Question

It is thought that basketball teams that make too many fouls in a game tend to lose the game even if they otherwise play well. Let x be the number of fouls more than (i.e., over and above) the opposing team. Let y be the percentage of times the team with the larger number of fouls wins the game.
X= 0 2 5 6
Y= 50 45 33 26

Complete parts (a) through (e), given Σx = 16, Σy = 154, Σx2 = 78, Σy2 = 6302, Σxy = 548, and r ≈ −0.941.

a. Draw a scatter diagram displaying the data.
b. Find X- bar, Y- bar, a and b. Then find the equation of the least squares line y=a+bx.
c. Graph the least squares line on your scatter diagram. Be sure to use the point ( X-bar, Y- bar) as one of the points on the line.
d. Find the value of the coefficient of determination r ^2. What percentage of the variation in y can be explained by the corresponding variation in x and the least squares line? What percentage is unexplained?
e. If a team had x = 4 fouls over and above the opposing tem, what does the leas squares equation forecast for y?

Answer 1

Answer:

a) First figure attached

b) $\bar x= \frac{\sum x_i}{n}=\frac{16}{4}=4$

$\bar y= \frac{\sum y_i}{n}=\frac{154}{4}=38.5$

c) $S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=78-\frac{16^2}{4}=14$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=548-\frac{16*154}{4}=-68$

And the slope would be:

$m=-\frac{68}{14}=-4.857$

And we can find the intercept using this:

$b=\bar y -m \bar x=38.5-(-4.857*4)=57.929$

So the line would be given by:

$y=-4.857 x +57.929$

d) $r^2 = (-0.941)^2= 0.885$

So then the porcentage of variation explained is 88.5%

And the percentage of unexplained variation would be 100-88.5=11.5%

e) Using the least squares regression:

$y(4) = -4.857*4 +57.929=38.501$

Using the excel equation with the $(\bar X, \bar Y)$ we have:

$y(4) = -3.8578*4 +51.616=36.185$

Step-by-step explanation:

We assume that the data is this one:

x: 0,2,5,6

y: 50,45,33,26

Part a

For this case on the figure attached we see the scatter plot of the data.

Part b

$\bar x= \frac{\sum x_i}{n}=\frac{16}{4}=4$

$\bar y= \frac{\sum y_i}{n}=\frac{154}{4}=38.5$

Find the least-squares line appropriate for this data.  

For this case we need to calculate the slope with the following formula:

$m=\frac{S_{xy}}{S_{xx}}$

Where:

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

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

So we can find the sums like this:

$\sum_{i=1}^n x_i =16$

$\sum_{i=1}^n y_i =154$

$\sum_{i=1}^n x^2_i =78$

$\sum_{i=1}^n y^2_i =6302$

$\sum_{i=1}^n x_i y_i =548$

With these we can find the sums:

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=78-\frac{16^2}{4}=14$

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i)}=548-\frac{16*154}{4}=-68$

And the slope would be:

$m=-\frac{68}{14}=-4.857$

And we can find the intercept using this:

$b=\bar y -m \bar x=38.5-(-4.857*4)=57.929$

So the line would be given by:

$y=-4.857 x +57.929$

Part c

For this case we add the point (4.38.5) for the data and we got the equation as we can see on the figure attached.

y = -3.8578x + 51.616 (Equation adjusted with Excel)

Part d

$r^2 = (-0.941)^2= 0.885$

So then the porcentage of variation explained is 88.5%

And the percentage of unexplained variation would be 100-88.5=11.5%

Part e

Using the least squares regression:

$y(4) = -4.857*4 +57.929=38.501$

Using the excel equation with the $(\bar X, \bar Y)$ we have:

$y(4) = -3.8578*4 +51.616=36.185$