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.

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.

who <- read.csv("who.csv", sep = ',')
ggplot(who, aes(x = TotExp, y = LifeExp)) + geom_point() + geom_smooth(method = lm, se=F)

lete.lm <- lm(LifeExp ~ TotExp, data = who)
summary(lete.lm)
## 
## 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 F-statistic, which is relatively large at 65.26, give a clear indication that there is a relationship between Total Expenditure and Life Expectancy. As for the standard error, the actual Life Expectancy can deviate from the true regression line by approximately 15.3795867 on average. According to the R^2, the model only accounts for 25.77% of the variance while the p-value is very small making the model significant.

Task 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 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?”

LifeExp2 <- who$LifeExp^4.6
TotExp2 <- who$TotExp^0.06
lete_df <- as.data.frame(cbind(LifeExp2, TotExp2))
ggplot(lete_df, aes(x = TotExp2, y = LifeExp2)) + geom_point() + geom_smooth(method = lm, se=F)

lete2.lm <- lm(LifeExp2 ~ TotExp2, lete_df)
summary(lete2.lm)
## 
## Call:
## lm(formula = LifeExp2 ~ TotExp2, data = lete_df)
## 
## 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 ***
## TotExp2      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

The F-statisticis very much larger at 507.7, give a clear indication of a stronger relationship between the two variables. As for the standard error, the actual Life Expectancy can deviate from the true regression line by approximately 90490000 on average. This is very high. According to the R^2, the model explains the 72.98% of the variance in Life Expectancy which is much better than the previous model. The p-value is smaller which makes the model even more significant.

Task 3

Using the results from 2, forecast life expectancy when \(TotExp^{.06} =1.5\). Then forecast life expectancy when \(TotExp^{.06}=2.5\).

From the second model, the formula is \(LifeExp = -736527910 + 620060216 \times TotExp\)

# Data has to be tramsformed 
lifeexp1 <- -736527910 + 620060216 * 1.5
lifeexp1 ^ (1/4.6)
## [1] 63.31153
lifeexp2 <- -736527910 + 620060216 * 2.5
lifeexp2 ^ (1/4.6)
## [1] 86.50645

Task 4

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

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

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

Task 5

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

62.77 + (1497 * 0.03) + (0.00007233 * 14) - (0.006026 * (0.03 * 14))
## [1] 107.6785

No, this model is not realistic considering the age being 107 years. Based on the second model we saw that as Total Expenditure increases, so does Life Expectancy. 107 with 14 makes it seem impossible.