| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2015 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Identify response/explanatory variables |
| Difficulty | Easy -1.3 This is a straightforward S1 regression question testing basic concepts: identifying variables (trivial), interpreting gradient (recall definition), calculating means (arithmetic), verifying a point lies on a line (substitution), making a prediction (substitution), and a standard normal distribution calculation. All parts are routine textbook exercises requiring no problem-solving or insight. |
| Spec | 2.02c Scatter diagrams and regression lines2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation |
| \(t\) | 20 | 35 | 40 | 25 | 45 | 70 | 75 | 90 |
| \(h\) | 88 | 85 | 77 | 75 | 71 | 66 | 60 | 54 |
| Answer | Marks | Guidance |
|---|---|---|
| Resting heart rate \(h\) is being measured (can't control it), so it is the response variable | B1, dB1 | 1st B1 for reason not using "response/explanatory"; 2nd dB1 dep. on 1st B1 for choosing \(h\) as response variable |
| Answer | Marks | Guidance |
|---|---|---|
| For every additional minute of exercise, heart rate decreases by 0.43 (bpm) | B1 | Need mention of "exercise" plus unit of time and "heart rate/beats" with correct value |
| Answer | Marks | Guidance |
|---|---|---|
| \(\bar{t} = 50\), \(\bar{h} = 72\) | B1 B1 | 1st B1 for 50, 2nd B1 for 72 |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = 93.5 - 0.43(50)\), so \(h=72\) | B1cso | Allow \(72 = 93.5 - 0.43\times50\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(h = 93.5 - 0.43(60) = \mathbf{67.7}\) | B1 | Allow 68 if correct expression seen |
| Answer | Marks | Guidance |
|---|---|---|
| Since 1 hour (60 minutes) is within the range of the \(t\)-values, the estimate is reliable | B1, dB1 | 1st B1 for reason; 2nd dB1 dep. on 1st B1 for saying it is reliable |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{a-73}{8} = -1.96\) or \(\frac{b-73}{8} = 1.96\); \(73\pm1.96\times8\); \((57.32,\ 88.68)\) | M1 B1, dM1, A1 | M1 for standardising; B1 for 1.96 used as \(z\)-value; dM1 for rearranging to find \(a\) or \(b\); A1 for both awrt 57.3 and awrt 88.7 |
# Question 5:
## Part (a)
| Resting heart rate $h$ is being measured (can't control it), so it is the response variable | B1, dB1 | 1st B1 for reason not using "response/explanatory"; 2nd dB1 dep. on 1st B1 for choosing $h$ as response variable |
## Part (b)
| For every additional minute of exercise, heart rate decreases by 0.43 (bpm) | B1 | Need mention of "exercise" plus unit of time and "heart rate/beats" with correct value |
## Part (c)
| $\bar{t} = 50$, $\bar{h} = 72$ | B1 B1 | 1st B1 for 50, 2nd B1 for 72 |
## Part (d)
| $h = 93.5 - 0.43(50)$, so $h=72$ | B1cso | Allow $72 = 93.5 - 0.43\times50$ |
## Part (e)
| $h = 93.5 - 0.43(60) = \mathbf{67.7}$ | B1 | Allow 68 if correct expression seen |
## Part (f)
| Since 1 hour (60 minutes) is within the range of the $t$-values, the estimate is reliable | B1, dB1 | 1st B1 for reason; 2nd dB1 dep. on 1st B1 for saying it is reliable |
## Part (g)
| $\frac{a-73}{8} = -1.96$ or $\frac{b-73}{8} = 1.96$; $73\pm1.96\times8$; $(57.32,\ 88.68)$ | M1 B1, dM1, A1 | M1 for standardising; B1 for 1.96 used as $z$-value; dM1 for rearranging to find $a$ or $b$; A1 for both awrt 57.3 and awrt 88.7 |
---\begin{enumerate}
\item The resting heart rate, $h$ beats per minute (bpm), and average length of daily exercise, $t$ minutes, of a random sample of 8 teachers are shown in the table below.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | l | l | l | l | l | l | l | l | }
\hline
$t$ & 20 & 35 & 40 & 25 & 45 & 70 & 75 & 90 \\
\hline
$h$ & 88 & 85 & 77 & 75 & 71 & 66 & 60 & 54 \\
\hline
\end{tabular}
\end{center}
(a) State, with a reason, which variable is the response variable.
The equation of the least squares regression line of $h$ on $t$ is
$$h = 93.5 - 0.43 t$$
(b) Give an interpretation of the gradient of this regression line.\\
(c) Find the value of $\bar { t }$ and the value of $\bar { h }$\\
(d) Show that the point $( \bar { t } , \bar { h } )$ lies on the regression line.\\
(e) Estimate the resting heart rate of a teacher with an average length of daily exercise of 1 hour.\\
(f) Comment, giving a reason, on the reliability of the estimate in part (e).
The resting heart rate of teachers is assumed to be normally distributed with mean 73 bpm and standard deviation 8 bpm .
The middle $95 \%$ of resting heart rates of teachers lies between $a$ and $b$\\
(g) Find the value of $a$ and the value of $b$.\\
\hfill \mbox{\textit{Edexcel S1 2015 Q5 [13]}}