Question

An economist is studying the job market in Denver area neighborhoods. Let x represent the total number of jobs in a given neighborhood, and let y represent the number of entry-level jobs in the same neighborhood. A sample of six Denver neighborhoods gave the following information (units in hundreds of jobs).
x 13 34 52 28 50 25

y 1 4 5 5 9 3

Complete parts (a) through (e), given Σx = 202, Σy = 27, Σx2 = 7938, Σy2 = 157, Σxy = 1074, and r ≈ 0.821.
a. Verify the given sums Σx, Σy, Σx2, Σy2, Σxy, and the value of the sample correlation coefficient r.
b. Find x, and y. Then find the equation of the least-squares line y = a + bx. (Round your answers for x and yto two decimal places. Round your answers for a and b to three decimal places.)
c. Find the value of the coefficient of determination r2. What percentage of the variation in y can be explainedby the corresponding variation in x and the least-squares line?
d. For a neighborhood with x = 34 hundred jobs, how many are predicted to be entry level jobs?

Answer 1

Answer:

a) $\sum X = 13+34+52+28+50+25 =202$

$\sum Y = 1+4+5+5+9+3= 27$

$\sum X^2 = 13^2+34^2+52^2+28^2+50^2+25^2 = 1074$

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

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

$r=\frac{6(1074)-(202)(27)}{\sqrt{[6(7938) -(202)^2][6(157) -(202)^2]}}=0.821$  

b) $m=\frac{165}{1137.33}=0.145$  

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

$\bar x= \frac{\sum x_i}{n}=\frac{202}{6}=33.67$  

$\bar y= \frac{\sum y_i}{n}=\frac{27}{6}=4.5$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=4.5-(0.145*33.67)=-0.384$  

So the line would be given by:  

$y=0.145 x -0.384$  

c) The coefficient of variation is given by:

$r^2 = 0.821^2 = 0.674$

And we can conclude that 67.4% of the variation in y can be explainedby the corresponding variation in x and the least-squares line

d) For this case we just need to replade x = 34 into our model and we got:

$y=0.145*34 -0.384= 3.097$  

Step-by-step explanation:

For this case we have the following data:

x 13 34 52 28 50 25

y 1 4 5 5 9 3

Part a

$\sum X = 13+34+52+28+50+25 =202$

$\sum Y = 1+4+5+5+9+3= 27$

$\sum X^2 = 13^2+34^2+52^2+28^2+50^2+25^2 = 1074$

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

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

$r=\frac{6(1074)-(202)(27)}{\sqrt{[6(7938) -(202)^2][6(157) -(202)^2]}}=0.821$  

Part 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:  

$S_{xx}=\sum_{i=1}^n x^2_i -\frac{(\sum_{i=1}^n x_i)^2}{n}=7938-\frac{202^2}{6}=1137.33$  

$S_{xy}=\sum_{i=1}^n x_i y_i -\frac{(\sum_{i=1}^n x_i)(\sum_{i=1}^n y_i){n}}=1074-\frac{202*27}{6}=165$  

And the slope would be:  

$m=\frac{165}{1137.33}=0.145$  

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

$\bar x= \frac{\sum x_i}{n}=\frac{202}{6}=33.67$  

$\bar y= \frac{\sum y_i}{n}=\frac{27}{6}=4.5$  

And we can find the intercept using this:  

$b=\bar y -m \bar x=4.5-(0.145*33.67)=-0.384$  

So the line would be given by:  

$y=0.145 x -0.384$  

Part c

The coefficient of variation is given by:

$r^2 = 0.821^2 = 0.674$

And we can conclude that 67.4% of the variation in y can be explainedby the corresponding variation in x and the least-squares line

Part d

For this case we just need to replade x = 34 into our model and we got:

$y=0.145*34 -0.384= 3.097$