Question

The data set below is on student absences and final grade. Sludem untudcat Student | Number of absences x | Final grade y (%) 82 86 43 74 58 90 78 15 12

A. Find the linear correlation coefficient.

B. Find the least-squares regression line.

C.Using the model above, predict grade for 22 absences.

Note: Please, if you use any values from the tables, indicate which table they come from. Also, please write clearly on a piece of paper.

Answer 1

A.  Using the following formula to find out the correlation coefficient:

$\dpi{80} \fn_jvn r =\frac{n.\sum xy - \sum x.\sum y}{\sqrt{\left [ n\sum x^{2}-(\sum x)^{2} \right ].\left [ n\sum y^{2}-(\sum y)^{2} \right ]}}$

  $\dpi{80} \fn_jvn =\frac{7*3745- 57*511}{\sqrt{\left [ 7*579- (57)^{2} \right ].\left [ 7*38993-(511)^{2} \right ]}}$

$\dpi{80} \fn_jvn = -0.9442$

B.  The independent variable is X, and the dependent variable is Y. The regression line is given by

y = a + b⋅x

In order to compute the regression coefficients, the following formula are used :

$\dpi{80} \fn_jvn a= \frac{\sum y.\sum x^{2}-\sum x.\sum xy}{n.\sum x^{2}-(\sum x)^{2}}$

  $\dpi{80} \fn_jvn = \frac{511*57^{2} -57*3745}{7*579 - 57^{2}}$

  $\dpi{80} \fn_jvn = 102.493$

$\dpi{80} \fn_jvn b= \frac{n.\sum xy-\sum x.\sum y}{n.\sum x^{2}-(\sum x)^{2}}$

$\dpi{80} \fn_jvn = \frac{7*3745 -57*511}{7*579 - 57^{2}}$

$\dpi{80} \fn_jvn = -3.622$

Substitute a and b in regression equation formula

y= 102.493 − 3.622⋅x

C. When, number of absence , x = 22

The grade y = 102.493 - 3.622*22

= 22.809