We are interested in how changes in these variables affect future temperatures, as well as how well these variables explain temperature changes so far. To do this, first read the dataset climate_change.csv into R.
Then, split the data into a training set, consisting of all the observations up to and including 2006, and a testing set consisting of the remaining years (hint: use subset). A training set refers to the data that will be used to build the model (this is the data we give to the lm() function), and a testing set refers to the data we will use to test our predictive ability.
Next, build a linear regression model to predict the dependent variable Temp, using MEI, CO2, CH4, N2O, CFC.11, CFC.12, TSI, and Aerosols as independent variables (Year and Month should NOT be used in the model). Use the training set to build the model.
Enter the model R2 (the “Multiple R-squared” value):
summary(lm(Temp ~ MEI+CO2+CH4+N2O+CFC.11+CFC.12+TSI+Aerosols,data = Training))
Call:
lm(formula = Temp ~ MEI + CO2 + CH4 + N2O + CFC.11 + CFC.12 +
TSI + Aerosols, data = Training)
Residuals:
Min 1Q Median 3Q Max
-0.25888 -0.05913 -0.00082 0.05649 0.32433
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.246e+02 1.989e+01 -6.265 1.43e-09 ***
MEI 6.421e-02 6.470e-03 9.923 < 2e-16 ***
CO2 6.457e-03 2.285e-03 2.826 0.00505 **
CH4 1.240e-04 5.158e-04 0.240 0.81015
N2O -1.653e-02 8.565e-03 -1.930 0.05467 .
CFC.11 -6.631e-03 1.626e-03 -4.078 5.96e-05 ***
CFC.12 3.808e-03 1.014e-03 3.757 0.00021 ***
TSI 9.314e-02 1.475e-02 6.313 1.10e-09 ***
Aerosols -1.538e+00 2.133e-01 -7.210 5.41e-12 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.09171 on 275 degrees of freedom
Multiple R-squared: 0.7509, Adjusted R-squared: 0.7436
F-statistic: 103.6 on 8 and 275 DF, p-value: < 2.2e-16Which variables are significant in the model? We will consider a variable signficant only if the p-value is below 0.05. (Select all that apply.)
MEI CO2 CH4 N2O CFC.11 CFC.12 TSI Aerosols
#MEI/N2O/CFC.11/CFC.12/TSI/AerosolsCurrent scientific opinion is that nitrous oxide and CFC-11 are greenhouse gases: gases that are able to trap heat from the sun and contribute to the heating of the Earth. However, the regression coefficients of both the N2O and CFC-11 variables are negative, indicating that increasing atmospheric concentrations of either of these two compounds is associated with lower global temperatures.
Which of the following is the simplest correct explanation for this contradiction?
Climate scientists are wrong that N2O and CFC-11 are greenhouse gases - this regression analysis constitutes part of a disproof. There is not enough data, so the regression coefficients being estimated are not accurate. All of the gas concentration variables reflect human development - N2O and CFC.11 are correlated with other variables in the data set.
#All of the gas concentration variables reflect human development - N2O and CFC.11 are correlated with other variables in the data set.Compute the correlations between all the variables in the training set. Which of the following independent variables is N2O highly correlated with (absolute correlation greater than 0.7)? Select all that apply.
MEI CO2 CH4 CFC.11 CFC.12 Aerosols TSI
cor(Training)
Year Month MEI CO2 CH4 N2O CFC.11
Year 1.00000000 -0.0279419602 -0.0369876842 0.98274939 0.91565945 0.99384523 0.56910643
Month -0.02794196 1.0000000000 0.0008846905 -0.10673246 0.01856866 0.01363153 -0.01311122
MEI -0.03698768 0.0008846905 1.0000000000 -0.04114717 -0.03341930 -0.05081978 0.06900044
CO2 0.98274939 -0.1067324607 -0.0411471651 1.00000000 0.87727963 0.97671982 0.51405975
CH4 0.91565945 0.0185686624 -0.0334193014 0.87727963 1.00000000 0.89983864 0.77990402
N2O 0.99384523 0.0136315303 -0.0508197755 0.97671982 0.89983864 1.00000000 0.52247732
CFC.11 0.56910643 -0.0131112236 0.0690004387 0.51405975 0.77990402 0.52247732 1.00000000
CFC.12 0.89701166 0.0006751102 0.0082855443 0.85268963 0.96361625 0.86793078 0.86898518
TSI 0.17030201 -0.0346061935 -0.1544919227 0.17742893 0.24552844 0.19975668 0.27204596
Aerosols -0.34524670 0.0148895406 0.3402377871 -0.35615480 -0.26780919 -0.33705457 -0.04392120
Temp 0.78679714 -0.0998567411 0.1724707512 0.78852921 0.70325502 0.77863893 0.40771029
CFC.12 TSI Aerosols Temp
Year 0.8970116635 0.17030201 -0.34524670 0.78679714
Month 0.0006751102 -0.03460619 0.01488954 -0.09985674
MEI 0.0082855443 -0.15449192 0.34023779 0.17247075
CO2 0.8526896272 0.17742893 -0.35615480 0.78852921
CH4 0.9636162478 0.24552844 -0.26780919 0.70325502
N2O 0.8679307757 0.19975668 -0.33705457 0.77863893
CFC.11 0.8689851828 0.27204596 -0.04392120 0.40771029
CFC.12 1.0000000000 0.25530281 -0.22513124 0.68755755
TSI 0.2553028138 1.00000000 0.05211651 0.24338269
Aerosols -0.2251312440 0.05211651 1.00000000 -0.38491375
Temp 0.6875575483 0.24338269 -0.38491375 1.00000000Which of the following independent variables is CFC.11 highly correlated with? Select all that apply.
