Question

A study was conducted to determine whether the final grade of a student in an introductory psychology course is linearly related to his or her performance on the verbal ability test administered before college entrance.
The verbal scores and final grades for 10 students are shown.

Student Verbal Score x=78,56,66,53,68,71,30.27,46,46
Final Grade Y= 87,67,80,73,99,84,74,72,61,80

Find the following:

a. The correlation coefficient: r=
b. The least squares line: y^=
c. Calculate the residual for the ninth student:

Answer 1

Answer:

a) In order to calculate the correlation coefficient we can use this formula:  

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

And replacing we got:

$r=0.598$

b) $S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=31911-\frac{541^2}{10}=2642.9$  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=43033-\frac{541*777}{10}=997.3$  

And the slope would be:  

$m=\frac{997.3}{2642.9}=0.377$  

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

$\bar x= \frac{\sum x_i}{n}=\frac{541}{10}=54.1$  

$\bar y= \frac{\sum y_i}{n}=\frac{777}{10}=77.7$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=77.7-(0.377*54.1)=57.30$  

So the line would be given by:  

$y=0.377 x +57.30$  

c) For this case if we use the score for the 9th student we got:

$y= 0.377* 46 +57.30 =74.642$

And the residual is given by:

$e= \hat y_i -y_i = 74.642-61=13.542$

Step-by-step explanation:

Part a

In order to calculate the correlation coefficient we can use this formula:  

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

And replacing we got:

$r=0.598$  

Part b

We need to find a model like this one:

$Y = mx +b$

$m=\frac{S_{xy}}{S_{xx}}$  

Where:  

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

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

With these we can find the sums:  

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=31911-\frac{541^2}{10}=2642.9$  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=43033-\frac{541*777}{10}=997.3$  

And the slope would be:  

$m=\frac{997.3}{2642.9}=0.377$  

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

$\bar x= \frac{\sum x_i}{n}=\frac{541}{10}=54.1$  

$\bar y= \frac{\sum y_i}{n}=\frac{777}{10}=77.7$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=77.7-(0.377*54.1)=57.30$  

So the line would be given by:  

$y=0.377 x +57.30$  

Part c

For this case if we use the score for the 9th student we got:

$y= 0.377* 46 +57.30 =74.642$

And the residual is given by:

$e= \hat y_i -y_i = 74.642-61=13.542$