Respuesta :
Answer:
a) [tex]r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857[/tex]
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) [tex]m=\frac{64.5}{2000}=0.03225[/tex]
Now we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425[/tex]
[tex]b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27[/tex]
So the line would be given by:
[tex]y=0.3225 x -0.27[/tex]
Step-by-step explanation:
Part a
The correlation coeffcient is given by this formula:
[tex]r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}[/tex]
For our case we have this:
n=4 [tex] \sum x = 200, \sum y = 5.37, \sum xy = 333, \sum x^2 =12000, \sum y^2 =9.3501[/tex]
[tex]r=\frac{4(333)-(200)(5.37)}{\sqrt{[4(12000) -(200)^2][4(9.3501) -(5.37)^2]}}=0.9857[/tex]
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
[tex]m=\frac{S_{xy}}{S_{xx}}[/tex]
Where:
[tex]S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}[/tex]
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}[/tex]
With these we can find the sums:
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=12000-\frac{200^2}{4}=2000[/tex]
[tex]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[/tex]
And the slope would be:
[tex]m=\frac{64.5}{2000}=0.03225[/tex]
Now we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{200}{4}=50[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{5.37}{4}=1.3425[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=1.3425-(0.03225*50)=-0.27[/tex]
So the line would be given by:
[tex]y=0.3225 x -0.27[/tex]
