| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2016 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate y on x from raw data table |
| Difficulty | Moderate -0.8 This is a straightforward S1 linear regression question requiring standard formula application with given summary statistics. Parts (a)-(d) are routine calculations using memorized formulas (Sdw, PMCC, regression line), while part (e) involves simple substitution and a basic interpolation/extrapolation comment. No problem-solving insight required, just methodical application of standard techniques. |
| Spec | 2.02c Scatter diagrams and regression lines2.02g Calculate mean and standard deviation5.08a Pearson correlation: calculate pmcc5.08d Hypothesis test: Pearson correlation5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Distance, \(\boldsymbol { d } \mathbf { m }\) | 30 | 50 | 80 | 100 | 150 | 400 | 500 | 650 | ||
| 0.725 | 1.210 | 1.775 | 2.250 | 3.518 | 6.382 | 8.185 | 9.555 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{dw} = 13833 - \frac{"1960" \times 33.6}{8}\) or \(13833 - \frac{65856}{8}\) | M1 | Clear attempt to find \(\Sigma d\) and use in correct formula. Accept \(1300 < \Sigma d < 2500\). Single slip condoned e.g. using 1383 instead of 13833 |
| \(= \mathbf{5601}\) | A1 cso | For correct \(\Sigma d\) and 5601 only. Must see formula, so M1 must be scored |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(w\), since the number of wiggles depends on the distance, or \(w\) depends on \(d\) | B1 | Must select \(w\) and reason based on idea that \(w\) is dependent on \(d\). Allow "changes according to"/"is determined/affected by". Must mention \(w\) and \(d\). B0 for "\(w\) is measured" or "\(d\) is explanatory" or "\(w\) can't be controlled" |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(r = \frac{5601}{\sqrt{394600 \times 80.481}}\) awrt 0.994 | M1, A1 | M1 for correct expression (allow ft of incorrect \(S_{dw}\)). A1 for awrt 0.994. Answer only 2/2. Answer of 0.99 scores M1A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(b = \frac{5601}{394600} = 0.014194\ldots\) (awrt 0.014) | M1, A1 | 1st M1 for correct expression for \(b\) (ft incorrect \(S_{dw}\)). 1st A1 for awrt 0.014. No fractions. Must come from correct formula, not gradient of line from e.g. (650, 9.555) to (30, 0.725) |
| \(a = \frac{33.6}{8} - "0.01419..." \times \frac{"1960"}{8} = 4.2 - "0.01419..." \times 245\) | M1 | 2nd M1 for correct method for \(a\), following through their \(b\) and \(\Sigma d\) |
| \(\mathbf{w = 0.722 + 0.0142d}\) | A1 | 2nd A1 for correct equation with \(a =\) awrt 0.722 and \(b =\) awrt 0.0142. No fractions. Equation in \(x\) and \(y\) is A0. Answer only 4/4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| (i) \([0.722 + 0.0142 \times 350 =]\) awrt 5.7 or 5.6 | B1 | 1st B1 for awrt 5.7 or awrt 5.6 |
| (ii) Reliable since 350 m is in the range of the data | B1 | 2nd B1 for reason citing 350m or mentioning \(d\) is in range of data and stating reliable. Allow "Interpolation (not extrapolation) therefore reliable". Saying "5.7 is in the range" is B0. "Accurate" instead of "reliable" is B0. "Strong correlation" without mention of interpolation is B0. Apply ISW if correct comment seen |
# Question 1:
## Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{dw} = 13833 - \frac{"1960" \times 33.6}{8}$ or $13833 - \frac{65856}{8}$ | M1 | Clear attempt to find $\Sigma d$ and use in correct formula. Accept $1300 < \Sigma d < 2500$. Single slip condoned e.g. using 1383 instead of 13833 |
