who <- "https://raw.githubusercontent.com/miasiracusa/Data607/master/data605hw12/who.csv"
who <- read.csv(who)
  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$LifeExp ~ who$TotExp, xlab = 'Total Expectancy', ylab = 'Life expectancy')

obs <- lm(who$LifeExp ~ who$TotExp)
summary(obs)
## 
## 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

F-statistic: 65.26 on 1 p-value: 7.714e-14 Multiple R-squared: 0.2577

The F-statistic is testing the model against the null model, and the p-value is the chance that, given the hypothesis is true, the data turned out the way it did. There is a very low chance of that, according to our p-value of 7.714e-14. The R-squared value indicates that about 25.77% of the variation in the response variable can be attributed to the independent variable.

  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 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?”
obs2 <- lm((who$LifeExp^4.6)~I(who$TotExp^.06))
summary(obs2)
## 
## Call:
## lm(formula = (who$LifeExp^4.6) ~ I(who$TotExp^0.06))
## 
## 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(who$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

This adjustment increased our \(R^2\) value to 72.83% and out F statistic to over 500, so it is already a better model than the first problem.

  1. Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
le <- function(fc)
{   y <- -736527910 + 620060216 * (fc)
    y <- y^(1/4.6)
    print(y)
}
le(1.5)
## [1] 63.31153
le(2.5)
## [1] 86.50645
  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

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

Since the p-value is smaller than .05, the model is statistically significant. The \(R^2\) value indicates that the independent variable accounts for 35.74% of the variability in the independent variable. The F value indicates that the model is strong.

  1. Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
le <- ( (6.277*10^1) + (1.497*10^3)*.03 + (7.233*10^(-5))*14 - ((6.026*10^(-3))*0.03*14) ) 
le
## [1] 107.6785

This forecast is unrealistic, considering the age of the person is 107 years old.