Library & Data Read in 

library(dplyr)
## 
## Attaching package: 'dplyr'
## The following objects are masked from 'package:stats':
## 
##     filter, lag
## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union
med_data <- read.csv("who.csv", header = TRUE)
summary(med_data)
##    Country             LifeExp      InfantSurvival   Under5Survival  
##  Length:190         Min.   :40.00   Min.   :0.8350   Min.   :0.7310  
##  Class :character   1st Qu.:61.25   1st Qu.:0.9433   1st Qu.:0.9253  
##  Mode  :character   Median :70.00   Median :0.9785   Median :0.9745  
##                     Mean   :67.38   Mean   :0.9624   Mean   :0.9459  
##                     3rd Qu.:75.00   3rd Qu.:0.9910   3rd Qu.:0.9900  
##                     Max.   :83.00   Max.   :0.9980   Max.   :0.9970  
##      TBFree           PropMD              PropRN             PersExp       
##  Min.   :0.9870   Min.   :0.0000196   Min.   :0.0000883   Min.   :   3.00  
##  1st Qu.:0.9969   1st Qu.:0.0002444   1st Qu.:0.0008455   1st Qu.:  36.25  
##  Median :0.9992   Median :0.0010474   Median :0.0027584   Median : 199.50  
##  Mean   :0.9980   Mean   :0.0017954   Mean   :0.0041336   Mean   : 742.00  
##  3rd Qu.:0.9998   3rd Qu.:0.0024584   3rd Qu.:0.0057164   3rd Qu.: 515.25  
##  Max.   :1.0000   Max.   :0.0351290   Max.   :0.0708387   Max.   :6350.00  
##     GovtExp             TotExp      
##  Min.   :    10.0   Min.   :    13  
##  1st Qu.:   559.5   1st Qu.:   584  
##  Median :  5385.0   Median :  5541  
##  Mean   : 40953.5   Mean   : 41696  
##  3rd Qu.: 25680.2   3rd Qu.: 26331  
##  Max.   :476420.0   Max.   :482750

Basic Model Building 

# Build a linear model for stopping distance as a function of speed
model <- lm(LifeExp  ~ TotExp, data = med_data)
summary(model)
## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = med_data)
## 
## 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

Looking at the summary, the R-Square of 0.2577 indicates that ~25% of variation in life expectancy can be attributed to the Total Expenses. In addition, the standard error for TotExp is small.

Analyzing the F-Statistic & p-value, we can test the null hypothesis. Since the p-value is less than 0.05, we can reject the null hypothesis, and conclude the model is significant.

Basic Visualization 

plot(LifeExp  ~ TotExp, data = med_data, main = "Life Expectancy vs Healthcare Expenditures", xlab = "Total Healthcare Expenses (Per Person)", ylab = "Life Expectancy")
abline(model, col = "blue")

Looking at the data, its clear that there is a wide amount of variability in healthcare costs to outcomes. I find it fascinating to see that there is a an asymptote at approximately 80 years old, and I wonder how much genetics & lifestyle of those countries play into that value.

par(mfrow = c(2, 2))
plot(model, which = 1:4)

Practically, we look for 4 key parts to analyze if a regression is reasonable.

Independence-This is assumed There is a linear relationship, which I don’t feel is solidily demonstrated in the scatter plot above, as to me it looks like this is a reciprocal function.
There is constant variance, which is shown by the uniform spread of residuals. There is a normality of Residuals across a reasonable range, which is not demonstrated by the chart above.

Also just looking at the normal qq plot, the curve would indicate the sample data is skewed.

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?

Right off the bat, I’m going to say its probably going to get better, but lets check it:

# Build a linear model for stopping distance as a function of speed
model <- lm(I(LifeExp^4.6)  ~ I(TotExp^0.06), data = med_data)
summary(model)
## 
## Call:
## lm(formula = I(LifeExp^4.6) ~ I(TotExp^0.06), data = med_data)
## 
## 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 ***
## I(TotExp^0.06)  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

Looking at the summary, the R-Square significantly improved, from 0.2577 to 0.7298 indicates that significantly more of the variation in life expectancy can be attributed to the Total Expenses. In addition, the standard error for TotExp is small.

Analyzing the F-Statistic & p-value, we can test the null hypothesis. Since the p-value is less than 0.05, we can reject the null hypothesis, and conclude the model is significant.

Basic Visualization 

plot(I(LifeExp^4.6)  ~ I(TotExp^0.06), data = med_data, main = "Life Expectancy vs Healthcare Expenditures", xlab = "Total Healthcare Expenses (Per Person)", ylab = "Life Expectancy^4.6")
abline(model, col = "blue")

Honestly, adding in these new exponents appeared to do an effective job of spreading and quashing the data, or make it behave in a more linear fashion.

par(mfrow = c(2, 2))
plot(model, which = 1:4)

Practically, we look for 4 key parts to analyze if a regression is reasonable.

Independence-This is assumed There is some variance, as the chart of residuals vs fitted is still skewed, but it is a significant improvement over the prior chart. There is a normality of Residuals across a reasonable range, which is somewhat demonstrated by the chart above. Finally, the normal qq plot has its lower tails deviate from linear, indicating that there are some extreme values. This observation is further reinforced by the chart of Cook’s distance.

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

TotExp^.06 =1.5 

predict(model, newdata = data.frame(TotExp = 1.5^(1/0.06)))^(1/4.6)
##        1 
## 63.31153

TotExp^.06 =2.5 

predict(model, newdata = data.frame(TotExp = 2.5^(1/0.06)))^(1/4.6)
##        1 
## 86.50645

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) 

model <- lm(LifeExp  ~ I(TotExp * PropMD) + TotExp + PropMD, data = med_data)
summary(model)
## 
## Call:
## lm(formula = LifeExp ~ I(TotExp * PropMD) + TotExp + PropMD, 
##     data = med_data)
## 
## 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 ***
## I(TotExp * PropMD) -6.026e-03  1.472e-03  -4.093 6.35e-05 ***
## TotExp              7.233e-05  8.982e-06   8.053 9.39e-14 ***
## PropMD              1.497e+03  2.788e+02   5.371 2.32e-07 ***
## ---
## 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
par(mfrow = c(2, 2))
plot(model, which = 1:4)

Firstly, we look for 3 key parts to analyze if a regression is reasonable.

Independence-This is assumed There is constant variance, which is shown by the uniform spread of residuals. There is a normality of Residuals across a reasonable range, which is not demonstrated by the chart above.

Also just looking at the normal qq plot, the curve would indicate the sample data is skewed.

Looking at the summary, the R-Square decreased, from 0.7298 to 0.3574. indicates that significantly more of the variation in life expectancy can be attributed to the Total Expenses rather than a combination of expenses and proportion of population that is doctors. In addition, the standard error for TotExp is small.

Analyzing the F-Statistic & p-value, we can test the null hypothesis. Since the p-value is less than 0.05, we can reject the null hypothesis, and conclude the model is significant, although not as good as a fit as the prior models.

Also just looking at the normal qq plot, the curve and the deviation of the tails would indicate the sample data is skewed, but it is not as bad as prior iterations.

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

predict(model, newdata = data.frame(I(14 * 0.03), TotExp = 14, PropMD=0.03))
##       1 
## 107.696

No this does not seem realistic as the expected age is 107, which is super unrealistic. As much as healthcare cost influences age, there are a slew of other factors, ranging from genetics to technology that will impact how long someone lives. Just from the technological standpoint, think of how much medicine has changed since the 1900s!

All in all, its a great predictor that indicates we shold try and improve the PropMD, but at the same time, throwing money at a problem doesn’t solve it.