Chapter 10 Homework Due in 4 hours, 46 minutes. Due Mon 12/11/2017 11:59 pm The

Question

Chapter 10 Homework Due in 4 hours, 46 minutes. Due Mon 12/11/2017 11:59 pm The following bivariate data set contains an outlier 65.6 781.7 92.7 230.5 86 -204,1 334 | 4294 819 65.2 48.7 1176 25 -176.7 52.3 2325 37.1 763 95.8 200.5 80.5-5736 788 -253.7 7.5 121.6 37.8 1090.5 08.2-173.3 Grade: 10/100 Print Version What is the correlation coefficient with the outlier? (Rround to two decimal places)- What is the correlation coefficient without the outlier (Rround to two decimal places) (hint: The last row is the outlier) Would inclusion of the outlier change the evidence for or against a significant lincar correlation? Yes. Including the outlier changes the evidence reparding a lincar correlation No. Including the outlier does not change the evidence regaeding a linear correlation MacBook Air F1O 7 4 6

Explanation / Answer

a.

calculation procedure for correlation

sum of (x) = x = 1211.3

sum of (y) = y = 2607.5

sum of (x^2)= x^2 = 161163.87

sum of (y^2)= y^2 = 4612492.09

sum of (x*y)= x*y = 69624.32

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

= 69624.32 - [ 15 * (1211.3/15) * (2607.5/15) ]/15- 1

= -9396

and now to calculate r( x,y) = -9396/ (SQRT(1/15*69624.32-(1/15*1211.3)^2) ) * ( SQRT(1/15*69624.32-(1/15*2607.5)^2)

=-9396 / (64.986*526.575)

=-0.275

value of correlation is =-0.275 =0.27

coeffcient of determination = r^2 = 0.075 =0.07

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 = -0.2746< 0, negative correlation

b.

calculation procedure for correlation

sum of (x) = x = 903.1

sum of (y) = y = 2780.8

sum of (x^2)= x^2 = 66176.63

sum of (y^2)= y^2 = 4582459.2

sum of (x*y)= x*y = 123035.38

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

= 123035.38 - [ 14 * (903.1/14) * (2780.8/14) ]/14- 1

= -4024.72

and now to calculate r( x,y) = -4024.72/ (SQRT(1/14*123035.38-(1/14*903.1)^2) ) * ( SQRT(1/14*123035.38-(1/14*2780.8)^2)

=-4024.72 / (23.79*536.53)

=-0.32

value of correlation is =-0.32

coeffcient of determination = r^2 = 0.1

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 = -0.3154< 0, negative correlation

Yes,the outlier changes the evdence of linear correlation  

( X) ( Y) X^2 Y^2 X*Y 65.6 781.7 4303.36 611054.89 51279.52 92.7 230.5 8593.29 53130.25 21367.35 86 -204.1 7396 41656.81 -17552.6 33.4 -429.4 1115.56 184384.36 -14341.96 81.9 65.2 6707.61 4251.04 5339.88 48.7 1176 2371.69 1382976 57271.2 25 -176.7 625 31222.89 -4417.5 52.3 232.5 2735.29 54056.25 12159.75 37.1 763 1376.41 582169 28307.3 95.8 200.5 9177.64 40200.25 19207.9 80.5 -573.6 6480.25 329016.96 -46174.8 78.8 -253.7 6209.44 64363.69 -19991.56 87.5 -121.6 7656.25 14786.56 -10640 38 1090.5 1428.84 1189190.25 41220.9 308 -173.3 94987.24 30032.89 -53411.06

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