DATA 605 HW 12

The attached who.csv dataset contains real-world data from 2008. The variables included follow.

Country: name of the country
LifeExp: average life expectancy for the country in years
InfantSurvival: proportion of those surviving to one year or more
Under5Survival: proportion of those surviving to five years or more
TBFree: proportion of the population without TB.
PropMD: proportion of the population who are MDs
PropRN: proportion of the population who are RNs
PersExp: mean personal expenditures on healthcare in US dollars at average exchange rate
GovtExp: mean government expenditures per capita on healthcare, US dollars at average exchange rate
TotExp: sum of personal and government expenditures.

library(knitr)
whodf <- read.csv(file="https://raw.githubusercontent.com/monuchacko/cuny_msds/master/data_605/who.csv", header=TRUE, sep=",")

kable(head(whodf), digits = 2, align = c(rep("l", 4), rep("c", 4), rep("r", 4)))

Country	LifeExp	InfantSurvival	Under5Survival	TBFree	PersExp	GovtExp	TotExp
Afghanistan	42	0.84	0.74	1	20	92	112
Albania	71	0.98	0.98	1	169	3128	3297
Algeria	71	0.97	0.96	1	108	5184	5292
Andorra	82	1.00	1.00	1	2589	169725	172314
Angola	41	0.85	0.74	1	36	1620	1656
Antigua and Barbuda	73	0.99	0.99	1	503	12543	13046

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).

plot(whodf$LifeExp ~ whodf$TotExp, main = "LifeExp vs TotExp", xlab = "Pers and gov expenditures", ylab = "Average life expectancy")
abline(lm(whodf$LifeExp ~ whodf$TotExp), col="red") # regression line (y~x)

# linear regression
m1 <- lm(LifeExp ~ TotExp, data = whodf)
summary(m1)

## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = whodf)
## 
## 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

From the data above we can say that there is high likelihood that the model is explaining the data failrly well. Here F-Stat is 65.26 and p-value is close to 0.

qqnorm(m1$residuals)
qqline(m1$residuals)

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?”

whodf2<-whodf
whodf2$LifeExp<-whodf2$LifeExp^4.6
whodf2$TotExp<-whodf2$TotExp^0.6

kable(head(whodf2), digits = 2, align = c(rep("l", 4), rep("c", 4), rep("r", 4)))

Country	LifeExp	InfantSurvival	Under5Survival	TBFree	PersExp	GovtExp	TotExp
Afghanistan	29305338	0.84	0.74	1	20	92	16.96
Albania	327935478	0.98	0.98	1	169	3128	129.08
Algeria	327935478	0.97	0.96	1	108	5184	171.46
Andorra	636126841	1.00	1.00	1	2589	169725	1386.09
Angola	26230450	0.85	0.74	1	36	1620	85.40
Antigua and Barbuda	372636298	0.99	0.99	1	503	12543	294.64

plot(whodf2$LifeExp ~ whodf2$TotExp, main = "LifeExpTransformed vs TotExpTransformed", xlab = "Pers and gov expenditures", ylab = "Average life expectancy")
abline(lm(whodf2$LifeExp ~ whodf2$TotExp), col="red") # regression line (y~x)

m2 <- lm(LifeExp ~ TotExp, data = whodf2)
summary(m2)

## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = whodf2)
## 
## Residuals:
##        Min         1Q     Median         3Q        Max 
## -257351739  -82599957   14030425   93896945  237720335 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 211907647   10234512   20.70   <2e-16 ***
## TotExp         238461      15021   15.88   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 113800000 on 188 degrees of freedom
## Multiple R-squared:  0.5728, Adjusted R-squared:  0.5705 
## F-statistic:   252 on 1 and 188 DF,  p-value: < 2.2e-16

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

TExp <- 1.5
LExp <- 238461*TExp + 211907647
round(LExp ^ (1/4.6),1)

## [1] 64.6

TExp <- 2.5
LExp <- 238461*TExp + 211907647
round(LExp ^ (1/4.6),1)

## [1] 64.6

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

m3 <- lm(LifeExp ~ PropMD + TotExp + PropMD*TotExp, data = whodf)
summary(m3)

## 
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = whodf)
## 
## 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

qqnorm(m3$residuals)
qqline(m3$residuals)

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

TExp <- 14
PrMD <- 0.03
LExp <- 6.277e+01 + 1.497e+03*PrMD + 7.233e-05*TExp -6.026e-03*PrMD*TExp
round(LExp,1)

## [1] 107.7

DATA 605 HW 12

Monu Chacko

4/26/2020