Question

A group of researchers conducted a study to determine whether the final grade in an honors section of introductory psychology was related to a student’s performance on a test of math ability administered for college entrance. The researchers looked at the test scores of 200 students (n= 200) and found a correlation of r= .45 between math ability scores and final course grade. The proportion of the variability seen in final grade performance that can be predicted by math ability scores is ____.

Answer 1

Answer:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

And for this case $r =0.45$

The % of variation is given by the determination coefficient given by $r^2$ and on this case $0.45^2 =0.2025$, so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.

Step-by-step explanation:

For this case we asume that we fit a linear model:

$y = mx+b$

Where y represent the final grade and x the math ability scores

$m=\frac{S_{xy}}{S_{xx}}$  

Where:  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}{n}$  

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}$  

$\bar x= \frac{\sum x_i}{n}$  

$\bar y= \frac{\sum y_i}{n}$  

And we can find the intercept using this:  

$b=\bar y -m \bar x$  

The correlation coeffcient is given by:

$r=\frac{n(\sum xy)-(\sum x)(\sum y)}{\sqrt{[n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]}}$

And for this case $r =0.45$

The % of variation is given by the determination coefficient given by $r^2$ and on this case $0.45^2 =0.2025$, so then the % of variation explained is 20.25%.

The proportion of the variability seen in final grade performance that can be predicted by math ability scores is 20.25%.