Question

The time it takes for a planet to complete its orbit around a particular star is called the? planet's sidereal year. The sidereal year of a planet is related to the distance the planet is from the star. The accompanying data show the distances of the planets from a particular star and their sidereal years. Complete parts? (a) through? (e).
I figured out what
(a) is already.
(b) Determine the correlation between distance and sidereal year.
(c) Compute the? least-squares regression line.
(d) Plot the residuals against the distance from the star.
(e) Do you think the? least-squares regression line is a good? model?
Planet
Distance from the? Star, x?(millions of? miles)
Sidereal? Year, y
Planet 1
36
0.22
Planet 2
67
0.62
Planet 3
93
1.00
Planet 4
142
1.86
Planet 5
483
11.8
Planet 6
887
29.5
Planet 7
? 1,785
84.0
Planet 8
? 2,797
165.0
Planet 9
?3,675
248.0

Answer 1

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  

$r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}$

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

$r = \dfrac{\sum{xy} - \sum{x} \sum{y}}{\sqrt{\left [n\sum{x}^{2}-\left (\sum{x}\right )^{2}\right]\left [n\sum{y}^{2} -\left (\sum{y}\right )^{2}\right]}}\\\\= \dfrac{9\times 1545776.75 - 9965\times 537.81}{\sqrt{[9\times 25569715 -9965^{2}][9\times 95799.63 - 537.81^{2}]}} \approx \mathbf{0.9879}$

(c) Regression line

The equation for the regression line is

y = a + bx where

$a = \dfrac{\sum y \sum x^{2} - \sum x \sum xy}{n\sum x^{2}- \left (\sum x\right )^{2}}\\\\= \dfrac{537.81\times 25569715 - 9965 \times 1545776.75}{9\times 25569715 - 9965^{2}} \approx \mathbf{-12.629}\\\\b = \dfrac{n \sum xy - \sum x \sum y}{n\sum x^{2}- \left (\sum x\right )^{2}} - \dfrac{9\times 1545776.75 - 9965 \times 537.81}{9\times 25569715 - 9965^{2}} \approx\mathbf{0.0654}\\\\\\\text{The equation for the regression line is \large \boxed{\mathbf{y = -12.629 + 0.0654x}} }$

(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.