| Exam Board | Edexcel |
|---|---|
| Module | S4 (Statistics 4) |
| Year | 2010 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Paired sample t-test |
| Difficulty | Standard +0.8 This is a Further Maths S4 paired t-test requiring students to calculate differences, find sample statistics, formulate one-sided hypotheses with a specific value (μ_d > 5, not just > 0), and conduct the test. The non-zero null hypothesis value adds complexity beyond standard paired t-tests, and S4 material is inherently more advanced, but the calculation itself follows a standard procedure once set up correctly. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Person | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| Heart rate lying down | 66 | 70 | 59 | 65 | 72 | 66 | 62 | 69 | 56 | 68 |
| Heart rate standing up | 75 | 76 | 63 | 67 | 80 | 75 | 65 | 74 | 63 | 75 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The differences in the mean heart rates are normally distributed | B1 | Must have "The differences in (mean heart rate) are normally distributed" (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(D =\) standing up \(-\) lying down | ||
| \(H_0: \mu_D = 5\), \(H_1: \mu_D > 5\) | B1 | Both correct; allow \(\mu_D - 5 > 0\) (\(\mu_D = -5\), \(H_1: \mu_D < -5\)) |
| \(d\): 9, 6, 4, 2, 8, 9, 3, 5, 7, 7 | M1 | M1 finding differences |
| \(\bar{d} = 6\); \(s_d = \sqrt{\frac{414 - 10 \times 36}{9}} = 2.45\) | M1; M1 | M1 finding \(\bar{d}\); M1 for \(\sqrt{\frac{\sum d^2 - 10\times(\bar{d})^2}{9}}\) o.e. |
| \(t_9 = \frac{6-5}{\frac{2.45}{\sqrt{10}}} = 1.29\) | M1A1 | M1 need to see full expression with numbers; A1 awrt \(\pm 1.29\) |
| \(t_9(5\%) = 1.833\) | B1 | B1 \(\pm 1.833\) only |
| Insignificant. There is no evidence to suggest that heart rate rises by more than 5 beats when standing up | A1ft | ft their CV and t; need context: heart rate and 5 beats (8) |
# Question 2:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| The differences in the mean heart rates are normally distributed | B1 | Must have "The differences in (mean heart rate) are normally distributed" (1) |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $D =$ standing up $-$ lying down | | |
| $H_0: \mu_D = 5$, $H_1: \mu_D > 5$ | B1 | Both correct; allow $\mu_D - 5 > 0$ ($\mu_D = -5$, $H_1: \mu_D < -5$) |
| $d$: 9, 6, 4, 2, 8, 9, 3, 5, 7, 7 | M1 | M1 finding differences |
| $\bar{d} = 6$; $s_d = \sqrt{\frac{414 - 10 \times 36}{9}} = 2.45$ | M1; M1 | M1 finding $\bar{d}$; M1 for $\sqrt{\frac{\sum d^2 - 10\times(\bar{d})^2}{9}}$ o.e. |
| $t_9 = \frac{6-5}{\frac{2.45}{\sqrt{10}}} = 1.29$ | M1A1 | M1 need to see full expression with numbers; A1 awrt $\pm 1.29$ |
| $t_9(5\%) = 1.833$ | B1 | B1 $\pm 1.833$ only |
| Insignificant. There is no evidence to suggest that heart rate rises by more than 5 beats when standing up | A1ft | ft their CV and t; need context: heart rate and 5 beats (8) |
---\begin{enumerate}
\item As part of an investigation, a random sample of 10 people had their heart rate, in beats per minute, measured whilst standing up and whilst lying down. The results are summarised below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | c | c | c | c | c | }
\hline
Person & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\
\hline
Heart rate lying down & 66 & 70 & 59 & 65 & 72 & 66 & 62 & 69 & 56 & 68 \\
\hline
Heart rate standing up & 75 & 76 & 63 & 67 & 80 & 75 & 65 & 74 & 63 & 75 \\
\hline
\end{tabular}
\end{center}
(a) State one assumption that needs to be made in order to carry out a paired $t$-test.\\
(b) Test, at the $5 \%$ level of significance, whether or not there is any evidence that standing up increases people's mean heart rate by more than 5 beats per minute. State your hypotheses clearly.
\hfill \mbox{\textit{Edexcel S4 2010 Q2 [9]}}