DATA 605 - Homework 12
Eddie Xu
2024-04-14
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.
Load Dependencies and Data
## # A tibble: 6 × 10
## Country LifeExp InfantSurvival Under5Survival TBFree PropMD PropRN PersExp
## <chr> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
## 1 Afghanis… 42 0.835 0.743 0.998 2.29e-4 5.72e-4 20
## 2 Albania 71 0.985 0.983 1.00 1.14e-3 4.61e-3 169
## 3 Algeria 71 0.967 0.962 0.999 1.06e-3 2.09e-3 108
## 4 Andorra 82 0.997 0.996 1.00 3.30e-3 3.5 e-3 2589
## 5 Angola 41 0.846 0.74 0.997 7.04e-5 1.15e-3 36
## 6 Antigua … 73 0.99 0.989 1.00 1.43e-4 2.77e-3 503
## # ℹ 2 more variables: GovtExp <dbl>, TotExp <dbl>Question 1
Provide a scatter plot 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.
ggplot(data = hw_dataset, aes(x = TotExp, y = LifeExp)) +
geom_point() +
theme_minimal()
exp_lm <- lm(LifeExp ~ TotExp, data = hw_dataset)
summary(exp_lm)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = hw_dataset)
##
## 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-14Question 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?”
new_dataset <- hw_dataset %>%
mutate(LifeExp_New = LifeExp^4.6,
TotExp_New = TotExp^.06)
# plot
ggplot(data = new_dataset, aes(x = TotExp_New, y = LifeExp_New)) +
geom_point() +
theme_minimal()
exp_lm_new <- lm(LifeExp_New ~ TotExp_New, data = new_dataset)
summary(exp_lm_new)##
## Call:
## lm(formula = LifeExp_New ~ TotExp_New, data = new_dataset)
##
## 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_New 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-16par(mfrow=c(2,2))
plot(exp_lm_new)
Question 3
Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
expected_TotExp1 <- 1.5
intercept_value <- abs(summary(exp_lm_new)$coefficients[1,1])
TotExp_estimate <- summary(exp_lm_new)$coefficients[2,1]
forcast_value_1 <- (intercept_value + TotExp_estimate * expected_TotExp1)^(1/4.6)
forcast_value_1## [1] 101.099expected_TotExp2 <- 2.5
forcast_value_2 <- (intercept_value + TotExp_estimate * expected_TotExp2)^(1/4.6)
forcast_value_2## [1] 108.2953Question 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
multi_lm <- lm(LifeExp ~ PropMD + TotExp + PropMD:TotExp, data = hw_dataset)
summary(multi_lm)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD:TotExp, data = hw_dataset)
##
## 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-16par(mfrow=c(2,2))
plot(multi_lm)## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
Question 5
Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
expected_PropMD <- .03
expected_TotExp3 <- 14
intercept_value_multi <- abs(summary(multi_lm)$coefficients[1,1])
Prop_MD_estimate_multi <- summary(multi_lm)$coefficients[2,1]
TotExp_estimate_multi <- summary(multi_lm)$coefficients[3,1]
Prop_MD_TotalExp_multi <- summary(multi_lm)$coefficients[4,1]
forcast_value_3 <- (intercept_value_multi + Prop_MD_estimate_multi * expected_PropMD + TotExp_estimate_multi * expected_TotExp3) - (Prop_MD_TotalExp_multi * expected_PropMD * expected_TotExp3)
forcast_value_3## [1] 107.7011