DATA605Assignment_12
IvanTikhonov
2022-11-09
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.
getURL <- "https://raw.githubusercontent.com/IvanGrozny88/DATA605Assignment_12/main/who.csv"
who_data <- read.csv(getURL, header = TRUE, sep = ",")
##Getting the basic statistics of the data set
summary(who_data)## Country LifeExp InfantSurvival Under5Survival
## Length:190 Min. :40.00 Min. :0.8350 Min. :0.7310
## Class :character 1st Qu.:61.25 1st Qu.:0.9433 1st Qu.:0.9253
## Mode :character Median :70.00 Median :0.9785 Median :0.9745
## Mean :67.38 Mean :0.9624 Mean :0.9459
## 3rd Qu.:75.00 3rd Qu.:0.9910 3rd Qu.:0.9900
## Max. :83.00 Max. :0.9980 Max. :0.9970
## TBFree PropMD PropRN PersExp
## Min. :0.9870 Min. :0.0000196 Min. :0.0000883 Min. : 3.00
## 1st Qu.:0.9969 1st Qu.:0.0002444 1st Qu.:0.0008455 1st Qu.: 36.25
## Median :0.9992 Median :0.0010474 Median :0.0027584 Median : 199.50
## Mean :0.9980 Mean :0.0017954 Mean :0.0041336 Mean : 742.00
## 3rd Qu.:0.9998 3rd Qu.:0.0024584 3rd Qu.:0.0057164 3rd Qu.: 515.25
## Max. :1.0000 Max. :0.0351290 Max. :0.0708387 Max. :6350.00
## GovtExp TotExp
## Min. : 10.0 Min. : 13
## 1st Qu.: 559.5 1st Qu.: 584
## Median : 5385.0 Median : 5541
## Mean : 40953.5 Mean : 41696
## 3rd Qu.: 25680.2 3rd Qu.: 26331
## Max. :476420.0 Max. :482750str(who_data)## 'data.frame': 190 obs. of 10 variables:
## $ Country : chr "Afghanistan" "Albania" "Algeria" "Andorra" ...
## $ LifeExp : int 42 71 71 82 41 73 75 69 82 80 ...
## $ InfantSurvival: num 0.835 0.985 0.967 0.997 0.846 0.99 0.986 0.979 0.995 0.996 ...
## $ Under5Survival: num 0.743 0.983 0.962 0.996 0.74 0.989 0.983 0.976 0.994 0.996 ...
## $ TBFree : num 0.998 1 0.999 1 0.997 ...
## $ PropMD : num 2.29e-04 1.14e-03 1.06e-03 3.30e-03 7.04e-05 ...
## $ PropRN : num 0.000572 0.004614 0.002091 0.0035 0.001146 ...
## $ PersExp : int 20 169 108 2589 36 503 484 88 3181 3788 ...
## $ GovtExp : int 92 3128 5184 169725 1620 12543 19170 1856 187616 189354 ...
## $ TotExp : int 112 3297 5292 172314 1656 13046 19654 1944 190797 193142 ...head(who_data)## 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- 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.
library(ggplot2)
ggplot(who_data, aes(x=who_data$TotExp, y=who_data$LifeExp)) + geom_point()## Warning: Use of `who_data$TotExp` is discouraged. Use `TotExp` instead.## Warning: Use of `who_data$LifeExp` is discouraged. Use `LifeExp` instead.
lifeexp.totexp.lm <- lm(LifeExp ~ TotExp, who_data)
lifeexp.totexp.lm##
## Call:
## lm(formula = LifeExp ~ TotExp, data = who_data)
##
## Coefficients:
## (Intercept) TotExp
## 6.475e+01 6.297e-05summary(lifeexp.totexp.lm)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = who_data)
##
## 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-14ggplot(who_data, aes(x=who_data$TotExp, y=who_data$LifeExp)) + geom_point(color = 'red') +
geom_line(aes(x = who_data$TotExp, y=predict(lifeexp.totexp.lm, newdata = who_data)), color = 'blue')## Warning: Use of `who_data$TotExp` is discouraged. Use `TotExp` instead.## Warning: Use of `who_data$LifeExp` is discouraged. Use `LifeExp` instead.## Warning: Use of `who_data$TotExp` is discouraged. Use `TotExp` instead.
Even though the p-value is extremely low, the R squared is also extremely low, as can be seen in the model summary above. It is also obvious from the graphic that this linear model does not match the data well.
- 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?”
ggplot(who_data, aes(x=(who_data$TotExp ^ 0.06), y=(who_data$LifeExp ^ 4.6))) + geom_point()## Warning: Use of `who_data$TotExp` is discouraged. Use `TotExp` instead.## Warning: Use of `who_data$LifeExp` is discouraged. Use `LifeExp` instead.
lifeexp.totexp.trf.lm <- lm(I(LifeExp ^ 4.6) ~ I(TotExp ^ 0.06), who_data)
lifeexp.totexp.trf.lm##
## Call:
## lm(formula = I(LifeExp^4.6) ~ I(TotExp^0.06), data = who_data)
##
## Coefficients:
## (Intercept) I(TotExp^0.06)
## -736527909 620060216summary(lifeexp.totexp.trf.lm)##
## Call:
## lm(formula = I(LifeExp^4.6) ~ I(TotExp^0.06), data = who_data)
##
## 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(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-16ggplot(who_data, aes(x=(who_data$TotExp ^ 0.06), y=(who_data$LifeExp ^ 4.6))) + geom_point(color = 'red') +
geom_line(aes(x = (who_data$TotExp ^ 0.06), y=predict(lifeexp.totexp.trf.lm, newdata = who_data)), color = 'blue')## Warning: Use of `who_data$TotExp` is discouraged. Use `TotExp` instead.## Warning: Use of `who_data$LifeExp` is discouraged. Use `LifeExp` instead.## Warning: Use of `who_data$TotExp` is discouraged. Use `TotExp` instead.
The linear link between these two converted variables appears to be extremely obvious from the aforementioned plot. The R squared values are high enough, as we can also see from the model, to assume that this is a good fit for the data. As a result, we can observe that the new model fits better than the one we created previously.
- Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
LifeExp.trf.3a <- predict(lifeexp.totexp.trf.lm, newdata = data.frame(TotExp = 1.5 ^ (1/0.06)))
LifeExp.3a <- LifeExp.trf.3a ^ (1/4.6)
print(LifeExp.3a)## 1
## 63.31153LifeExp.trf.3b <- predict(lifeexp.totexp.trf.lm, newdata = data.frame(TotExp = 2.5 ^ (1/0.06)))
LifeExp.3b <- LifeExp.trf.3b ^ (1/4.6)
print(LifeExp.3b)## 1
## 86.50645- 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
LifeExp.prob4.lm <- lm(LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = who_data)
summary(LifeExp.prob4.lm)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = who_data)
##
## 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-16The above model’s R-squared value shows that it is extremely low. As a result, it is clear that this does not fit.
- Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
LifeExp.trf.5 <- predict(LifeExp.prob4.lm, newdata = data.frame(TotExp = 14, PropMD = 0.03))
print(LifeExp.trf.5)## 1
## 107.696The forecast appears to be unrealistic because it is an extremely high value that defies logic.