Respuesta :
Answer:
[tex]S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=632-\frac{70^2}{10}=142[/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}=8128-\frac{70*1080}{10}=568[/tex]
And the slope would be:
[tex]m=\frac{568}{142}=4[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{70}{10}=7[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{1080}{10}=108[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=108-(4*7)=80[/tex]
So the line would be given by:
[tex]y=4 x +80[/tex]
And in order to find the predicted value for 9 years we can use the model with x =9 and we got:
[tex] y = 4*9 +80=116[/tex]
Step-by-step explanation:
Data given
x: 1,3,4,4,6,8,10,10,11,13
y: 80,97, 92,102,103,111,119,123,117,136
For this case we need to calculate the slope with the following formula:
[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]
So we can find the sums like this:
[tex]\sum_{i=1}^n x_i =70[/tex]
[tex]\sum_{i=1}^n y_i =1080[/tex]
[tex]\sum_{i=1}^n x^2_i =632[/tex]
[tex]\sum_{i=1}^n y^2_i =119082[/tex]
[tex]\sum_{i=1}^n x_i y_i =8128[/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}=632-\frac{70^2}{10}=142[/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}=8128-\frac{70*1080}{10}=568[/tex]
And the slope would be:
[tex]m=\frac{568}{142}=4[/tex]
Nowe we can find the means for x and y like this:
[tex]\bar x= \frac{\sum x_i}{n}=\frac{70}{10}=7[/tex]
[tex]\bar y= \frac{\sum y_i}{n}=\frac{1080}{10}=108[/tex]
And we can find the intercept using this:
[tex]b=\bar y -m \bar x=108-(4*7)=80[/tex]
So the line would be given by:
[tex]y=4 x +80[/tex]
And in order to find the predicted value for 9 years we can use the model with x =9 and we got:
[tex] y = 4*9 +80=116[/tex]