Chapter 3.1, Problem 21E

(a)

To determine

Scatterplot for all five specimens.

Whether all five specimens come from same species.

Answer to Problem 21E

Yes, all five specimens come from same species.

Explanation of Solution

Given information:

Data on the lengths (in centimeters) of the femur (a leg bone) and the humerus (a bone in the upper arm):

  

For scatterplot:

On horizontal x − axis:

Femur (in centimeters)

On vertical y − axis:

Humerus (in centimeters):

  

Note that

All points form a linear pattern.

Thus,

We can say

All five species come from the same species.

(b)

To determine

Correlation r step by step and matching the value for r with graph in Part (a).

Answer to Problem 21E

Correlation,

  $r\approx 0.9941$

Explanation of Solution

Given information:

Data on the lengths (in centimeters) of the femur (a leg bone) and the humerus (a bone in the upper arm):

  

From the given table,

Find xy , x² , y² :

x	y	xy	x2	y2
38 56 59 64 74	41 63 70 72 84	1558 3528 4130 4608 6216	1444 3136 3481 4096 5476	1681 3969 4900 5184 7056

Now,

Sum up every column:

  ${\displaystyle \sum x=291}$ , ${\displaystyle \sum y=330}$ , ${\displaystyle \sum xy=20040}$ , ${\displaystyle \sum x^2=17633}$ , ${\displaystyle \sum y^2=22790}$

Formula to estimate the correlation coefficient:

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

Substitute the values:

  $r=\frac{5\times 20040-291\times 330}{\sqrt{\left[5\times 17633-{\left(291\right)}^2\right]\cdot \left[5\times 22790-{\left(330\right)}^2\right]}}\approx 0.9941$

Since all five specimens come from the same species and all the points are lying on the same line.

Thus,

The value of correlation matches with the graph in Part (a).