Question

Five observations taken for two variables follow.

X 6 11 15 21 27
Y 6 9 6 17 12

a. Develop a scatter diagram for these data.
b. What does the scatter diagram indicate about a relationship between x and y?
c. Compute and interpret the sample covariance.
d. Compute and interpret the sample correlation coefficient.

Answer 1

Answer:

a) Figure attached

b) If we see the scatter plot we can conclude that the possible relation between x and y is linear and with a positive correlation since when the values of x increases the values for y increases as well.

c) $Cov (X,Y) = \frac{\sum_{i=1}^n (x_i -\bar X)(y_i -\bar Y)}{n-1}$

We can find the numerator like this:

$\sum_{i=1}^5 (6-16)(6-10)+(11-16)(9-10)+(15-16)(6-10)+(21-16)(17-10)+(27-16)(12-10)=106$

And then:

$Cov(X,Y) = \frac{106}{5-1}=26.5$

d) $r=\frac{5(906)-(80)(50)}{\sqrt{[5(1552) -(80)^2][5(586) -(50)^2]}}=0.693$  

Step-by-step explanation:

Part a

For this part we use excel in order to create the scatterplot and we got the result on the figure attached.

Part b

If we see the scatter plot we can conclude that the possible relation between x and y is linear and with a positive correlation since when the values of x increases the values for y increases as well.

Part c

The sample covariance is defined as:

$Cov (X,Y) = \frac{\sum_{i=1}^n (x_i -\bar X)(y_i -\bar Y)}{n-1}$

We can find the numerator like this:

$\sum_{i=1}^5 (6-16)(6-10)+(11-16)(9-10)+(15-16)(6-10)+(21-16)(17-10)+(27-16)(12-10)=106$

And then:

$Cov(X,Y) = \frac{106}{5-1}=26.5$

Part d

The correlation coefficient is a "statistical measure that calculates the strength of the relationship between the relative movements of two variables". It's denoted by r and its always between -1 and 1.

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

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

For our case we have this:

n=5 $\sum x = 80, \sum y = 50, \sum xy = 906, \sum x^2 =1552, \sum y^2 =586$  

$r=\frac{5(906)-(80)(50)}{\sqrt{[5(1552) -(80)^2][5(586) -(50)^2]}}=0.693$  

So then the correlation coefficient would be r =0.693