Homework 12
2024-03-13
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.
Read data
country <- read.csv('https://raw.githubusercontent.com/Kingtilon1/DATA607/main/who.csv')
glimpse(country)## 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…scatterplot <- ggplot(country, aes(x = TotExp, y = LifeExp)) +
geom_point() +
xlab("Total Expenditures (US dollars)") +
ylab("Life Expectancy (years)") +
ggtitle("Scatterplot of Life Expectancy vs. Total Expenditures")
print(scatterplot)
model <- lm(LifeExp ~ TotExp, data = country)
summary(model)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = country)
##
## 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-14The F-statistic is 65.26 with a very low p-value (7.714e-14), indicating that the model is statistically significant, suggesting that at least one predictor variable has a non-zero effect on the response variable.
The R-squared value is 0.2577, meaning that approximately 25.77% of the variability in life expectancy is explained by total expenditures on healthcare.
The standard error of the residuals is 9.371, which represents the average deviation of the observed values from the predicted values.
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 model using the transformed variables. Provide and interpret the F statistics, R^2, standard error, and p-values. Which model is “better?”
country_transformed <- country %>%
mutate(LifeExp_transformed = LifeExp^4.6,
TotExp_0.06 = TotExp^0.06)
scatterplot_transformed <- ggplot(country_transformed, aes(x = TotExp_0.06, y = LifeExp_transformed)) +
geom_point() +
xlab("Total Expenditures Transformed") +
ylab("Life Expectancy Transformed") +
ggtitle("Scatterplot of Transformed Life Expectancy vs. Total Expenditures")
print(scatterplot_transformed)
model_transformed <- lm(LifeExp_transformed ~ TotExp_0.06, data = country_transformed)
summary(model_transformed)##
## Call:
## lm(formula = LifeExp_transformed ~ TotExp_0.06, data = country_transformed)
##
## 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_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-16The F-statistic, which is 507.7 with a very low p-value (< 2.2e-16), shows the statistical significance of the model. This means that the transformed model provides a better fit to the data compared to the original one. Also, with an R-squared value of 0.7298, 73% of the variability in transformed life expectancy can be explained by transformed total expenditures, indicating a significant improvement in explanatory power. Additionally, the standard error of the residuals, at 90490000, indicates a higher level of variability compared to the original model, implying greater dispersion in prediction accuracy.
3: Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
totexp_1 <- 1.5
totexp_2 <- 2.5
lifeexp_transformed_1 <- -736527910 + 620060216 * totexp_1
lifeexp_transformed_2 <- -736527910 + 620060216 * totexp_2
lifeexp_1 <- lifeexp_transformed_1^(1/4.6)
lifeexp_2 <- lifeexp_transformed_2^(1/4.6)
lifeexp_1## [1] 63.31153lifeexp_2## [1] 86.506454: . Build the following multiple regression model and interpret the F Statistics, R^2, standard error, and p-values. How good is the model?
model_multiple <- lm(LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = country)
summary(model_multiple)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = country)
##
## 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-16In comparison to the previous model (LifeExp_transformed ~ TotExp_0.06), the current model reveals a strong statistical significance with a high F-statistic of 507.7 and a very low p-value, suggesting a robust relationship between the transformed variables. Additionally, the model exhibits a high R-squared value of 0.7298, indicating that approximately 73% of the variability in transformed life expectancy can be explained by TotExp_0.06. This underscores the effectiveness of the model in capturing the relationship between healthcare expenditures and life expectancy, further highlighting its importance in healthcare research and policy formulation.
5: Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
# Coefficients from the multiple regression model
b0 <- 6.277e+01
b1 <- 1.497e+03
b2 <- 7.233e-05
b3 <- -6.026e-03
PropMD <- 0.03
TotExp <- 14
LifeExp <- b0 + b1 * PropMD + b2 * TotExp + b3 * PropMD * TotExp
LifeExp## [1] 107.6785This value is a lot higher than the life expectancy values we see so I don’t think this value is realistic seeing that the max life expectancy is 83 years old