This week, we are covering relationships that can be approximated by linear equations. For instance, y = 453x + 3768 represents the increase in cost of tuition and fees at public four-year universities. Now think about how this might apply in your chosen field and answer the following question: What is an example of a relationship (in your chosen field) that can be represented by a linear equation? Show the equation and explain the relationship. Please provide a reference for your example (website, book, etc.)

Where X represent the cost for scholarships in thousands of dollars and y represent the cost of life for an academic semester (The data comes from the web)

We can find the least-squares line appropriate for this data.

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 = 30+30+30+50+50+50+70+70+70+90+90+90=720[/tex]

[tex]\sum_{i=1}^n y_i =38+43+29+32+26+33+19+27+23+14+19+21=324[/tex]

[tex]\sum_{i=1}^n x^2_i =30^2+30^2+30^2+50^2+50^2+50^2+70^2+70^2+70^2+90^2+90^2+90^2=49200[/tex]

[tex]\sum_{i=1}^n y^2_i =38^2+43^2+29^2+32^2+26^2+33^2+19^2+27^2+23^2+14^2+19^2+21^2=9540[/tex]

[tex]\sum_{i=1}^n x_i y_i =30*38+30*43+30*29+50*32+50*26+50*33+70*19+70*27+70*23+90*14+90*19+90*21=17540[/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}=49200-\frac{720^2}{12}=6000[/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)}=17540-\frac{720*324}{12}{12}=-1900[/tex]

And the slope would be:

[tex]m=-\frac{1900}{6000}=-0.317[/tex]

Nowe we can find the means for x and y like this:

[tex]\bar x= \frac{\sum x_i}{n}=\frac{720}{12}=60[/tex]

[tex]\bar y= \frac{\sum y_i}{n}=\frac{324}{12}=27[/tex]

And we can find the intercept using this:

[tex]b=\bar y -m \bar x=27-(-0.317*60)=46.02[/tex]

So the line would be given by:

[tex]y=-0.317 x +46.02[/tex]

We have an inverse linear relationship since the slope is negative between the variables of interest.