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

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=(n(\sum xy)-(\sum x)(\sum y))/(√([n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]))`

`r=(6(1074)-(202)(27))/(√([6(7938) -(202)^2][6(157) -(202)^2]))=0.821`

b)
`m=(165)/(1137.33)=0.145`

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

`\bar x= (\sum x_i)/(n)=(202)/(6)=33.67`

`\bar y= (\sum y_i)/(n)=(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`

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=(n(\sum xy)-(\sum x)(\sum y))/(√([n\sum x^2 -(\sum x)^2][n\sum y^2 -(\sum y)^2]))`

`r=(6(1074)-(202)(27))/(√([6(7938) -(202)^2][6(157) -(202)^2]))=0.821`

Part b

`m=(S_(xy))/(S_(xx))`

Where:

`S_(xy)=\sum_(i=1)^n x_i y_i -((\sum_(i=1)^n x_i)(\sum_(i=1)^n y_i))/(n)`

`S_(xx)=\sum_(i=1)^n x^2_i -((\sum_(i=1)^n x_i)^2)/(n)`

With these we can find the sums:

`S_(xx)=\sum_(i=1)^n x^2_i -((\sum_(i=1)^n x_i)^2)/(n)=7938-(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-(202*27)/(6)=165`

And the slope would be:

`m=(165)/(1137.33)=0.145`

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

`\bar x= (\sum x_i)/(n)=(202)/(6)=33.67`

`\bar y= (\sum y_i)/(n)=(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`