Respuesta :
Answer:
a) [tex] \sum X = 13+34+52+28+50+25 =202 [/tex]
[tex]\sum Y = 1+4+5+5+9+3= 27 [/tex]
[tex] \sum X^2 = 13^2+34^2+52^2+28^2+50^2+25^2 = 1074[/tex]
And 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]
[tex]r=\frac{6(1074)-(202)(27)}{\sqrt{[6(7938) -(202)^2][6(157) -(202)^2]}}=0.821[/tex]
b) [tex]m=\frac{165}{1137.33}=0.145[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{202}{6}=33.67[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{27}{6}=4.5[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=4.5-(0.145*33.67)=-0.384[/tex]
So the line would be given by:
[tex]y=0.145 x -0.384[/tex]
c) The coefficient of variation is given by:
[tex] r^2 = 0.821^2 = 0.674[/tex]
And we can conclude that 67.4% of the variation in y can be explainedby the corresponding variation in x and the least-squares line
d) For this case we just need to replade x = 34 into our model and we got:
[tex]y=0.145*34 -0.384= 3.097[/tex]
Step-by-step explanation:
For this case we have the following data:
x 13 34 52 28 50 25
y 1 4 5 5 9 3
Part a
[tex] \sum X = 13+34+52+28+50+25 =202 [/tex]
[tex]\sum Y = 1+4+5+5+9+3= 27 [/tex]
[tex] \sum X^2 = 13^2+34^2+52^2+28^2+50^2+25^2 = 1074[/tex]
And 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]
[tex]r=\frac{6(1074)-(202)(27)}{\sqrt{[6(7938) -(202)^2][6(157) -(202)^2]}}=0.821[/tex]
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}=7938-\frac{202^2}{6}=1137.33[/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}}=1074-\frac{202*27}{6}=165[/tex]
And the slope would be:
[tex]m=\frac{165}{1137.33}=0.145[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{202}{6}=33.67[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{27}{6}=4.5[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=4.5-(0.145*33.67)=-0.384[/tex]
So the line would be given by:
[tex]y=0.145 x -0.384[/tex]
Part c
The coefficient of variation is given by:
[tex] r^2 = 0.821^2 = 0.674[/tex]
And we can conclude that 67.4% of the variation in y can be explainedby the corresponding variation in x and the least-squares line
Part d
For this case we just need to replade x = 34 into our model and we got:
[tex]y=0.145*34 -0.384= 3.097[/tex]
