Load in the Data and Libraries

df <- read.csv("C:/Users/carnout/Documents/who.csv",header =TRUE)
library(ggplot2)
library(tidyverse)

## -- Attaching packages ------------------------------------------------------------------------------------------------- tidyverse 1.2.1 --

## v tibble  1.4.2     v purrr   0.2.5
## v tidyr   0.8.2     v dplyr   0.7.8
## v readr   1.3.1     v stringr 1.3.1
## v tibble  1.4.2     v forcats 0.3.0

## -- Conflicts ---------------------------------------------------------------------------------------------------- tidyverse_conflicts() --
## x dplyr::filter() masks stats::filter()
## x dplyr::lag()    masks stats::lag()

Question 1

ggplot(data = df, aes(x = TotExp, y = LifeExp)) + 
  geom_point(color='blue') +
  geom_smooth(method = "lm", se = FALSE)

model <- lm(LifeExp~TotExp,df)
plot(model)

summary(model)

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

model

## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = df)
## 
## Coefficients:
## (Intercept)       TotExp  
##   6.475e+01    6.297e-05

Results from First Model

The Straight up Linear Model between Life Expectancy and Total Expenditure shouws that for every additional 15880 (in units of tens of millions of dollars) spent, the average life expenctancy will go up one year.

The Standard Error (S) is 9.371, which tells us that the average distance of the data points from the fitted line is about 9.371%

The P-value between the two variables is 7.714e-14 which shows that there is a statistically significant relationship between government expenditure and the life expectancy of it’s citizensd

The F- statistic also corresponds to a statistically significant relationship between the two variabels

The Adjusted R-squared value is .2537 which means that roughly 25% of the variability in life expectancy can be explained by the country’s expenditures

Question 2 - Manipulating the Variables for a Second Regression

df2 <- df

df2$LifeExp <- df2$LifeExp **(4.6)
df2$TotExp <- df2$TotExp **(.06)


ggplot(data = df2, aes(x = TotExp, y = LifeExp)) + 
  geom_point(color='blue') +
  geom_smooth(method = "lm", se = FALSE)

transformed_model <- lm(LifeExp~TotExp,df2)
plot(transformed_model)

summary(transformed_model)

## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = df2)
## 
## 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       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

transformed_model

## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = df2)
## 
## Coefficients:
## (Intercept)       TotExp  
##  -736527909    620060216

THe transformed model has an even more statistically signifcant F statistic and P-value and the R-squared value is no .728 which means that almost 73% of the variability of the response variable can be explained by the predictor variable.

Question 3 - Predict the life expectancy when TotExp^.06 = 1.5.

forecasted_Life_expectancy <- function(x){
  ans <- transformed_model$coefficients[1] + transformed_model$coefficients[2] * x
  return(ans ** (1/4.6)) # Must transform the data into a sensible number
}

print(paste0("What is the forecasted life expectancy when TotExp^.06 = 1.5? ", round(forecasted_Life_expectancy(1.5),2), " years."))

## [1] "What is the forecasted life expectancy when TotExp^.06 = 1.5? 63.31 years."

print(paste0("What is the forecasted life expectancy when TotExp^.06 = 2.5? ", round(forecasted_Life_expectancy(2.5),2), " years."))

## [1] "What is the forecasted life expectancy when TotExp^.06 = 2.5? 86.51 years."

Question 4 Build the Following Model

\[LifeExp = B_0 + B_1 * PropMD + B_2 * TotExp + B_3 * PropMD * TotExp\]

df$PropMD_TotExp <- df$PropMD*df$TotExp

multiple_model <- lm(LifeExp~ PropMD + TotExp + PropMD_TotExp,df)
plot(multiple_model)

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

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

summary(multiple_model)

## 
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD_TotExp, data = df)
## 
## 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

transformed_model

## 
## Call:
## lm(formula = LifeExp ~ TotExp, data = df2)
## 
## Coefficients:
## (Intercept)       TotExp  
##  -736527909    620060216

This model has an F-statistic of 34.49 and a statistically significant p-value < 2.2e-16, a residual standard error of 8.765, and adjusted R-squared 0.3471. This model is better than the first but worse than the second. The only thing improved here compared to the first model is that the precision is slightly better with the standard error being at 8.765. The adjusted R-squared suggests that 34.71% of the variation in this data could be interpreted from this model.

Question 5 - Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this seem realistic

forecast_revised.lm <- function(propmd, totexp){
  ans <- summary(multiple_model)$coefficients[1] + summary(multiple_model)$coefficients[2] * propmd + summary(multiple_model)$coefficients[3] * totexp + summary(multiple_model)$coefficients[4] * propmd * totexp
  return(ans)
}

print(paste0("What is the forecasted life expectancy when PropMD = 0.3 and TotExp = 14? ", round(forecast_revised.lm(0.3,14),2), " years."))

## [1] "What is the forecasted life expectancy when PropMD = 0.3 and TotExp = 14? 512 years."

The answer is 512 years which does not seem like a reasonable life expectancy