| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Two-sample t-test equal variance |
| Difficulty | Standard +0.3 This is a standard S4 two-sample hypothesis testing question requiring an F-test for equal variances followed by a confidence interval for difference in means. The procedures are routine and well-practiced at this level, though the calculations involve multiple steps. Part (c) requires basic understanding of experimental design. Slightly easier than average due to being a textbook application of standard techniques with clear guidance. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution |
| Sample size \(n\) | Standard deviation \(s\) | Mean \(\bar { x }\) | |
| With background music | 8 | 4.1 | 15.9 |
| Without background music | 7 | 5.2 | 17.9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_0: \sigma_1^2 = \sigma_2^2\), \(H_1: \sigma_1^2 \neq \sigma_2^2\) | B1 | Allow \(\sigma_1 = \sigma_2\) and \(\sigma_1 \neq \sigma_2\) |
| Critical values \(F_{6,7} = 3.87\) \(\left(\frac{1}{F_{6,7}} = 0.258\right)\) | B1 | Must match their F |
| \(\frac{s_2^2}{s_1^2} = \frac{5.2^2}{4.1^2} = 1.61\) \(\left(\frac{s_1^2}{s_2^2} = \frac{4.1^2}{5.2^2} = 0.622\right)\) | M1; A1 | M1 for \(\frac{s_2^2}{s_1^2}\) or other way up; A1 awrt 1.61 (0.622) |
| Since 1.61 (0.622) is not in the critical region, accept \(H_0\); no evidence the two variances are different | A1ft | (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S_p^2 = \frac{7 \times 4.1^2 + 6 \times 5.2^2}{7+6} = 21.53\ldots\) | M1A1 | M1 A1 \(S_p^2\) may be seen in part (a) |
| \(t_{13} = 3.012\) | B1 | B1 3.012 only |
| \(99\%\) CI \(= (17.9 - 15.9) \pm 3.012 \times \sqrt{21.53} \times \sqrt{\frac{1}{8}+\frac{1}{7}}\) | M1A1ft | M1 for \((17.9-15.9) \pm t\text{ value} \times \sqrt{S_p^2} \times \sqrt{\frac{1}{8}+\frac{1}{7}}\); A1ft their \(S_p^2\) |
| \(= \pm(9.23,\ -5.233)\) [or accept: \([0, 9.23]\) or \([-9.23, 0]\)] awrt 9.23, \(-5.23\) | A1A1 | (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A person will be quicker at the task second time through / times not independent / familiar with the task / groups are not independent | B1 | Any correct sensible comment (1) |
# Question 1:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_0: \sigma_1^2 = \sigma_2^2$, $H_1: \sigma_1^2 \neq \sigma_2^2$ | B1 | Allow $\sigma_1 = \sigma_2$ and $\sigma_1 \neq \sigma_2$ |
| Critical values $F_{6,7} = 3.87$ $\left(\frac{1}{F_{6,7}} = 0.258\right)$ | B1 | Must match their F |
| $\frac{s_2^2}{s_1^2} = \frac{5.2^2}{4.1^2} = 1.61$ $\left(\frac{s_1^2}{s_2^2} = \frac{4.1^2}{5.2^2} = 0.622\right)$ | M1; A1 | M1 for $\frac{s_2^2}{s_1^2}$ or other way up; A1 awrt 1.61 (0.622) |
| Since 1.61 (0.622) is not in the critical region, accept $H_0$; no evidence the two variances are different | A1ft | (5) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S_p^2 = \frac{7 \times 4.1^2 + 6 \times 5.2^2}{7+6} = 21.53\ldots$ | M1A1 | M1 A1 $S_p^2$ may be seen in part (a) |
| $t_{13} = 3.012$ | B1 | B1 3.012 only |
| $99\%$ CI $= (17.9 - 15.9) \pm 3.012 \times \sqrt{21.53} \times \sqrt{\frac{1}{8}+\frac{1}{7}}$ | M1A1ft | M1 for $(17.9-15.9) \pm t\text{ value} \times \sqrt{S_p^2} \times \sqrt{\frac{1}{8}+\frac{1}{7}}$; A1ft their $S_p^2$ |
| $= \pm(9.23,\ -5.233)$ [or accept: $[0, 9.23]$ or $[-9.23, 0]$] awrt 9.23, $-5.23$ | A1A1 | (7) |
## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| A person will be quicker at the task second time through / times not independent / familiar with the task / groups are not independent | B1 | Any correct sensible comment (1) |
---\begin{enumerate}
\item A teacher wishes to test whether playing background music enables students to complete a task more quickly. The same task was completed by 15 students, divided at random into two groups. The first group had background music playing during the task and the second group had no background music playing.\\
The times taken, in minutes, to complete the task are summarised below.
\end{enumerate}
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
& Sample size $n$ & Standard deviation $s$ & Mean $\bar { x }$ \\
\hline
With background music & 8 & 4.1 & 15.9 \\
\hline
Without background music & 7 & 5.2 & 17.9 \\
\hline
\end{tabular}
\end{center}
You may assume that the times taken to complete the task by the students are two independent random samples from normal distributions.\\
(a) Stating your hypotheses clearly, test, at the $10 \%$ level of significance, whether or not the variances of the times taken to complete the task with and without background music are equal.\\
(b) Find a 99\% confidence interval for the difference in the mean times taken to complete the task with and without background music.
Experiments like this are often performed using the same people in each group.\\
(c) Explain why this would not be appropriate in this case.\\
\hfill \mbox{\textit{Edexcel S4 2010 Q1 [13]}}