DATA605-assignment12
Harpreet Shoker
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.
who <- data.frame(read.csv("who.csv"))
head(who) # look at the top of the df## 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 13046summary(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
## fit1 <- lm(LifeExp ~ TotExp, data=who)
summary(fit1)##
## 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(LifeExp~TotExp, data=who,
xlab="Total Expenditures", ylab="Life Expectancy",
main="Life Expectancy vs Total Expenditures")
abline(fit1)
hist(resid(fit1), main = "Histogram of Residuals", xlab = "residuals")
plot(fitted(fit1), resid(fit1))
qqnorm(fit1$residuals)
qqline(fit1$residuals)
The p-value suggests a statistically significant correlation between total expenditures and life expectancy, since p<<0.05. The R^2 of 0.2577 means that about 25.77% of the variability of life expectancy about the mean is explained by the model. This is a moderately weak correlation. The F-statistic tells us that adding the variable ‘total expenditures’ to the model improves the model compared to only having an intercept Looking at residuals plots it is clear that there is no constant variability and that residuals are not normally distributed. This is not a good model to describe the relationship
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?”
life4.6 <- who$LifeExp^4.6
texp <- who$TotExp^0.06
fit2 <- lm(life4.6 ~ texp)
summary(fit2)##
## Call:
## lm(formula = life4.6 ~ texp)
##
## 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 ***
## texp 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(life4.6~texp,
xlab="Total Expenditures", ylab="Life Expectancy",
main="Life Expectancy vs Total Expenditures (Transformed)")
abline(fit2)
hist(resid(fit2), main = "Histogram of Residuals", xlab = "residuals")
plot(fitted(fit2), resid(fit2))
qqnorm(fit2$residuals)
qqline(fit2$residuals)
Here we can conclude that Model-2 is better compared to Model-1. Adjusted Rsquare is 0.729 that means 72.98% of variability of life expectancy about the mean is explained whereas in model-1 is only 25%.There seems to be a good correlation. p-value is less in Model2 compared to Model1. F-stat is 507 in model2 whereas only 65 in Model1. Looking at residuals plots, variability is fairly constant with a few outliers and distribution of residuals is nearly normal making it a useful model.
3. Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
putting in equation \(y = mx + b\) \(y=−736527910+620060216x\)
\(lifeexpectancy=y^{(1/4.6)}\)
Substituting x = 1.5 and 2.5
e <- 1/4.6
le <- function(x)
{
y <- -736527910 + 620060216 *x
y <- y^e
print(y)
}
le(1.5)## [1] 63.31153le(2.5)## [1] 86.506454.4. 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
fit3 <- lm(LifeExp ~ PropMD + TotExp + PropMD*TotExp, data = who)
summary(fit3)##
## 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(fit3), xlab = "residuals")
plot(fitted(fit3), resid(fit3))
p-value is less than .05. model is statistically significant. F-statistic is 34.49 by adding 3 variables. Based on Rsquare only 35% of the variability can be explained by 3 variables.residuals have a strong right skew and do not show constant variance. Therefore, the linear model is not valid in this case.
5. Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
\(LifeExp=6.277∗101+1.497∗103∗PropMD+7.233∗10−5TotExp−6.026∗10−3∗PropMD∗TotExp\)
LifeE <- ( (6.277*10^1) + (1.497*10^3)*.03 + (7.233*10^(-5))*14 - ((6.026*10^(-3))*0.03*14) )
LifeE## [1] 107.6785lifeexpectancy of 107 years does not sense real.it seems to be an outlier. Total expenditure also seems to be very low here. This doesnot seem realistic