An economist is studying the job market in Denver area neighborhoods. Let x repr

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).

Complete parts (a) through (e), given x = 204, y = 27, x2 = 8032, y2 = 157, xy = 1077, and r 0.806.

(a) Verify the given sums x, y, x2, y2, xy, and the value of the sample correlation coefficient r. (Round your value for r to three decimal places.)

+ x

(c) Find the value of the coefficient of determination r2. 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? (Round your answer for r2 to three decimal places. Round your answers for the percentages to one decimal place.)

(d) For a neighborhood with x = 36 hundred jobs, how many are predicted to be entry level jobs? (Round your answer to two decimal places.)
hundred jobs

x 15 33 53 28 50 25 y 1 4 5 5 9 3

Explanation / Answer

a.

calculation procedure for correlation

sum of (x) = x = 204

sum of (y) = y = 27

sum of (x^2)= x^2 = 8032

sum of (y^2)= y^2 = 157

sum of (x*y)= x*y = 1077

to caluclate value of r( x,y) = covariance ( x,y ) / sd (x) * sd (y)

covariance ( x,y ) = [ x*y - N *(x/N) * (y/N) ]/n-1

= 1077 - [ 6 * (204/6) * (27/6) ]/6- 1

= 26.5

and now to calculate r( x,y) = 26.5/ (SQRT(1/6*1077-(1/6*204)^2) ) * ( SQRT(1/6*1077-(1/6*27)^2)

=26.5 / (13.515*2.432)

=0.806

value of correlation is =0.806

coeffcient of determination = r^2 = 0.65

properties of correlation

1. If r = 1 Corrlation is called Perfect Positive Corrlelation

2. If r = -1 Correlation is called Perfect Negative Correlation

3. If r = 0 Correlation is called Zero Correlation

& with above we conclude that correlation ( r ) is = 0.8062> 0 ,perfect positive correlation

b.

calculation procedure for regression

mean of X = X / n = 34

mean of Y = Y / n = 4.5

(Xi - Mean)^2 = 1096

(Yi - Mean)^2 = 35.5

(Xi-Mean)*(Yi-Mean) = 159

b1 = (Xi-Mean)*(Yi-Mean) / (Xi - Mean)^2

= 159 / 1096

= 0.145

bo = Y / n - b1 * X / n

bo = 4.5 - 0.145*34 = -0.432

value of regression equation is, Y = bo + b1 X

Y'=-0.432+0.145* X

c.

coeffcient of determination = r^2 = 0.65 = 65%

percentage of the variation in y can be explained by the corresponding variation in x and the least-squares line

14.5%

unexplained =85.5%

d.

Y'=-0.432+0.145* X

x = 36

then predicted entry level jobs

Y' = -0.432+0.145*36

Y'=4.78

( X) ( Y) X^2 Y^2 X*Y 15 1 225 1 15 33 4 1089 16 132 53 5 2809 25 265 28 5 784 25 140 50 9 2500 81 450 25 3 625 9 75

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