Chi Squared Tests for Standard Distributions (Edexcel A-Level Further Mathematics): Revision Notes

Example Given that the following data is known to be linearly correlated, calculate the regression line of y on $x$:

$(x)$ Age of tree (years)	7	24	12	19
$(y)$ Height of tree (m)	2.4	6	5	5.8

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Step 1: $y$ on $x$ means "$y$ depends on $x$." Decide which variable depends on the other and assign these to be $y$ and $x$, respectively.

$$\text{Height depends on age} \quad \Rightarrow \quad y \text{ on } x$$

The reason for this is that there is an assumption that any variability (deviation from the line of best fit) only occurs for the dependent $(y)$ variable.

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Step 2: Assigning the relevant variables to be $y$ and $x$, we use the least squares regression line formula given in the booklet:

The regression coefficient of $y$ on $x$ is: $b = \dfrac{S_{xy}}{S_{xx}}$

Least squares regression line of $y$ on $x$ is: $y = a + bx \quad \text{where} \quad a = \bar{y} - b\bar{x}$

$$\begin{array}{c|c|c|c|c} x & y & x^2 & y^2 & xy \\ \hline 7 & 2.4 & 49 & 5.76 & 16.8 \\ 24 & 6 & 576 & 36 & 144 \\ 12 & 5 & 144 & 25 & 60 \\ 19 & 5.8 & 361 & 33.64 & 110.2 \\ \end{array}$$

Summing up the values:

$$\sum x = 62, \quad \sum y = 19.2, \quad \sum x^2 = 1130, \quad \sum y^2 = 100.4, \quad \sum xy = 331$$

Now:

$$S_{xy} = 331 - \frac{62 \times 19.2}{4} = 33.4$$$$S_{xx} = 1130 - \frac{62^2}{4} =169$$

So:

$$b = \frac{33.4}{169} = \frac{167}{845}$$

Now for $a$:

$$\overline {x}= \frac {62}{4}=15.5 \quad | \quad \overline {y}= \frac {19.2}{4}=4.8$$$$\Rightarrow a = 4.8 - \frac{167}{845} \times 15.5 = \frac{587}{338}$$

Thus the equation of the regression line is:

$$\therefore y=a+bx$$$$\boxed {y = \frac{587}{338} + \frac{167}{845}x}$$

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Least Squares Regression Line on a Calculator:

$$\Rightarrow y = 1.74 + 0.198x$$

Unless required to show full working, do this.

Example: Research was done to see if there is a relationship between finger dexterity and the ability to do work on a production line. The data is shown in the table.

Dexterity score, $x$	2.5	3	3.5	4	5	5	5.5	6.5	7	8
Productivity, $y$	80	130	100	220	190	210	270	290	350	400

The equation of the regression line for these data is $y = -59 + 57x$.

a. Use the equation to estimate the productivity of someone with a dexterity of $6$.

b. Give an interpretation of the value of $57$ in the equation of the regression line.

c. State, giving in each case a reason, whether or not it would be reasonable to use this equation to work out the productivity of someone with dexterity of:

i) 2 ii) 14

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Answers:

a. Let x = 6

$$\Rightarrow y = -59 + 57(6) = 283$$

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b. As the dexterity increases by $1$, the productivity increases by $57$. (Give answer in context.)

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c. i) Reasonable as it is close to the range of data we have.

ii) Unreasonable as it is well outside of the range of data we have. (Extrapolation is unreliable.)