Question

Consider the given data. x 0 2 4 6 9 11 12 15 17 19 y 5 6 7 6 9 8 8 10 12 12 Use the least-squares regression to fit a straight line to the given data. Along with the slope and intercept, compute the standard error of the estimate and the correlation coefficient. Repeat the problem, but regress x versus y—that is, switch the variables. Interpret your results. (Round the final answers to four decimal places.) Least-squares regression: y versus x B

Answer 1

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)

$\bf y=m_{yx}x+b_{yx}$

where

$\bf m_{yx}$ is the slope of the line

$\bf b_{yx}$ is the y-intercept

In this case we use these formulas:

$\bf m_{yx}=\frac{(\sum y)(\sum x)^2-(\sum x)(\sum xy)}{n\sum x^2-(\sum x)^2}$

$\bf b_{yx}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum x^2)-(\sum x)^2}$

n = 10 is the number of observations taken (pairs x,y)

Note: Be careful not to confuse  

$\bf \sum x^2$ with $\bf (\sum x)^2$

Performing our calculations we get:

$\bf m_{yx}=\frac{(83)(95)^2-(95)(923)}{10*1277-(95)^2}=176.6061$

$\bf b_{yx}=\frac{10*923-(95)(83)}{10(1277)-(95)^2}=0.3591$

So the equation of the line that relates y as a function of x is

y = 176.6061x + 0.3591

In order to compute the standard error $\bf S_{yx}$, we must use Table 2 (See Table 2 attached) and use the definition

$\bf s_{yx}=\sqrt{\frac{(y-y_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{yx}=\sqrt{\frac{39515985}{10}}=1987.8628$

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)

$\bf x=m_{xy}y+b_{xy}$

where

$\bf m_{xy}$ is the slope of the line

$\bf b_{xy}$ is the x-intercept

In this case we use these formulas:

$\bf m_{xy}=\frac{(\sum x)(\sum y)^2-(\sum y)(\sum xy)}{n\sum y^2-(\sum y)^2}$

$\bf b_{xy}=\frac{n\sum xy-(\sum x)(\sum y)}{n(\sum y^2)-(\sum y)^2}$

n = 10 is the number of observations taken (pairs x,y)

Note: Be careful not to confuse  

$\bf \sum y^2$ with $\bf (\sum y)^2$

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

$\bf m_{xy}=\frac{(95)(83)^2-(83)(923)}{10*743-(83)^2}=1068.1072$

$\bf b_{xy}=\frac{10*923-(95)(83)}{10(743)-(83)^2}=2.4861$

and the line that relates x as a function of y is

x = 1068.1072y + 2.4861

To find the standard error $\bf S_{xy}$ we use Table 3 (See Table 3 attached) and the formula

$\bf s_{xy}=\sqrt{\frac{(x-x_{est})^2}{n}}$

and we have that standard error when y is a function of x is

$\bf s_{xy}=\sqrt{\frac{846507757}{10}}=9200.5856$

In both cases the correlation coefficient r is the same and it can be computed with the formula:

$\bf r=\frac{\sum xy}{\sqrt{(\sum x^2)(\sum y^2)}}$

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 :

$\bf r=\frac{923}{\sqrt{(1277)(743)}}=0.9478$