Answer:
(a) See below
(b) r = 0.9879
(c) y = -12.629 + 0.0654x
(d) See below
(e) No.
Step-by-step explanation:
(a) Plot the data
I used Excel to plot your data and got the graph in Fig 1 below.
(b) Correlation coefficient
One formula for the correlation coefficient is
[tex]r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}[/tex]
The calculation is not difficult, but it is tedious.
(i) Calculate the intermediate numbers
We can display them in a table.
x y xy x² y²
36 0.22 7.92 1296 0.05
67 0.62 42.21 4489 0.40
93 1.00 93.00 20164 3.46
433 11.8 5699.4 233289 139.24
887 29.3 25989.1 786769 858.49
1785 82.0 146370 3186225 6724
2797 163.0 455911 7823209 26569
3675 248.0 911400 13505625 61504
9965 537.81 1545776.75 25569715 95799.63
(ii) Calculate the correlation coefficient
[tex]r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}[/tex]
(c) Regression line
The equation for the regression line is
y = a + bx where
[tex]a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} - \dfrac{9\times 1545776.75 - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is $\large \boxed{\mathbf{y = -12.629 + 0.0654x}}$}[/tex]
(d) Residuals
Insert the values of x into the regression equation to get the estimated values of y.
Then take the difference between the actual and estimated values to get the residuals.
x y Estimated Residual
36 0.22 -10 10
67 0.62 -8 9
93 1.00 -7 8
142 1.86 -3 5
433 11.8 19 - 7
887 29.3 45 -16
1785 82.0 104 -22
2797 163.0 170 - 7
3675 248.0 228 20
(e) Suitability of regression line
A linear model would have the residuals scattered randomly above and below a horizontal line.
Instead, they appear to lie along a parabola (Fig. 2).
This suggests that linear regression is not a good model for the data.
