Chapter 10, Problem 20E

(a)

To determine

To Explain: the distances to construct a model for the Distance from the sum on the basis of the planet’s position on the basis of revied information.

Explanation of Solution

Given:

  

Calculation:

Re-express the data on the basis of given information

Position	Distance from sun	Log (distance)
1	36	1.556302501
2	67	1.826074803
3	93	1.968482949
4	142	2.152288344
6	484	2.684845362
7	887	2.94792362
8	1784	3.25139485
9	2796	3.446537167
10	3707	3.569022586

Graph:

Scatter plot:

  

Residual Plot:

  

From the result of the equation of the regression line is

  $\log \left(\text{Distance}\right)=1.304+0.233\text{Position}$ , Therefore, this is the required model. R-squared of the model is 0.994, it implies that about 99.4 percent of the variability in the log (Distance) is estimated for by the model.

(b)

To determine

To Explain: on the basis of model, would agree with this international Astronomical union that Pluto is not Planet on the basis of revised information.

Explanation of Solution

Given:

Re-express the data on the basis of given information

Position	Distance from sun
1	36
2	67
3	93
4	142
6	484
7	887
8	1784
9	2796

Calculation:

Part (a) of the model does not provide enough evidence to support the argument of the declaration.About astronomy. Pluto seems to match the pattern well, but the distance of Pluto from the sun is slightly less.The distance was planned. Let us discover the distance between Pluto and the Sun predicted by the previous about model.

  $\begin{array}{c}\log \left(\text{Distance}\right)=1.304+0.233\text{Position}\\ \text{=}1.304+0.233\left(10\right)\\ =1.304+2.33\\ =3.634\end{array}$

Therefore, the expected distance of the Pluto is

  $\begin{array}{c}{\text{Distance=10}}^{3.634}\\ =4305.27\end{array}$

Creating without Pluto

  

  

Regression equation without the Pluto is

  $\log \left(\text{Distance}\right)=1.285+0.239\text{Position}$

Now estimated the distance of the Pluto is

  $\begin{array}{c}\log \left(\text{Distance}\right)=1.285+0.239\text{Position}\\ \text{=}1.285+0.239\left(10\right)\\ =1.285+2.39\\ =3.675\end{array}$

  $\begin{array}{c}{\text{Distance=10}}^{3.675}\\ =4731.5\end{array}$

In the solar system, Pluto does not match the pattern for location and distance. Currently, the model created. It's not nice with Pluto included, because Pluto affects those predictions. The ModelWithout Pluto, Pluto's distance from the Sun is predicted to be 4731.51 million miles. There is proof, therefore, in its relationship to location and distance, Pluto does not function like a planet if the IAU is right.