Question

The accompanying data on x = current density (mA/cm2) and y = rate of deposition (m/min)μ appeared in a recent study.

Answer 1

Answer:

a) $r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

b) $m=\frac{64.5}{2000}=0.03225$  

Now we can find the means for x and y like this:  

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$  

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$  

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$  

So the line would be given by:  

$y=0.3225 x -0.27$  

Step-by-step explanation:

Part a

The correlation coeffcient is given by 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 $\sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501$  

$r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857$  

The correlation coefficient for this case is very near to 1 so then we can ensure that we have linear correlation between the two variables

Part b

$m=\frac{S_{xy}}{S_{xx}}$  

Where:  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$  

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$  

With these we can find the sums:  

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=12000-\frac{200^2}{4}=2000$  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=333-\frac{200*5.37}{4}=64.5$  

And the slope would be:  

$m=\frac{64.5}{2000}=0.03225$  

Now we can find the means for x and y like this:  

$\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50$  

$\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27$  

So the line would be given by:  

$y=0.3225 x -0.27$