The number of years (xx) an employee has worked for the company and the corresponding salary increment, in dollars (yy), received by the employee are given in the table.
Years, xx67911131518
Amount, yy155170211230248260265

  1. (i)

    Draw the scatter diagram for these values, labelling the axes clearly.

  2. (ii)

    It is thought that the salary increment, $y, can be modelled by one of the formulae
    y=ax+bory=cln⁡x+dy = ax + b \quad \text{or} \quad y = c \ln x + d
    where aa, bb, cc and dd are constants.

    Find, correct to 4 decimal places, the value of the product moment correlation coefficient between
    (a) xx and yy,
    (b) ln⁡x\ln x and yy.

  3. (iii)

    Use your answers to parts (i) and (ii) to explain which of y=ax+by = ax + b or y=cln⁡x+dy = c \ln x + d is the better model.

  4. (iv)

    It is required to estimate the value of xx for which y=200y = 200.

    Explain why neither the regression line of xx on yy nor the regression line of ln⁡x\ln x on yy should be used.

  5. (v)

    Find the equation of a suitable regression line and use it to find the required estimate, commenting on its reliability.

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Worked solution

7(i)

1. Set up and label the axes with appropriate scales

Draw a horizontal axis labelled 'Years, xx' covering values from at least 66 to 1818, and a vertical axis labelled 'Amount, yy' covering values from at least 155155 to 265265. A scale break may be used on the yy-axis.

2. Plot the given data points

Accurately plot the following coordinates on the grid:
(6,155)(6, 155)
(7,170)(7, 170)
(9,211)(9, 211)
(11,230)(11, 230)
(13,248)(13, 248)
(15,260)(15, 260)
(18,265)(18, 265)

Answer

[A scatter diagram with the horizontal axis labelled 'Years, xx' and the vertical axis labelled 'Amount, yy', containing 7 accurately plotted points.]

7(ii)

1. Find the product moment correlation coefficient for $x$ and $y$.

By entering the given values of xx and yy into the statistics mode of a graphing calculator,
r=0.94827…r = 0.94827\dots
r≈0.9483r \approx 0.9483 (correct to 4 d.p.)

2. Find the product moment correlation coefficient for $\ln x$ and $y$.

By applying the natural logarithm to the values of xx, enter the paired values of ln⁡x\ln x and yy into the graphing calculator,
r=0.98494…r = 0.98494\dots
r≈0.9849r \approx 0.9849 (correct to 4 d.p.)

Answer

(a) 0.94830.9483
(b) 0.98490.9849

7(iii)

1. Compare the product moment correlation coefficients calculated in part (ii).

Calculations from part (ii):
Using the data table and a calculator, the product moment correlation coefficients are:

  • For xx and yy: r≈0.9483r \approx 0.9483
  • For ln⁡x\ln x and yy: r≈0.9849r \approx 0.9849

The correlation coefficient rr for ln⁡x\ln x and yy is closer to 11 than that for xx and yy, indicating a stronger positive linear relationship between ln⁡x\ln x and yy.

2. Analyse the shape of the scatter diagram from part (i) to conclude the best model.

Observation from part (i):
Based on the table of values, plotting yy against xx shows that as xx increases, the rate of increase of yy visibly decreases (the trend curves downwards rather than maintaining a constant gradient). A logarithmic function models this shape much better than a straight line.

Since both the scatter diagram and the correlation coefficients support a non-linear log relationship, y=cln⁡x+dy = c \ln x + d is the better model.

Answer

y=cln⁡x+dy = c \ln x + d

7(iv)

1. State the relationship between the variables and explain why the given regression lines are inappropriate

The number of years, xx, is the independent (or controlled) variable, and the salary increment, yy, is the dependent variable. Therefore, the regression line of the dependent variable on the independent variable (i.e. yy on xx, or yy on ln⁡x\ln x) must be used, even when estimating xx from yy.

Answer

xx is the independent variable.

7(v)

1. Identify and find the equation of the suitable regression line.

Since xx (years) is the predetermined independent variable, we must regress the dependent variable yy on xx. Based on part (iii), the logarithmic model is the better fit. Therefore, the suitable regression line is the regression line of yy on ln⁡x\ln x.

Using a graphing calculator with the given data points:
y=106.559ln⁡x−31.263y = 106.559\ln x - 31.263
y=107ln⁡x−31.3 (to 3 s.f.)y = 107 \ln x - 31.3 \text{ (to 3 s.f.)}

2. Use the equation to find the required estimate for $x$.

Substitute y=200y = 200 into the unrounded regression equation:
200=106.559ln⁡x−31.263200 = 106.559 \ln x - 31.263
106.559ln⁡x=231.263106.559 \ln x = 231.263
ln⁡x≈2.1703\ln x \approx 2.1703
x=e2.1703≈8.76 (to 3 s.f.)x = e^{2.1703} \approx 8.76 \text{ (to 3 s.f.)}

3. Comment on the reliability of the estimate.

The value y=200y = 200 falls within the given data range for yy (155≤y≤265155 \le y \le 265), meaning this is an interpolation. Furthermore, the correlation coefficient r≈0.985r \approx 0.985 is close to 1, indicating a strong fit for the model. Therefore, the estimate is reliable.

Answer

Equation: y=107ln⁡x−31.3y = 107 \ln x - 31.3
Estimate: x≈8.76x \approx 8.76
Reliability: The estimate is reliable because y=200y=200 is within the data range (interpolation) and there is a strong correlation.

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