolumn{1}{c|}{} & Sample size & Sample mean & \begin{tabular}{ c }
Standard deviation \\
$S$ \\
\end{tabular} \\
\hline
$X$ & 9 & 14.8 & 6.76 \\
\hline
$Y$ & 6 & 7.2 & 5.42 \\
\hline
\end{tabular}
\end{center}
You may assume that the response times are from independent normal distributions.\\
Stating your hypotheses and showing your working clearly
\begin{enumerate}[label=(\alph*)]
\item test, at the $10 \%$ level of significance, whether or not the variances of the populations from which the response times are drawn are the same,\\
(5)
\item test, at the $5 \%$ level of significance, whether or not the mean response time for the fire brigade in city $X$ is more than 5 minutes longer than the mean response time for the fire brigade in city $Y$.
\item Explain why your result in part (a) enables you to carry out the test in part (b).
\end{enumerate}
\hfill \mbox{\textit{Edexcel S4 2016 Q5 [14]}}