Respuesta :
Answer:
a) In order to calculate the correlation coefficient we can use 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]
And replacing we got:
[tex] r=0.598[/tex]
b) [tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=31911-\frac{541^2}{10}=2642.9[/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}}=43033-\frac{541*777}{10}=997.3[/tex]
And the slope would be:
[tex]m=\frac{997.3}{2642.9}=0.377[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{541}{10}=54.1[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{777}{10}=77.7[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=77.7-(0.377*54.1)=57.30[/tex]
So the line would be given by:
[tex]y=0.377 x +57.30[/tex]
c) For this case if we use the score for the 9th student we got:
[tex] y= 0.377* 46 +57.30 =74.642[/tex]
And the residual is given by:
[tex] e= \hat y_i -y_i = 74.642-61=13.542[/tex]
Step-by-step explanation:
Part a
In order to calculate the correlation coefficient we can use 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]
And replacing we got:
[tex] r=0.598[/tex]
Part b
We need to find a model like this one:
[tex] Y = mx +b[/tex]
[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}=31911-\frac{541^2}{10}=2642.9[/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}}=43033-\frac{541*777}{10}=997.3[/tex]
And the slope would be:
[tex]m=\frac{997.3}{2642.9}=0.377[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{541}{10}=54.1[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{777}{10}=77.7[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=77.7-(0.377*54.1)=57.30[/tex]
So the line would be given by:
[tex]y=0.377 x +57.30[/tex]
Part c
For this case if we use the score for the 9th student we got:
[tex] y= 0.377* 46 +57.30 =74.642[/tex]
And the residual is given by:
[tex] e= \hat y_i -y_i = 74.642-61=13.542[/tex]
