Question

In: Math

1. Obtain a linear regression equation for the data to predict the mean temperature values for...

1. Obtain a linear regression equation for the data to predict the mean temperature values for any given CO2 level. How good is the linear fit for this data? Explain using residual plot and R-square value. To draw residual plot, compute the estimated temperatures for every value of the CO2 level using the regression equation. Then compute the difference between observed (y) and estimated temperature values (called residual; ). Plot the residuals versus CO2 level (called a residual plot).

320.09 8
321.44 9.29
322.17 9.39
323.09 8.61
324.68 8.95
325.74 8.36
326.33 9.11
327.52 8.43
329.78 8.39
330.24 8.18
331.18 9.06
332.09 9.12
333.88 8.11
335.43 7.51
336.83 7.42
338.78 7.78
340.17 8.2
340.99 8.6
342.97 8.9
344.23 8.04
345.94 7.18
347.26 7.89
349.06 7.66
351.56 8.9
352.91 9.68
354.21 9.98
355.54 8.88
356.29 9.46
356.97 8.83
358.69 10.29
360.71 10.27
362.41 8.01
363.53 9.28
366.64 9.3
368.16 9.78
369.45 9.88
371.12 9.6
373.24 9.73
375.88 10.35
377.6 9.48
379.87 9.53
381.89 9.94
383.79 10.59

Solutions

Expert Solution

i have considered regression equation with no constant or intercept

x=carbondioxide values;dependent variable

y=temparature;independent variable

my estimated regression line is

  

R square value is 0.993853

The definition of R-squared is fairly straight-forward; it is the percentage of the response variable variation that is explained by a linear model. Or:

R-squared = Explained variation / Total variation

R-squared is always between 0 and 100%:

  • 0% indicates that the model explains none of the variability of the response data around its mean.
  • 100% indicates that the model explains all the variability of the response data around its mean.
  • here our model explains 99.385% of variability and hence good fit
Predicted Y Residuals
8.199514 -0.19951
8.234096 1.055904
8.252796 1.137204
8.276363 0.333637
8.317093 0.632907
8.344246 0.015754
8.35936 0.75064
8.389843 0.040157
8.447736 -0.05774
8.459519 -0.27952
8.483599 0.576401
8.506909 0.613091
8.552763 -0.44276
8.592468 -1.08247
8.628331 -1.20833
8.678282 -0.89828
8.713889 -0.51389
8.734894 -0.13489
8.785615 0.114385
8.817891 -0.77789
8.861695 -1.68169
8.895508 -1.00551
8.941618 -1.28162
9.005658 -0.10566
9.04024 0.63976
9.073542 0.906458
9.107611 -0.22761
9.126823 0.333177
9.144242 -0.31424
9.188302 1.101698
9.240047 1.029953
9.283595 -1.27359
9.312285 -0.03229
9.391952 -0.09195
9.430889 0.349111
9.463934 0.416066
9.506713 0.093287
9.561019 0.168981
9.628646 0.721354
9.672706 -0.19271
9.730855 -0.20086
9.7826 0.1574
9.831271 0.758729

here the residuals are scattered around without any pattern this indicates that our fit is good

if we consider with a constant

estimated equation:

R square value is 0.3398

our model poorly explains 33.98% of variability

Predicted Y Residuals
8.160634 -0.16063
8.196994 1.093006
8.216655 1.173345
8.241433 0.368567
8.284256 0.665744
8.312805 0.047195
8.328696 0.781304
8.360746 0.069254
8.421614 -0.03161
8.434004 -0.254
8.459321 0.600679
8.48383 0.63617
8.53204 -0.42204
8.573786 -1.06379
8.611492 -1.19149
8.664011 -0.88401
8.701448 -0.50145
8.723533 -0.12353
8.77686 0.12314
8.810796 -0.7708
8.856851 -1.67685
8.892403 -1.0024
8.940882 -1.28088
9.008214 -0.10821
9.044574 0.635426
9.079587 0.900413
9.115407 -0.23541
9.135607 0.324393
9.153922 -0.32392
9.200246 1.089754
9.254651 1.015349
9.300437 -1.29044
9.330602 -0.0506
9.414363 -0.11436
9.455301 0.324699
9.490045 0.389955
9.535023 0.064977
9.592121 0.137879
9.663224 0.686776
9.709548 -0.22955
9.770686 -0.24069
9.825091 0.114909
9.876263 0.713737

here the residuals are scattered around without any pattern this indicates that our fit is good

milcah answered 6 months ago
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