VStoyanova_Assign12
Violeta Stoyanova
November 18, 2018
Homework 12
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.
Loading the data into a data frame
rm(list=ls())
who<-read.csv('https://raw.githubusercontent.com/VioletaStoyanova/Data605/master/who.csv', header=TRUE)
head(who)## 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 13046Question 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.
summary(who)## 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
## pairs(who[,-1], gap = 0.5, col = "blue")
Creating the linear regression model
mod1<- lm(LifeExp ~ TotExp, data = who)
summary(mod1)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = who)
##
## 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-14plot(who$TotExp, who$LifeExp, xlab = "Total Expenditures ($)" ,ylab = "Life Expectancy (yrs)", col = "blue")
abline(mod1, col="red")
hist(resid(mod1), main = "Histogram of Residuals", xlab = "residuals")
plot(fitted(mod1), resid(mod1))
p value is smaller than .05 therefore there’s a significant relationship between total expenditures and life expectancy The R2 of 0.2577 means that about 25.77% of the variability of life expectancy about the mean is explained by the model. The correlation in this case is moderately weak.F???Statistic is 65.26 and the Standard Error is 9.371
The linear model, when plotted over the data, does not match the data very closely. Furthermore, the residual analysis shows that the residuals have a strong right skew and do not show constant variance. Therefore, the linear model is not valid in this case.
Question 2
Raise life expectancy to the 4.6 power (i.e., LifeExp^4.6). Raise total expenditures to the 0.06 power (nearly a log transform, TotExp^.06). Plot LifeExp^4.6 as a function of TotExp^.06, and r re-run the simple regression model using the transformed variables. Provide and interpret the F statistics, R^2, standard error, and p-values. Which model is “better?”
le_4.6 <- who$LifeExp^4.6
te_0.06 <- who$TotExp^0.06
mod2 <- lm(le_4.6 ~ te_0.06)
summary(mod2)##
## Call:
## lm(formula = le_4.6 ~ te_0.06)
##
## Residuals:
## Min 1Q Median 3Q Max
## -308616089 -53978977 13697187 59139231 211951764
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -736527910 46817945 -15.73 <2e-16 ***
## te_0.06 620060216 27518940 22.53 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 90490000 on 188 degrees of freedom
## Multiple R-squared: 0.7298, Adjusted R-squared: 0.7283
## F-statistic: 507.7 on 1 and 188 DF, p-value: < 2.2e-16plot(who$TotExp^0.06, who$LifeExp^4.6, xlab = "Total Expenditures^0.06 ($^0.06)" ,ylab = "Life Expectancy^4.6 (yrs^0.06)", col = "blue")
abline(mod2, col="red")
hist(resid(mod2), main = "Histogram of Residuals", xlab = "residuals")
plot(fitted(mod2), resid(mod2))
The p???value is again smaller than .05 F???Statistic is 507.7 and the Standard Error is 90490000. The correlation is 0.8543 which is much better than the previous case and R2 is 0.7298. This transformed model seems to perform better than the previous one.
Question 3
Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
x = 1.5
y = round(mod2$coefficients[1], 4) + round(mod2$coefficients[2], 4) * x
le = round(y^(1/4.6), 2)
print(le)## (Intercept)
## 63.31The forecast life expectancy is 63.31 years when TotExp^.06=1.6
x = 2.5
y = round(mod2$coefficients[1], 4) + round(mod2$coefficients[2], 4) * x
le = round(y^(1/4.6), 2)
print(le)## (Intercept)
## 86.51The forecast life expctancy is 86.51 years when TotExp^.06=2.5 #Question 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, data = who)
summary(mod3)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = who)
##
## 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-16hist(resid(mod3), main = "Histogram of Residuals", xlab = "residuals")
plot(fitted(mod3), resid(mod3))
F???Statistic is 34.49 and the Standard Error is 8.765. The p???value is smaller than .05. The R2 is 0.3574. The model explains only 35.74% of variability. The residuals are not normally distributed. This model is not a good model to describe the relationships between variables TotExp, PropMd and LifeExp.
Question 5
Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not? LifeExp=6.277???101+1.497???103???PropMD+7.233???10???5TotExp???6.026???10???3???PropMD???TotExp LifeExp=6.277???101+1.497???103???0.03+7.233???10???5???14???6.026???10???3???0.03???14 LifeExp=107.6808
When PropMd=0.03 and TotExp=14, the forecast value of LifeExp is 107.69 years which is unrealistic because the highest life expectancy in the dataset is 83 years. Thus this forecast seems unrealistic.