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dsp73
3 years ago
11

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

Mathematics
1 answer:
levacccp [35]3 years ago
5 0

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.

<u>y as a function of x </u>

<em>(y=response variable, x=explanatory variable) </em>

\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)

<u>Note:</u> <em>Be careful not to confuse  </em>

\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

<h3>y = 176.6061x + 0.3591 </h3>

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)

<u>Note:</u> <em>Be careful not to confuse  </em>

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

Remark:<em> </em><em>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) </em>

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

<h3>x = 1068.1072y + 2.4861 </h3>

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

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

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

Remark: <em>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. </em>

Computing the correlation coefficient :

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

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