| Exam Board | Edexcel |
|---|---|
| Module | S1 (Statistics 1) |
| Year | 2017 |
| Session | October |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear regression |
| Type | Calculate variance from summations |
| Difficulty | Moderate -0.8 This is a routine S1 statistics question testing standard formulas for mean, variance, correlation coefficient, and linear regression prediction. All parts involve direct substitution into well-rehearsed formulas with no conceptual challenges or problem-solving required. The calculations are straightforward with summations provided. |
| Spec | 2.02g Calculate mean and standard deviation2.02h Recognize outliers2.05f Pearson correlation coefficient2.05g Hypothesis test using Pearson's r5.09a Dependent/independent variables5.09b Least squares regression: concepts5.09c Calculate regression line5.09e Use regression: for estimation in context |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\bar{x} = \left[\frac{283}{10}\right] = \mathbf{28.3}\) | B1 |
| \(\sigma_x^2 = \frac{9011}{10} - 28.3^2\) | M1 | M1 for a correct expression for variance (no \(\sqrt{}\)) but ft their 28.3 |
| \(= 100.21\) accept awrt \(\mathbf{100}\) | A1 | A1 for awrt 100 |
| (b) | \(\bar{y} = \mathbf{30.61}\) (allow 30.6) | B1 |
| \(\sigma_y^2 = \mathbf{54.63}\) (allow 54.6) | B1 | |
| (c) | \(0.659 = \frac{S_{xy}}{S_{xx}}\) and \(S_{xx} = 10\sigma_x^2 [= 1002.1]\) | M1M1 |
| \(S_{xy} = 0.659 \times 1002.1 [= 660.3839\) (659-660.4)] | A1 | 1st A1 for a value for \(S_{xy}\) in (659, 660.4) or \(0.659 \times 1002.1\) |
| \(r = \frac{660.3839}{\sqrt{1002.1 \times 546.3}}\) | M1 | 3rd M1 for a correct expression for \(r\) (ft their \(S_{xx}\) and their \(S_{xy}\)) |
| \(= \) awrt \(\mathbf{0.892-0.893}\) | A1 | 2nd A1 for awrt 0.892 or 0.893 |
| (d) | Value of \(r\) is close to 1 so it does support the use of a linear regression model | B1 |
| (e) | \(y = 12.0 + 0.659 \times 35 = [35.065]\) Proposed salary is \(\mathbf{\\) 35 065}\( (awrt \)\mathbf{\\( 35 100}\)) | M1, A1 |
| **(a)** | $\bar{x} = \left[\frac{283}{10}\right] = \mathbf{28.3}$ | B1 | |
|---|---|---|---|
| | $\sigma_x^2 = \frac{9011}{10} - 28.3^2$ | M1 | M1 for a correct expression for variance (no $\sqrt{}$) but ft their 28.3 |
| | $= 100.21$ accept awrt $\mathbf{100}$ | A1 | A1 for awrt 100 |
| **(b)** | $\bar{y} = \mathbf{30.61}$ (allow 30.6) | B1 | |
| | $\sigma_y^2 = \mathbf{54.63}$ (allow 54.6) | B1 | |
| **(c)** | $0.659 = \frac{S_{xy}}{S_{xx}}$ and $S_{xx} = 10\sigma_x^2 [= 1002.1]$ | M1M1 | 1st M1 for a correct expression using $b = 0.659$ that connects this value with $S_{xy}$ and $S_{xy}$ 2nd M1 for correct method for $S_{xx}$ (value not required) |
| | $S_{xy} = 0.659 \times 1002.1 [= 660.3839$ (659-660.4)] | A1 | 1st A1 for a value for $S_{xy}$ in (659, 660.4) or $0.659 \times 1002.1$ |
| | $r = \frac{660.3839}{\sqrt{1002.1 \times 546.3}}$ | M1 | 3rd M1 for a correct expression for $r$ (ft their $S_{xx}$ and their $S_{xy}$) |
| | $= $ awrt $\mathbf{0.892-0.893}$ | A1 | 2nd A1 for awrt 0.892 or 0.893 |
| **(d)** | Value of $r$ is close to 1 so it **does support** the use of a linear regression model | B1 | if $0.5 < \mid r \mid < 1$ for saying that it **does support** and giving a suitable comment, e.g. "strong" correlation, about $r$ being close to 1 (or $-1$) $\mid r \mid > 1$ is B0 If $\mid r \mid < 0.5$ allow it does not support with supporting comment about $r$ close to 0 |
| **(e)** | $y = 12.0 + 0.659 \times 35 = [35.065]$ Proposed salary is $\mathbf{\$ 35 065}$ (awrt $\mathbf{\$ 35 100}$) | M1, A1 | M1 for substituting 35 into the given regression equation (may be implied by 35.065) A1 for 35 065 (i.e. a correct value and multiplying by 1000) **must be at least 3sf** Accept "35.065 thousand dollars" $ 35 000 is A0 (could have come from the 35 in the question) |
**[Total 13]**
---\begin{enumerate}
\item A company wants to pay its employees according to their performance at work. Last year's performance score $x$ and annual salary $y$, in thousands of dollars, were recorded for a random sample of 10 employees of the company.
\end{enumerate}
The performance scores were
$$\begin{array} { l l l l l l l l l l }
15 & 24 & 32 & 39 & 41 & 18 & 16 & 22 & 34 & 42
\end{array}$$
(You may use $\sum x ^ { 2 } = 9011$ )\\
(a) Find the mean and the variance of these performance scores.
The corresponding $y$ values for these 10 employees are summarised by
$$\sum y = 306.1 \quad \text { and } \quad \mathrm { S } _ { y y } = 546.3$$
(b) Find the mean and the variance of these $y$ values.
The regression line of $y$ on $x$ based on this sample is
$$y = 12.0 + 0.659 x$$
(c) Find the product moment correlation coefficient for these data.\\
(d) State, giving a reason, whether or not the value of the product moment correlation coefficient supports the use of a regression line to model the relationship between performance score and annual salary.
The company decides to use this regression model to determine future salaries.\\
(e) Find the proposed annual salary, in dollars, for an employee who has a performance score of 35
\hfill \mbox{\textit{Edexcel S1 2017 Q5 [13]}}