DATA 605 - Assignment 13

The scatter plot of Total Expenditures and Life Expectancy does not scream linear relationship so it difficult to say all the assumptions of a simple linear regression are met. The intercept of 64.7 means that with no expenditures one could expect to live 64 years. The coefficient for TotExp is very small, however, this likely reflects the scale of the expenditures. The p-values (0 values) for the intercept and coefficient indicate the variables are statistically significant. The R-Squared and Adj R-Squared around .25 indicates that Tot Exp explains approximately 25% of the variance - additional variable may improve the fit. The F-statistic of 65.26 and the p-value of zero mean that we can reject the hypothesis that the model is not better than the zero beta model.

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 r re-run the simple regression mod1 using the transformed variables. Provide and interpret the F statistics, R^2, standard error, and p-values. Which mod1 is “better?”

The scatter plot for Mod2 shows a linear relationship and right away we see a big jump in the R-squared and Adj. R-Squared compared to the previous model - 25% vs 72%. Similar to mod1 both the intercept and coefficient are statistically significant. We have all seen a substantial increase in the F-statistic (507 vs 65). The interesting thing about this model is that while the transforms improved the linear relationship and model fit, they also make the model more difficult to interpret.

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

Lifeat46 <- -736527909 +  620060216 * (1.5)
Lifeat15 <- exp(log(Lifeat46)/4.6)
Lifeat15

## [1] 63.31153

Lifeat46 <- -736527909 +  620060216 * (2.5)
Lifeat25 <- exp(log(Lifeat46)/4.6)
Lifeat25

## [1] 86.50645

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

mod3 <- lm(LifeExp ~ PropMD + TotExp + PropMD * TotExp, whodat)

gf <- tidy(mod3)

gf %>% 
  kable() %>% 
  kable_styling()

The R-Squared and Adj R-Squared of .357 and .347 indicate that the model is only capturing 34% to 36% percent of the variance. This could be an indication that we are missing some important explanatory variables. The F Statistic of 34.48 and a p-value of 0 mean we can reject the null hypothesis that our model is no better than the zero beta model. Additionally, the p-values of the coefficients are all small and close to zero, thus indicating the coefficients are statistically significant. This is also consistent with our F-Statistic. The sigma or residual standard error of 8.765 indicates that this is the average prediction error in a lifespan of 80 years that would be within about 10% - not bad.

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

LE <- 62.772 + .0000723 * 14 + 1497.493 * 0.03 + -0.0060257 * 14 * 0.03
LE

## [1] 107.6953

While this result is possible, it does not seem realistic. The model would seem to indicate we can that if we reduce spending, but increase doctors we can increase life expectancy. I suspect that there is a correlation between our two explanatory variables and this may undermining our results.