DATA605_HW12_Multiple Regression Analysis
library(tidyverse)Explore the Dataset
The attached who.csv dataset contains real-world data from 2008. The variables included follow.
Country: name of the countryLifeExp: average life expectancy for the country in yearsInfantSurvival: proportion of those surviving to one year or moreUnder5Survival: proportion of those surviving to five years or moreTBFree: proportion of the population without TBPropMD: proportion of the population who are MDsPropRN: proportion of the population who are RNsPersExp: mean personal expenditures on healthcare in US dollars at average exchange rateGovtExp: mean government expenditures per capita on healthcare, US dollars at average exchange rateTotExp: sum of personal and government expenditures
Data Import
# Import data
who <- read.csv("https://raw.githubusercontent.com/omocharly/DATA605/main/who.csv")
head(who)## 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 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.
Solutions 1
Data Exploration
summary(who)## 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. :482750who.lm <- lm(LifeExp~TotExp, data=who)
summary(who.lm)##
## 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-14par(mfrow=c(2,2)) #prints out two rows, two columns of plots
plot(who.lm)
hist(who.lm$residuals)
F statistic: 65.26 on 1 and 188 DF, P-value: 7.714e-14, given the small pvalue, which is much below 0.05 indicates the model has some level of validity.
Multiple R2: 0.2577, Adjusted R2: 0.2537 - The model only accounts for roughly 25% of the data’s variation.
Residual standard error: In our example, the total expenditure required can deviate from the true regression line by approximately 15.3795867, on average.
Are the assumptions of simple linear regression met?
No, the conditions are not met.
Question 2
Raise life expectancy to the 4.6 power (exponential increase) (i.e., LifeExp4.6). Raise total expenditures to the 0.06 power (exponential decrease) (nearly a log transform, TotExp.06). Plot LifeExp4.6 as a function of TotExp.06, and re-run the simple regression model using the transformed variables. Provide and interpret the F statistics, R2, standard error, and p-values. Which model is “better?”
Solutions 2
who2 <- who %>%
mutate(LifeExp2 = LifeExp^4.6,
TotExp2 = TotExp^.06)
who.lm2 <- lm(LifeExp2~TotExp2, data=who2)
summary(who.lm2)##
## Call:
## lm(formula = LifeExp2 ~ TotExp2, data = who2)
##
## Residuals:
## Min 1Q Median 3Q Max
## -308616089 -53978977 13697187 59139231 211951764
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -736527910 46817945 -15.73 <2e-16 ***
## TotExp2 620060216 27518940 22.53 <2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 90490000 on 188 degrees of freedom
## Multiple R-squared: 0.7298, Adjusted R-squared: 0.7283
## F-statistic: 507.7 on 1 and 188 DF, p-value: < 2.2e-16ggplot(who2, aes(x =TotExp2 , y = LifeExp2)) +
geom_point()+
geom_smooth(method = "lm", formula = y ~ x, se = FALSE, col="red")+
ggtitle("Plot of Life Expectancy by Total Expenditure") +
xlab("Total Expenditure ^0.06") + ylab("Life Expectancy ^4.6")
par(mfrow=c(2,2)) #prints out two rows, two columns of plots
plot(who.lm2)
hist(who.lm2$residuals)
F-statistic: 507.7 on 1 and 188 DF, p-value: < 2.2e-16, given the small pvalue, which is much below 0.05 indicates the model has some level of validity.
Multiple R-squared: 0.7298, Adjusted R-squared: 0.7283 - The model only accounts for roughly 73% of the data’s variation.
Residual standard error: In our example, the total expenditure required can deviate from the true regression line by approximately 90490000, on average.
Are the assumptions of simple linear regression met?
Yes, conditions are met because the three conditions below are satisfied.
Linearity: The relationship between X and the mean of Y is linear. Based on the Residuals vs. Fitted plot, the the red line exhibits an almost linear relationship.
Homoscedasticity: The variance of residual is the same for any value of X. The Scale-Location plot shows the residuals are almost spread equally along the ranges of predictor.
Independence: Observations are independent of each other. Upon examining the Residuals vs. Fitted plot, we can see the there is no correlation between the points, and the red line is fairly flat.
Normality: For any fixed value of X, Y is normally distributed. The nearly normal residual condition is closely met, although the distribution is slightly left skewed.
Which model is “better?”
Of the models in exercise 1 and 2, the model of exercise 2 is better based solely on the statistics and Linearity criteria.
Questions 3
Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^ = 2.5.
Solutions 3
#y = 620060216x - 736527910
x = 1.5
y <- (-736527910 + 620060216*x)
y## [1] 193562414(y)^(1/4.6)## [1] 63.31153Life expectancy when TotExp.06 = 1.5 is approximately 63.3
x = 2.5
y <- (-736527910 + 620060216*x)
y## [1] 813622630(y)^(1/4.6)## [1] 86.50645Life expectancy when TotExp.06=2.5 is approximately 86.5
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
Solution 4
who.lm3 <- lm(LifeExp~TotExp + PropMD + PropMD * TotExp, data=who)
summary(who.lm3)##
## Call:
## lm(formula = LifeExp ~ TotExp + PropMD + 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 ***
## TotExp 7.233e-05 8.982e-06 8.053 9.39e-14 ***
## PropMD 1.497e+03 2.788e+02 5.371 2.32e-07 ***
## TotExp:PropMD -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-16par(mfrow=c(2,2))
plot(who.lm3)## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
par(mfrow=c(1,1))
# residuals histogram
hist(who.lm3$residuals,
xlab = "Residuals", ylab = "",
main = "Histogram of Residuals Distribution")
F-statistic: 34.49 on 3 and 186 DF, p-value: <2.2e−16, given the small pvalue, which is much below 0.05 indicates the model has some level of validity.
Multiple R-squared: 0.3574, Adjusted R-squared: 0.3471 - The model only accounts for roughly 36% of the data’s variation.
Are the assumptions of simple linear regression met?
Linearity: The relationship between X and the mean of Y is not linear. Based on the Residuals vs. Fitted plot, the the red line exhibits a quadratic relationship and is not linear.
Homoscedasticity: The variance of residual is not the same for any value of X. The Scale-Location plot shows the residuals are not spread equally along the ranges of predictor.
Independence: Observations are not independent of each other. Upon examining the Residuals vs. Fitted plot, we can see a correlation between the variables.
Multivariate Normality: The nearly normal residual condition doesn’t seem to be met based on the histogram of residuals shown below show the residuals are heavily left skew.
How good is the model? Are the assumptions of simple linear regression met?
The model is not that great given the criteria mentioned above because the conditions are not met
Question 5
Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
Solution 5
b0 <- 6.277 * 10^1
b1 <- 1.497 * 10^3
b2 <- 7.233 * 10^-5
b3 <- -6.026 * 10^-3
#PropMD
x1 <- .03
#TotExp
x2 <- 14 y = b0 + (b1 * x1) + (b2 * x2) + (b3 * x1 * x2)
y## [1] 107.6785max(who$LifeExp)## [1] 83mean(who$LifeExp)## [1] 67.37895The forecasted life expectancy using the linear model from exercise 4 is 107.7. This forecast is unrealistic with the actual data because the maximum life expectancy from the WHO data is 83 and the mean of all life expectancies is 67.4.