From the given table, the equation of the line of best fit is:
[tex]y=13.98x+478.73[/tex]
where x denotes the years and y the number of students.
Now, the mean of the y-values is
[tex]\mu_y=\frac{5556}{10}=555.6[/tex]
The variance of y-values is given by
[tex]\text{SSy}=\Sigma(y-\mu_y)^2[/tex]
By expanding this sum, we have
[tex]\text{SSy}=(492-555.6)^2+(507-555.6)^2+(520-555.6)^2+\ldots+(618-555.6)^2[/tex]
which gives
[tex]\text{SSy}=16118.4[/tex]
By doind the same process for the varible x, we have that the correlation coefficient R is given by:
[tex]R=0.9999[/tex]
since this result is near to 1, there is a strong positive relationship between the year and the number of students