MEI CO2 CH4 N2O CFC.12 Aerosols TSI
#CH4/CFC.12Given that the correlations are so high, let us focus on the N2O variable and build a model with only MEI, TSI, Aerosols and N2O as independent variables. Remember to use the training set to build the model.
Enter the coefficient of N2O in this reduced model:
summary(lm(Temp ~ MEI+N2O+TSI+Aerosols,data = Training))
Call:
lm(formula = Temp ~ MEI + N2O + TSI + Aerosols, data = Training)
Residuals:
Min 1Q Median 3Q Max
-0.27916 -0.05975 -0.00595 0.05672 0.34195
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.162e+02 2.022e+01 -5.747 2.37e-08 ***
MEI 6.419e-02 6.652e-03 9.649 < 2e-16 ***
N2O 2.532e-02 1.311e-03 19.307 < 2e-16 ***
TSI 7.949e-02 1.487e-02 5.344 1.89e-07 ***
Aerosols -1.702e+00 2.180e-01 -7.806 1.19e-13 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.09547 on 279 degrees of freedom
Multiple R-squared: 0.7261, Adjusted R-squared: 0.7222
F-statistic: 184.9 on 4 and 279 DF, p-value: < 2.2e-16(How does this compare to the coefficient in the previous model with all of the variables?)
Enter the model R2:
# 0.7261We have many variables in this problem, and as we have seen above, dropping some from the model does not decrease model quality. R provides a function, step, that will automate the procedure of trying different combinations of variables to find a good compromise of model simplicity and R2. This trade-off is formalized by the Akaike information criterion (AIC) - it can be informally thought of as the quality of the model with a penalty for the number of variables in the model.
The step function has one argument - the name of the initial model. It returns a simplified model. Use the step function in R to derive a new model, with the full model as the initial model (HINT: If your initial full model was called “climateLM”, you could create a new model with the step function by typing step(climateLM). Be sure to save your new model to a variable name so that you can look at the summary. For more information about the step function, type ?step in your R console.)
Enter the R square value of the model produced by the step function:
summary(step(climateLM))
Start: AIC=-1348.16
Temp ~ MEI + CO2 + CH4 + N2O + CFC.11 + CFC.12 + TSI + Aerosols
Df Sum of Sq RSS AIC
- CH4 1 0.00049 2.3135 -1350.1
<none> 2.3130 -1348.2
- N2O 1 0.03132 2.3443 -1346.3
- CO2 1 0.06719 2.3802 -1342.0
- CFC.12 1 0.11874 2.4318 -1335.9
- CFC.11 1 0.13986 2.4529 -1333.5
- TSI 1 0.33516 2.6482 -1311.7
- Aerosols 1 0.43727 2.7503 -1301.0
- MEI 1 0.82823 3.1412 -1263.2
Step: AIC=-1350.1
Temp ~ MEI + CO2 + N2O + CFC.11 + CFC.12 + TSI + Aerosols
Df Sum of Sq RSS AIC
<none> 2.3135 -1350.1
- N2O 1 0.03133 2.3448 -1348.3
- CO2 1 0.06672 2.3802 -1344.0
- CFC.12 1 0.13023 2.4437 -1336.5
- CFC.11 1 0.13938 2.4529 -1335.5
- TSI 1 0.33500 2.6485 -1313.7
- Aerosols 1 0.43987 2.7534 -1302.7
- MEI 1 0.83118 3.1447 -1264.9
Call:
lm(formula = Temp ~ MEI + CO2 + N2O + CFC.11 + CFC.12 + TSI +
Aerosols, data = Training)
Residuals:
Min 1Q Median 3Q Max
-0.25770 -0.05994 -0.00104 0.05588 0.32203
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -1.245e+02 1.985e+01 -6.273 1.37e-09 ***
MEI 6.407e-02 6.434e-03 9.958 < 2e-16 ***
CO2 6.402e-03 2.269e-03 2.821 0.005129 **
N2O -1.602e-02 8.287e-03 -1.933 0.054234 .
CFC.11 -6.609e-03 1.621e-03 -4.078 5.95e-05 ***
CFC.12 3.868e-03 9.812e-04 3.942 0.000103 ***
TSI 9.312e-02 1.473e-02 6.322 1.04e-09 ***
Aerosols -1.540e+00 2.126e-01 -7.244 4.36e-12 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 0.09155 on 276 degrees of freedom
Multiple R-squared: 0.7508, Adjusted R-squared: 0.7445
F-statistic: 118.8 on 7 and 276 DF, p-value: < 2.2e-16Which of the following variable(s) were eliminated from the full model by the step function? Select all that apply.
MEI CO2 CH4 N2O CFC.11 CFC.12 TSI Aerosols
#CH4It is interesting to note that the step function does not address the collinearity of the variables, except that adding highly correlated variables will not improve the R2 significantly. The consequence of this is that the step function will not necessarily produce a very interpretable model - just a model that has balanced quality and simplicity for a particular weighting of quality and simplicity (AIC).
We have developed an understanding of how well we can fit a linear regression to the training data, but does the model quality hold when applied to unseen data?
Using the model produced from the step function, calculate temperature predictions for the testing data set, using the predict function.
Enter the testing set R2:
R2
[1] 0.6286051