Task:

  1. Provide a scatterplot of LifeExp~TotExp, and run simple linear regression. Do not transform the variables. Provide and interpret the F statistics, R^2, standard error,and p-values only. Discuss whether the assumptions of simple linear regression met.

  2. Raise life expectancy to the 4.6 power (i.e., LifeExp^4.6). Raise total expenditures to the 0.06 power (nearly a log transform, TotExp^.06). Plot LifeExp^4.6 as a function of TotExp^.06, and r re-run the simple regression model using the transformed variables. Provide and interpret the F statistics, R^2, standard error, and p-values. Which model is “better?”

  3. Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.

  4. Build the following multiple regression model and interpret the F Statistics, R^2, standard error, and p-values. How good is the model?

\(LifeExp = b0+b1 x PropMd + b2 x TotExp +b3 x PropMD x TotExp\)

  1. Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?

Read Data

who <- read.csv("who.csv")
head(who)
##               Country LifeExp InfantSurvival Under5Survival  TBFree
## 1         Afghanistan      42          0.835          0.743 0.99769
## 2             Albania      71          0.985          0.983 0.99974
## 3             Algeria      71          0.967          0.962 0.99944
## 4             Andorra      82          0.997          0.996 0.99983
## 5              Angola      41          0.846          0.740 0.99656
## 6 Antigua and Barbuda      73          0.990          0.989 0.99991
##        PropMD      PropRN PersExp GovtExp TotExp
## 1 0.000228841 0.000572294      20      92    112
## 2 0.001143127 0.004614439     169    3128   3297
## 3 0.001060478 0.002091362     108    5184   5292
## 4 0.003297297 0.003500000    2589  169725 172314
## 5 0.000070400 0.001146162      36    1620   1656
## 6 0.000142857 0.002773810     503   12543  13046

Summary of Data

summary(who)
##                 Country       LifeExp      InfantSurvival  
##  Afghanistan        :  1   Min.   :40.00   Min.   :0.8350  
##  Albania            :  1   1st Qu.:61.25   1st Qu.:0.9433  
##  Algeria            :  1   Median :70.00   Median :0.9785  
##  Andorra            :  1   Mean   :67.38   Mean   :0.9624  
##  Angola             :  1   3rd Qu.:75.00   3rd Qu.:0.9910  
##  Antigua and Barbuda:  1   Max.   :83.00   Max.   :0.9980  
##  (Other)            :184                                   
##  Under5Survival       TBFree           PropMD              PropRN         
##  Min.   :0.7310   Min.   :0.9870   Min.   :0.0000196   Min.   :0.0000883  
##  1st Qu.:0.9253   1st Qu.:0.9969   1st Qu.:0.0002444   1st Qu.:0.0008455  
##  Median :0.9745   Median :0.9992   Median :0.0010474   Median :0.0027584  
##  Mean   :0.9459   Mean   :0.9980   Mean   :0.0017954   Mean   :0.0041336  
##  3rd Qu.:0.9900   3rd Qu.:0.9998   3rd Qu.:0.0024584   3rd Qu.:0.0057164  
##  Max.   :0.9970   Max.   :1.0000   Max.   :0.0351290   Max.   :0.0708387  
##                                                                           
##     PersExp           GovtExp             TotExp      
##  Min.   :   3.00   Min.   :    10.0   Min.   :    13  
##  1st Qu.:  36.25   1st Qu.:   559.5   1st Qu.:   584  
##  Median : 199.50   Median :  5385.0   Median :  5541  
##  Mean   : 742.00   Mean   : 40953.5   Mean   : 41696  
##  3rd Qu.: 515.25   3rd Qu.: 25680.2   3rd Qu.: 26331  
##  Max.   :6350.00   Max.   :476420.0   Max.   :482750  
##

Task 1:

library(ggplot2)
ggplot(data = who, aes(x=who$LifeExp, y=who$TotExp)) + geom_point(size = 2, alpha = .4) + geom_smooth(method = "lm", se = FALSE)

linearmodel = lm(LifeExp~ TotExp, data = who)

summary(linearmodel)
## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = who)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -24.764  -4.778   3.154   7.116  13.292 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 6.475e+01  7.535e-01  85.933  < 2e-16 ***
## TotExp      6.297e-05  7.795e-06   8.079 7.71e-14 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 9.371 on 188 degrees of freedom
## Multiple R-squared:  0.2577, Adjusted R-squared:  0.2537 
## F-statistic: 65.26 on 1 and 188 DF,  p-value: 7.714e-14

The linear model is as follow:

\(LifeExp=6.475e+6.297e-05*TotExp\)

It is of statistically significant that the predictor of evaluation score with p-value less than 0.05.