| $= \mathbf{5601}$ | A1 cso | For correct $\Sigma d$ and 5601 only. Must see formula, so M1 must be scored |
## Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $w$, since the number of wiggles depends on the distance, or $w$ depends on $d$ | B1 | Must select $w$ and reason based on idea that $w$ is dependent on $d$. Allow "changes according to"/"is determined/affected by". Must mention $w$ and $d$. B0 for "$w$ is measured" or "$d$ is explanatory" or "$w$ can't be controlled" |
## Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $r = \frac{5601}{\sqrt{394600 \times 80.481}}$ **awrt 0.994** | M1, A1 | M1 for correct expression (allow ft of incorrect $S_{dw}$). A1 for awrt 0.994. Answer only 2/2. Answer of 0.99 scores M1A0 |
## Part (d)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $b = \frac{5601}{394600} = 0.014194\ldots$ (awrt 0.014) | M1, A1 | 1st M1 for correct expression for $b$ (ft incorrect $S_{dw}$). 1st A1 for awrt 0.014. No fractions. Must come from correct formula, not gradient of line from e.g. (650, 9.555) to (30, 0.725) |
| $a = \frac{33.6}{8} - "0.01419..." \times \frac{"1960"}{8} = 4.2 - "0.01419..." \times 245$ | M1 | 2nd M1 for correct method for $a$, following through their $b$ and $\Sigma d$ |
| $\mathbf{w = 0.722 + 0.0142d}$ | A1 | 2nd A1 for correct equation with $a =$ awrt 0.722 and $b =$ awrt 0.0142. No fractions. Equation in $x$ and $y$ is A0. Answer only 4/4 |
## Part (e)
| Answer | Marks | Guidance |
|--------|-------|----------|
| (i) $[0.722 + 0.0142 \times 350 =]$ **awrt 5.7 or 5.6** | B1 | 1st B1 for awrt 5.7 or awrt 5.6 |
| (ii) Reliable since 350 m is in the range of the data | B1 | 2nd B1 for reason citing 350m or mentioning $d$ is in range of data and stating reliable. Allow "Interpolation (not extrapolation) therefore reliable". Saying "5.7 is in the range" is B0. "Accurate" instead of "reliable" is B0. "Strong correlation" without mention of interpolation is B0. Apply ISW if correct comment seen |\begin{enumerate}
\item A biologist is studying the behaviour of bees in a hive. Once a bee has located a source of food, it returns to the hive and performs a dance to indicate to the other bees how far away the source of the food is. The dance consists of a series of wiggles. The biologist records the distance, $d$ metres, of the food source from the hive and the average number of wiggles, $w$, in the dance.
\end{enumerate}
\begin{center}
\begin{tabular}{ | l | c | c | c | c | c | c | c | c | }
\hline
Distance, $\boldsymbol { d } \mathbf { m }$ & 30 & 50 & 80 & 100 & 150 & 400 & 500 & 650 \\
\hline
\begin{tabular}{ l }
Average number \\
of wiggles, $\boldsymbol { w }$ \\
\end{tabular} & 0.725 & 1.210 & 1.775 & 2.250 & 3.518 & 6.382 & 8.185 & 9.555 \\
\hline
\end{tabular}
\end{center}
[You may use $\sum w = 33.6 \sum d w = 13833 \mathrm {~S} _ { d d } = 394600 \mathrm {~S} _ { w w } = 80.481$ (to 3 decimal places)]\\
(a) Show that $\mathrm { S } _ { d w } = 5601$\\
(b) State, giving a reason, which is the response variable.\\
(c) Calculate the product moment correlation coefficient for these data.\\
(d) Calculate the equation of the regression line of $w$ on $d$, giving your answer in the form $w = a + b d$
A new source of food is located 350 m from the hive.\\
(e) (i) Use your regression equation to estimate the average number of wiggles in the corresponding dance.\\
(ii) Comment, giving a reason, on the reliability of your estimate.
\hfill \mbox{\textit{Edexcel S1 2016 Q1 [11]}}