library(tidyverse)
data = as.data.frame(read.csv('who.csv'))
glimpse(data)
## Rows: 190
## Columns: 10
## $ Country        <chr> "Afghanistan", "Albania", "Algeria", "Andorra", "Angola~
## $ LifeExp        <int> 42, 71, 71, 82, 41, 73, 75, 69, 82, 80, 64, 74, 75, 63,~
## $ InfantSurvival <dbl> 0.835, 0.985, 0.967, 0.997, 0.846, 0.990, 0.986, 0.979,~
## $ Under5Survival <dbl> 0.743, 0.983, 0.962, 0.996, 0.740, 0.989, 0.983, 0.976,~
## $ TBFree         <dbl> 0.99769, 0.99974, 0.99944, 0.99983, 0.99656, 0.99991, 0~
## $ PropMD         <dbl> 0.000228841, 0.001143127, 0.001060478, 0.003297297, 0.0~
## $ PropRN         <dbl> 0.000572294, 0.004614439, 0.002091362, 0.003500000, 0.0~
## $ PersExp        <int> 20, 169, 108, 2589, 36, 503, 484, 88, 3181, 3788, 62, 1~
## $ GovtExp        <int> 92, 3128, 5184, 169725, 1620, 12543, 19170, 1856, 18761~
## $ TotExp         <int> 112, 3297, 5292, 172314, 1656, 13046, 19654, 1944, 1907~

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.

  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.
plot(data$LifeExp~data$TotExp)

lm <- lm(data = data, data$LifeExp~data$TotExp)
summary <- summary(lm)
summary
## 
## Call:
## lm(formula = data$LifeExp ~ data$TotExp, data = 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 ***
## data$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
plot(lm)

F-Statistic: The large F-statistic (65.26) indicates a strong relationship between the dependent and independent variables.

\(R^2\): The multiple R-squared value of 0.2577, implies that 25.77% of the variation in the data can be explained by the model.

Standard Error: The high standard error tells us that the residuals are more spread out in relation to the fitted line.

P-Values: The p-value is less than 0.05, the significance threshold. This implies that the data is significant and not due to chance.

Assumption for Linear Regression: The Normal Q-Q plots shows the residuals do not follow the normal distribution line. The histogram shows the residuals are not normally distributed. The variation of the residuals in not constant. The assumptions for linear regression are not met.

  1. 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?”
data$LifeExp_4.6 <- data$LifeExp^4.6
data$TotExp_.06 <- data$TotExp^0.06

plot(data$LifeExp~data$TotExp)

lm2 <- lm(data = data, data$LifeExp_4.6~data$TotExp_.06)
summary2 <- summary(lm2)
summary2
## 
## Call:
## lm(formula = data$LifeExp_4.6 ~ data$TotExp_.06, data = 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 ***
## data$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(lm2)

F-Statistic: The large F-Statistic (507.7) indicates a strong relationship between Life Expectancy and Total Expenditure.

\(R^2\): The high R-squared values (0.7298) indicates that the model can explain 72.89% of the variation in the data.

Standard Error: The standard error, 90490000, is very high.

P-Value: The p-value is less than 0.05, the significance threshold. This implies that the data is significant and not due to chance.

Assumptions for Linear Regression: The scatter plot shows a positive linear trend. The residual v. fitted plot shows constant variability. The Normal Q-Q plot shows the residuals following the normal distribution line more closely. The assumptions for linear regression are met for this model.

  1. Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
exp(log(-736527910 + 620060216*(1.5))/4.6)
## [1] 63.31153
exp(log(-736527910 + 620060216*(2.5))/4.6)
## [1] 86.50645

The life expectancy when TotExp^.06=1.5 is approximately 63 years. The life expectancy when TotExp^.06=2.5 is approximately 86 years.

  1. 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

lm3 <- lm(data$LifeExp~data$TotExp + data$PropMD + data$PropMD * data$TotExp)
summary3 <- summary(lm3)
summary3
## 
## Call:
## lm(formula = data$LifeExp ~ data$TotExp + data$PropMD + data$PropMD * 
##     data$TotExp)
## 
## 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 ***
## data$TotExp              7.233e-05  8.982e-06   8.053 9.39e-14 ***
## data$PropMD              1.497e+03  2.788e+02   5.371 2.32e-07 ***
## data$TotExp:data$PropMD -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
plot(lm3)

F-Statistic: The F-Statistic is not very high (8.767), but does indicate a relationship between the dependent variables and independent variable.

\(R^2\): The R-squared value of 0.3574, indicates that 35.74% of the variation in the data can be explained by the model.

Standard Error: The standard error 8.765, tells us that the residuals are on average 8.765 years away from the fitted model.

P-value: The p-value is less than 0.05, the significance threshold. This implies that the data is significant and not due to chance.

Assumptions for Linear Regression: The residuals v. fitted plot does not show constant variability. The Normal Q-Q plot shows the residuals following the theoretical normal line. The histogram shows a large proportion of the residuals on the right, with the outliers that skew the distribution to the left. The assumptions for linear regression were not met for this model.

  1. Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
6.277e+01 + 7.233e-05*(14) +  1.497e+03*(0.03) + -6.026e-03*(14)*(0.03)
## [1] 107.6785

The forcast shows a life expectancy of approximately 107 years. This does not seem realistic, because the increase in the proportion of doctors is not matched by the set expenditure.