Answer:
See below
Step-by-step explanation:
By using the table 1 attached (See Table 1 attached)
We can perform all the calculations to express both, y as a function of x or x as a function of y.
Let's make first the line relating y as a function of x.
y as a function of x
(y=response variable, x=explanatory variable)
[tex] \bf y=m_{yx}x+b_{yx}[/tex]
where
[tex] \bf m_{yx}[/tex] is the slope of the line
[tex] \bf b_{yx}[/tex] is the y-intercept
In this case we use these formulas:
[tex] \bf m_{yx}=\frac{(\sum y)(\sum x)^2-(\sum x)(\sum xy)}{n\sum x^2-(\sum x)^2}[/tex]
[tex] \bf b_{yx}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}[/tex]
n = 10 is the number of observations taken (pairs x,y)
Note: Be careful not to confuse
[tex] \bf \sum x^2[/tex] with [tex] \bf (\sum x)^2[/tex]
Performing our calculations we get:
[tex] \bf m_{yx}=\frac{(83)(95)^2-(95)(923)}{10*1277-(95)^2}=176.6061[/tex]
[tex] \bf b_{yx}=\frac{10*923-(95)(83)}{10(1277)-(95)^2}=0.3591[/tex]
So the equation of the line that relates y as a function of x is
In order to compute the standard error [tex] \bf S_{yx}[/tex], we must use Table 2 (See Table 2 attached) and use the definition
[tex] \bf s_{yx}=\sqrt{\frac{(y-y_{est})^2}{n}}[/tex]
and we have that standard error when y is a function of x is
[tex] \bf s_{yx}=\sqrt{\frac{39515985}{10}}=1987.8628[/tex]
Now, to find the line that relates x as a function of y, we simply switch the roles of x and y in the formulas.
So now we have:
x as a function of y
(x=response variable, y=explanatory variable)
[tex] \bf x=m_{xy}y+b_{xy}[/tex]
where
[tex] \bf m_{xy}[/tex] is the slope of the line
[tex] \bf b_{xy}[/tex] is the x-intercept
In this case we use these formulas:
[tex] \bf m_{xy}=\frac{(\sum x)(\sum y)^2-(\sum y)(\sum xy)}{n\sum y^2-(\sum y)^2}[/tex]
[tex] \bf b_{xy}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum y^2)-(\sum y)^2}[/tex]
n = 10 is the number of observations taken (pairs x,y)
Note: Be careful not to confuse
[tex] \bf \sum y^2[/tex] with [tex] \bf (\sum y)^2[/tex]
Remark: If you wanted to draw this line in the classical style (the independent variable on the horizontal axis), you would have to swap the axis X and Y)
Computing our values, we get
[tex] \bf m_{xy}=\frac{(95)(83)^2-(83)(923)}{10*743-(83)^2}=1068.1072[/tex]
[tex] \bf b_{xy}=\frac{10*923-(95)(83)}{10(743)-(83)^2}=2.4861[/tex]
and the line that relates x as a function of y is
To find the standard error [tex] \bf S_{xy}[/tex] we use Table 3 (See Table 3 attached) and the formula
[tex] \bf s_{xy}=\sqrt{\frac{(x-x_{est})^2}{n}}[/tex]
and we have that standard error when y is a function of x is
[tex] \bf s_{xy}=\sqrt{\frac{846507757}{10}}=9200.5856[/tex]
In both cases the correlation coefficient r is the same and it can be computed with the formula:
[tex] \bf r=\frac{\sum xy}{\sqrt{(\sum x^2)(\sum y^2)}}[/tex]
Remark: This formula for r is only true if we assume the correlation is linear. The formula does not hold for other kind of correlations like parabolic, exponential,..., etc.
Computing the correlation coefficient :
[tex] \bf r=\frac{923}{\sqrt{(1277)(743)}}=0.9478[/tex]
