SElshahawy_Assign12
Salma Elshahawy
4/23/2020
Dataset importing
dataset = read.csv('who.csv', header = TRUE)
head(dataset)The dataset is 190 instances(observations) and 10 attributes(features) as the following:
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.
str(dataset)## '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 ...summary(dataset)## 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
## 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.
library(ggplot2)
ggplot(dataset, aes(x = TotExp, y = LifeExp)) +
geom_point()
Simple linear regression
regressor = lm(LifeExp ~ TotExp, data = dataset)
summary(regressor)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = dataset)
##
## 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-14P-value: is less than 0.05 significance level (2e-16) and F-Statistic is 65.26, so we can be sure that TotExp independent variable affecting the LifeExp. We can reject the null hypothesis. However, the R-squared is low (25.3%), which means that the model would explain only 25% of the model variability.
Std. Error is an estimate of the standard deviation of the coefficient, the amount it varies across cases.
This model can not be presented by a simple linear regression. The Adjusted R-squared percent is too low, may be if we included more independent variables we can get a higher value.
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?”
# raise LifeExp to 4.6 power
dataset$LifeExp_4.6 <- (dataset$LifeExp)^4.6
# raise TotExp to 0.06 power
dataset$TotExp_0.06 <- (dataset$TotExp)^0.06
head(dataset)ggplot(dataset, aes(x = TotExp_0.06, y = LifeExp_4.6)) +
geom_point() 
regressor_2 = lm(LifeExp_4.6 ~ TotExp_0.06, data = dataset)
summary(regressor_2)##
## Call:
## lm(formula = LifeExp_4.6 ~ TotExp_0.06, data = dataset)
##
## 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-16ggplot(data = dataset, aes(x = TotExp_0.06, y = LifeExp_4.6)) +
geom_point() +
stat_smooth(method = "lm", col = "dodgerblue3") +
theme(panel.background = element_rect(fill = "white"),
axis.line.x=element_line(),
axis.line.y=element_line()) +
ggtitle("Linear Model Fitted to Data")
plot(regressor_2$fitted.values, regressor_2$residuals,
xlab = 'fitted values',
ylab = 'residuals',
main = 'Residual plot')
abline(h=0)
# Residuals Q-Q plot
qqnorm(regressor_2$residuals)
qqline(regressor_2$residuals)
P-value: is still less than 0.05 significance level (2e-16) ,but F-Statistic got higher 507.7, so we can be sure that TotExp independent variable affecting the LifeExp. We can reject the null hypothesis. Moreover, the R-squared got better as well (72.8%), which means that the model would explain about 72.8% of the model variability.
Std. Error is an estimate of the standard deviation of the coefficient, the amount it varies across cases.
This model Definitly is better than the previous one.
3.
Using the results from 2, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
predict(regressor_2, data.frame(TotExp_0.06=c(1.5, 2.5)), interval = 'predict')^(1/4.6)## fit lwr upr
## 1 63.31153 35.93545 73.00793
## 2 86.50645 81.80643 90.434144.
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
regressor_3 <- lm(LifeExp ~ PropMD + TotExp + TotExp:PropMD, data = dataset)
summary(regressor_3)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + TotExp:PropMD, data = dataset)
##
## 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-16plot(regressor_3$fitted.values, regressor_3$residuals,
xlab = 'fitted values',
ylab = 'residuals',
main = 'Residual plot')
abline(h=0)
# Residuals Q-Q plot
qqnorm(regressor_2$residuals)
qqline(regressor_2$residuals)
Residual standard error is 8.765 and F-statistic is 34.49. Considering that average life expectancy is 67.38, the SE is not terrible and F-statistics is fairly high (but lower than in the first model). R2 is only 0.3574, so the model explains only 35.74% of variability, which is not high. P-value is nearly 0, so the relationship is not due to random variation.
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. Kind of similar to the first model.
5.
Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
predict(regressor_3, data.frame(PropMD = 0.03, TotExp = 14), interval = 'predict')## fit lwr upr
## 1 107.696 84.24791 131.1441The prediction is 107.696 years with 95% confidence interval between 84.247 and 131.144. The prediction could be realistic but it is very rare. I don’t think that it is a good model, The Adjusted R-squared in way too low to explain the variability of the model.