Question

The accompanying data on x = current density (mA/cm2) and y = rate of deposition (µm/min) appeared in an article. Do you agree with the claim by the article's author that "a linear relationship was obtained from the tin-lead rate of deposition as a function of current density"?

col1 x 20 40 60 80
col2 y 0.24 1.10 1.71 2.12
Find the value of r2. (Round your answer to three decimal places.)

Answer 1

Answer:

$r=\frac{4(321)-(200)(5.17)}{\sqrt{[4(12000) -(200)^2][4(8.6861) -(5.17)^2]}}=0.98726$  

So then the correlation coefficient would be r =0.98726

And the determination coeffcient would be:

$r^2 = 0.98726^2 = 0.975$

And that means that the a linear model will explain about 97.5% of the variability in the data. So for this case we can conclude that we have a strong linear relationship.

Step-by-step explanation:

Data given

col1 x 20 40 60 80

col2 y 0.24 1.10 1.71 2.12

Solution to the problem

We can find the correlation coefficient we can use this formula:  

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

For our case we have this:

n=4 $\sum x = 200, \sum y = 5.17, \sum xy = 321, \sum x^2 =12000, \sum y^2 =8.6861$  

$r=\frac{4(321)-(200)(5.17)}{\sqrt{[4(12000) -(200)^2][4(8.6861) -(5.17)^2]}}=0.98726$  

So then the correlation coefficient would be r =0.98726

And the determination coeffcient would be:

$r^2 = 0.98726^2 = 0.975$

And that means that the a linear model will explain about 97.5% of the variability in the data. So for this case we can conclude that we have a strong linear relationship.