However, the Multiple R-squared and R-squared, of the model is around 25% fits the data.

qqnorm(linearmodel$residuals)
qqline(linearmodel$residuals)

Task 2:

raiseLifeExp = who$LifeExp^4.6
raiseTotExp = who$TotExp^0.06

linearmodel2 =lm(raiseLifeExp~ raiseTotExp, data = who)
summary(linearmodel2)
## 
## Call:
## lm(formula = raiseLifeExp ~ raiseTotExp, data = who)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -308616089  -53978977   13697187   59139231  211951764 
## 
## Coefficients:
##               Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -736527910   46817945  -15.73   <2e-16 ***
## raiseTotExp  620060216   27518940   22.53   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 90490000 on 188 degrees of freedom
## Multiple R-squared:  0.7298, Adjusted R-squared:  0.7283 
## F-statistic: 507.7 on 1 and 188 DF,  p-value: < 2.2e-16

linear model is:

\(LifeExp = ((−736527910+620060216)(TotExpf))\)

The predictor of evaluation score with p-value less than 0.05.

In addition, for Multiple R-squared and R-squared, the model is only around 72% fits the data.

This model using the transformed variables has a larger Multiple R-squared and R-squared, making it a better model than the first.

qqnorm(linearmodel2$residuals)
qqline(linearmodel2$residuals)

Task 3:

\(LifeExp = ((−736527910+620060216)(TotExpf))\)

TotExp = 1.5
raiseLife = -736527910 + 620060216*TotExp
raiseLife
## [1] 193562414
LifeExp = raiseLife ^(1/4.6)
LifeExp
## [1] 63.31153

Forecast life expectancey: 63

Task 4:

\(LifeExp=((b0+b1)(PropMd))+((b2)(TotExp))+(((b3)(PropMD)(TotExp)))\)

mmodel = lm(LifeExp ~ PropMD + TotExp + PropMD*TotExp, data = who)
summary(mmodel)
## 
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = who)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -27.320  -4.132   2.098   6.540  13.074 
## 
## Coefficients:
##                 Estimate Std. Error t value Pr(>|t|)    
## (Intercept)    6.277e+01  7.956e-01  78.899  < 2e-16 ***
## PropMD         1.497e+03  2.788e+02   5.371 2.32e-07 ***
## TotExp         7.233e-05  8.982e-06   8.053 9.39e-14 ***
## PropMD:TotExp -6.026e-03  1.472e-03  -4.093 6.35e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 8.765 on 186 degrees of freedom
## Multiple R-squared:  0.3574, Adjusted R-squared:  0.3471 
## F-statistic: 34.49 on 3 and 186 DF,  p-value: < 2.2e-16

The linear model is :

\(LifeExp=(((6.27710+1.497103)(PropMD))+((7.23310−5)(TotExp))−((6.02610−3)(PropMDTotExp))\)

F-statistic with a P-value < 0.05 is a a statistically significant predictor of evaluation score.

The Multiple R-squared and adjusted R-squared is 34%-35% of what the data fits.

The F-Statistic of 34.49 with a relatively small p-value shows that the currect model is better than the model with one less predictor.

The R2 value of .357 means the model explains 35.7% of the data which may not be a good fit.

Task 5:

PropMD = 0.03
TotExp = 14
LifeExp =  (6.277*10) + (1.497*10^3)*PropMD + (7.233*10^-5)*TotExp - (6.026*10^-3) *PropMD*TotExp
LifeExp
## [1] 107.6785

Life expectancy with given PropMD and TotExp is 107. This forecast does not look realistic as this is more than the max life expectancy which is 83.