Data 605 Assignment-12
Deepak Mongia
April 24, 2019
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/deepakmongia/Data605-Spring2019/master/who.csv"
who_data <- read.csv(getURL, header = TRUE, sep = ",")
##Getting the basic statistics of the data set
summary(who_data)## Country LifeExp InfantSurvival
## Afghanistan : 1 Min. :40.00 Min. :0.8350
## Albania : 1 1st Qu.:61.25 1st Qu.:0.9433
## Algeria : 1 Median :70.00 Median :0.9785
## Andorra : 1 Mean :67.38 Mean :0.9624
## Angola : 1 3rd Qu.:75.00 3rd Qu.:0.9910
## Antigua and Barbuda: 1 Max. :83.00 Max. :0.9980
## (Other) :184
## Under5Survival TBFree PropMD PropRN
## Min. :0.7310 Min. :0.9870 Min. :0.0000196 Min. :0.0000883
## 1st Qu.:0.9253 1st Qu.:0.9969 1st Qu.:0.0002444 1st Qu.:0.0008455
## Median :0.9745 Median :0.9992 Median :0.0010474 Median :0.0027584
## Mean :0.9459 Mean :0.9980 Mean :0.0017954 Mean :0.0041336
## 3rd Qu.:0.9900 3rd Qu.:0.9998 3rd Qu.:0.0024584 3rd Qu.:0.0057164
## Max. :0.9970 Max. :1.0000 Max. :0.0351290 Max. :0.0708387
##
## PersExp GovtExp TotExp
## Min. : 3.00 Min. : 10.0 Min. : 13
## 1st Qu.: 36.25 1st Qu.: 559.5 1st Qu.: 584
## Median : 199.50 Median : 5385.0 Median : 5541
## Mean : 742.00 Mean : 40953.5 Mean : 41696
## 3rd Qu.: 515.25 3rd Qu.: 25680.2 3rd Qu.: 26331
## Max. :6350.00 Max. :476420.0 Max. :482750
## str(who_data)## 'data.frame': 190 obs. of 10 variables:
## $ Country : Factor w/ 190 levels "Afghanistan",..: 1 2 3 4 5 6 7 8 9 10 ...
## $ 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
## 1 Afghanistan 42 0.835 0.743 0.99769
## 2 Albania 71 0.985 0.983 0.99974
## 3 Algeria 71 0.967 0.962 0.99944
## 4 Andorra 82 0.997 0.996 0.99983
## 5 Angola 41 0.846 0.740 0.99656
## 6 Antigua and Barbuda 73 0.990 0.989 0.99991
## PropMD PropRN PersExp GovtExp TotExp
## 1 0.000228841 0.000572294 20 92 112
## 2 0.001143127 0.004614439 169 3128 3297
## 3 0.001060478 0.002091362 108 5184 5292
## 4 0.003297297 0.003500000 2589 169725 172314
## 5 0.000070400 0.001146162 36 1620 1656
## 6 0.000142857 0.002773810 503 12543 130461. 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()
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$TotExp, y=predict(lifeexp.totexp.lm, newdata = who_data)), color = 'blue')
As we see above in the summary of the model, even though the p-value is very leass, but R squared is also very less. Also from the plot, it is clear that this linear model is not a good fit.
2. Raise life expectancy to the 4.6 power (i.e., LifeExp^4.6). Raise total expenditures to the 0.06 power (nearly a log transform, TotExp^.06). Plot LifeExp^4.6 as a function of TotExp^.06, and r re-run the simple regression model using the transformed variables. Provide and interpret the F statistics, R^2, standard error, and p-values. Which model is “better?”
ggplot(who_data, aes(x=(who_data$TotExp ^ 0.06), y=(who_data$LifeExp ^ 4.6))) + geom_point()
lifeexp.totexp.trf.lm <- lm(I(LifeExp ^ 4.6) ~ I(TotExp ^ 0.06), who_data)
lifeexp.totexp.trf.lm##
## Call:
## lm(formula = I(LifeExp^4.6) ~ I(TotExp^0.06), data = who_data)
##
## Coefficients:
## (Intercept) I(TotExp^0.06)
## -736527909 620060216summary(lifeexp.totexp.trf.lm)##
## Call:
## lm(formula = I(LifeExp^4.6) ~ I(TotExp^0.06), data = who_data)
##
## Residuals:
## Min 1Q Median 3Q Max
## -308616089 -53978977 13697187 59139231 211951764
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -736527910 46817945 -15.73 <2e-16 ***
## I(TotExp^0.06) 620060216 27518940 22.53 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 90490000 on 188 degrees of freedom
## Multiple R-squared: 0.7298, Adjusted R-squared: 0.7283
## F-statistic: 507.7 on 1 and 188 DF, p-value: < 2.2e-16ggplot(who_data, aes(x=(who_data$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')
From this above plot, the linear relationship looks very clear between these 2 transformed variables.
Also as we see from the model, the R squared values are high enough to assume that this is a good fit for the data.
Hence we see that the new model fits good as compared to the model we built earlier.
3. 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
## 86.506454. 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-16As we see from the above model, R-squared value is very less. Hence we see that this is not a very good fit.
5. 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 does not seem to be realistic as it is a very high value which is not reasonable.