Load & View the Data

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.

who <- read.csv('who.csv')
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

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(who$TotExp,who$LifeExp)

Linear Regression - Who Data

  • The standard error tells how much variability the coefficients of the model can have. Ideally the standard error values are 5-10 times smaller than the coefficient.
    • The standard error for TotExp is 8 times smaller than the coefficient for TotExp (6.297e-05/7.795e-06)
    • The standard error for the intercept is 86 times smaller than the coefficient for intercept (6.475e+01/7.535e-01)
  • The p-value for TotExp and the intercept are both below 0.05 so they are statistically significant.
    • The probability that TotExp is not relevant is 7.71e-14, which is very low
    • The probability that the intercept is not relevant is 2e-16, which is very low
  • The R^2 value is 0.2577, which tells us the linear model describes 25.77% of the variability in the data.
  • The F statistic is 65.26. This statistic tells how the current model compares to the model with 1 less variables. From our book we know that, “Because the one-factor model already has only a single parameter, this test is not particularly useful in this case.” (page 24).

This doesn’t appear to be a great model but it is better than nothing.

who_lm <- lm(who$LifeExp ~ who$TotExp)
summary(who_lm)
## 
## Call:
## lm(formula = who$LifeExp ~ who$TotExp)
## 
## 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 ***
## who$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

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?”

who2 <- who
who2$LifeExp4Point6  <- who2$LifeExp^4.6
who2$TotExpPoint6 <- who2$TotExp^0.06
plot(who2$TotExpPoint6,who2$LifeExp4Point6)

Summary Who2

  • The standard error tells how much variability the coefficients of the model can have. Ideally the standard error values are 5-10 times smaller than the coefficient.
    • The standard error for TotExp is ~23 times smaller than the coefficient for TotExp (620060216/27518940)
    • The standard error for the intercept is ~16 times smaller than the coefficient for intercept (-736527910/46817945)
  • The p-value for TotExp and the intercept are both below 0.05 so they are statistically significant.
    • The probability that TotExp is not relevant is <2e-16, which is very low
    • The probability that the intercept is not relevant is <2e-16, which is very low
  • The R^2 value is 0.7298, which tells us the linear model describes 72.98% of the variability in the data, which is pretty good
  • The F statistic is 507.7 and is even higher than before. Since this number is so large, this statistic tells us the model is good.

This model predicts the data fairly well and much better than the first model. The R-squared value alone increased by almost 50%.

who2_lm <- lm(who2$LifeExp4Point6 ~ who2$TotExpPoint6)
summary(who2_lm)
## 
## Call:
## lm(formula = who2$LifeExp4Point6 ~ who2$TotExpPoint6)
## 
## 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 ***
## who2$TotExpPoint6  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

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

Our function is the following: LifeExp4Point6 = -736,527,910 + 620,060,216*TotalExpPoint6

TotalExpPoint6 = 1.5
LifeExp4Point6 <- -736527910 + 620060216 * TotalExpPoint6
LifeExp <- LifeExp4Point6^(1/4.6)
paste("When total exp ^ 0.06 is 1.5 then Life Exp is ",LifeExp)
## [1] "When total exp ^ 0.06 is 1.5 then Life Exp is  63.3115334478635"
TotalExp = 2.5
LifeExp4Point6 <- (-736527910 + 620060216 * TotalExp)
LifeExp <- LifeExp4Point6^(1/4.6)
paste("When total exp ^ 0.06 is 2.5 then Life Exp is ",LifeExp)
## [1] "When total exp ^ 0.06 is 2.5 then Life Exp is  86.5064484928337"

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 x PropMd + b2 x TotExp +b3 x PropMD x TotExp

who4 <- who
who4$PropMDTotExp <- who4$PropMD * who4$TotExp

Summary of Model:

  • The standard error tells how much variability the coefficients of the model can have. Ideally the standard error values are 5-10 times smaller than the coefficient.
    • The standard error for PropMD is ~5 times smaller than the coefficient for PropMD (1.497e+03/2.788e+02)
    • The standard error for TotExp is ~8 times smaller than the coefficient for TotExp (7.233e-05/8.982e-06)
    • The standard error for PropMDTotExp is ~4 times smaller than the coefficient for PropMDTotExp (-6.026e-03/1.472e-03)
    • The standard error for the intercept is ~80 times smaller than the coefficient for intercept (6.277e+01/7.956e-01)
  • The p-values are all below 0.05 so they are statistically significant and relevant variables.
  • The R^2 value is 0.3574, which tells us the linear model describes 35.74% of the variability in the data, which is a decent model of the data but not as good as the previous model.
  • The F statistic is 14.39 and is still significant but not nearly as high as previous models.
lm_who4 <- lm(who4$LifeExp ~ who4$PropMD + who4$TotExp + who4$PropMDTotExp)
summary(lm_who4)
## 
## Call:
## lm(formula = who4$LifeExp ~ who4$PropMD + who4$TotExp + who4$PropMDTotExp)
## 
## 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 ***
## who4$PropMD        1.497e+03  2.788e+02   5.371 2.32e-07 ***
## who4$TotExp        7.233e-05  8.982e-06   8.053 9.39e-14 ***
## who4$PropMDTotExp -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

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

Our function is the following: LifeExp = 6.277e+01 + 1.497e+03 PropMD + 7.233e-05 TotalExp -6.026e-03 PropMDTotExp

PropMD = 0.03
TotalExp = 14
PropMDTotExp = PropMD*TotalExp
LifeExp <- 6.277e+01 + 1.497e+03 * PropMD + 7.233e-05 * TotalExp -6.026e-03 * PropMDTotExp
paste("When total exp ^ 0.06 is 1.5 then Life Exp is ",LifeExp)
## [1] "When total exp ^ 0.06 is 1.5 then Life Exp is  107.6784817"

This doesn’t make sense. If the government and people are together only spending on average $14 per person and only 3% of the population are doctors you wouldn’t expect people to live to 100 years old. They would likely die at a younger age.