library(tidyverse)

Dataset

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.

#Data Load

who = read.csv("https://raw.githubusercontent.com/zahirf/Data605/master/who.csv", stringsAsFactors = F)
head(who)
##               Country LifeExp InfantSurvival Under5Survival  TBFree      PropMD
## 1         Afghanistan      42          0.835          0.743 0.99769 0.000228841
## 2             Albania      71          0.985          0.983 0.99974 0.001143127
## 3             Algeria      71          0.967          0.962 0.99944 0.001060478
## 4             Andorra      82          0.997          0.996 0.99983 0.003297297
## 5              Angola      41          0.846          0.740 0.99656 0.000070400
## 6 Antigua and Barbuda      73          0.990          0.989 0.99991 0.000142857
##        PropRN PersExp GovtExp TotExp
## 1 0.000572294      20      92    112
## 2 0.004614439     169    3128   3297
## 3 0.002091362     108    5184   5292
## 4 0.003500000    2589  169725 172314
## 5 0.001146162      36    1620   1656
## 6 0.002773810     503   12543  13046

Question 01

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.

Solution:

Simple regression model

lm1 <- lm(LifeExp~TotExp, data=who)
summary(lm1)
## 
## 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

Scatterplot

plot(who$LifeExp~who$TotExp, xlab='Average Life Expectancy', ylab='Sum of Government Expenditures', 
     main='LifeExp vs TotExp')
abline(lm1)

The linear model is a statistically significant of the evaluation score with p-value less than 0.05.

he \(R^2\) indicates that .2577 of variation in life expectancy is explained by the total expenditure. The Standard Error is smaller then the coefficient. The P-value is small which indicates that expenditure is a significant variable and that it is likely to impact life expectancy. The F-Statistic is large which indicates a strong relationship between the independent and dependent variables.

Residuals

plot(lm1)

The residual plot shows there is no constant variability and that the residuals are not normally distributed. This model does not fit the data well to describe the relationship between total expenditure and life expectancy.

Question 02

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?”

Solution

LifeExp_46 = who$LifeExp^4.6
TotExP_06 = who$TotExp^0.06
lm2 = lm(LifeExp_46 ~ TotExP_06, who)
summary(lm2)
## 
## Call:
## lm(formula = LifeExp_46 ~ TotExP_06, 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 ***
## TotExP_06    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
plot(LifeExp_46~TotExP_06, 
     xlab="Total Expenditures", ylab="Life Expectancy",
     main="Life Expectancy vs Total Expenditures (Transformed)")
abline(lm2)

The second linear model is a statistically significant predictor of evaluation score with p-value less than 0.05. For Multiple R-squared and R-squared, the model is only around 73% fits the data.

The \(R^2\) indicates that .7298 of variation in life expectancy is explained by the total expenditure. The Standard Error is smaller then the coefficient. The P-value is small which indicates that expenditure is a significant variable and that it is likely to impact life expectancy. The F-Statistic is large which indicates a strong relationship between the independent and dependent variables.

plot(lm2)

The residual plot shows there is constant variability and that the residuals are more normally distributed. This model fits the data well enough to describe the relationship between total expenditure and life expectancy.

Question 03

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

Solution

TotExp^.06 =1.5

x <-1.5
y <- lm2$coefficients[1] + lm2$coefficients[2]* x
life_exp <- round(y^(1/4.6), 2)
print(life_exp)
## (Intercept) 
##       63.31

For TotExp^0.06=1.5, the forecasted life expectancy is 63.31 years.

TotExp^0.06=2.5

x <-2.5
y <- lm2$coefficients[1] + lm2$coefficients[2]* x
life_exp <- round(y^(1/4.6), 2)
print(life_exp)
## (Intercept) 
##       86.51

For TotExp^0.06=2.5, the forecasted life expectancy is 86.51

Question 04

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

\[LifeExp = b0+b1 * PropMd + b2 * TotExp +b3 * PropMD * TotExp\]

Solution

lm3 <- lm(LifeExp ~ PropMD+TotExp+(PropMD*TotExp), data=who)
summary(lm3)
## 
## 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 lm 3 seems to be a better model than lm1. This model is a statistically significant predictor of evaluation score. For Multiple R-squared and R-squared, the model is only around 34% to 35% fits the data.

The \(R^2\) indicates that .3574 of variation in life expectancy is explained by the total expenditure. The Standard Error is smaller then the coefficient. The P-value is small which indicates that expenditure is a significant variable and that it is likely to impact life expectancy. The F-Statistic is not large which does not indicates a strong relationship between the independent and dependent variables.

plot(lm3)

## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced

## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced

The residual plot shows there is no constant variability and that the residuals are not normally distributed. This model does not fit the data well enough to describe the relationship between total expenditure and life expectancy.The data seems to show a quadratic relationship.

Question 05

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

Solution

propmd = 0.03
totexp = 14

LifeExp<- lm3$coefficients[1] + lm3$coefficients[2] * propmd +
    lm3$coefficients[3] * totexp + lm3$coefficients[4] * propmd * totexp
print(LifeExp)
## (Intercept) 
##     107.696

The life expectancy predicted by the model is 107.7 years, which seems unrealistic as the maximum life expectancy in the dataset is 83 years. The model seems to overpredict.

print(max(who$LifeExp))
## [1] 83