Assignment12
sudhan_maharjan
April 19, 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.
#get the data who.csv
who <- read.csv("https://raw.githubusercontent.com/maharjansudhan/DATA605/master/who.csv", sep = ",", 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 13046- 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.
Answer:
scatter_plot <- lm(LifeExp ~ TotExp, data = who)
plot(who$LifeExp ~ who$TotExp)
abline(scatter_plot)
summary(scatter_plot)##
## 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-14F statistics : 65.26
R^2 : 0.2577 which means the model describes 25.77% of the whole data.
p-value : 7.714 * 10^-14 means the TotExp is not relevant.
standard error: 6.297 * 10^-5
The assumption is not met. The data is not linear.
# to check the linearity assumptions by plotting the residuals vs. TotExp
plot(scatter_plot$residuals ~ who$TotExp)
abline(h = 0, lty = 3)
- 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?”
Answer:
# raise life expectancy to 4.6
LifeExp_4.6 <- who$LifeExp^4.6
TotExp_0.06 <- who$TotExp^0.06
scatter_plot2 <- lm(LifeExp_4.6 ~ TotExp_0.06)
plot(LifeExp_4.6 ~ TotExp_0.06)
abline(scatter_plot2)
summary(scatter_plot2)##
## Call:
## lm(formula = LifeExp_4.6 ~ TotExp_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 ***
## 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-16F statistics : 507.7
R^2 : 0.7298 which means the model describes 72.98% of the whole data.
p-value : 2.2 * 10^-16 means the TotExp is not relevant.
standard error: 620060216
Since, the data is linear the model looks much better.
- Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
Answer:
Life expectancy when TotExp^0.06 = 1.5 is 193562414.
Life expectancy when TotExp^0.06 = 2.5 is 813622630.
-736527910 + 620060216 * 1.5## [1] 193562414-736527910 + 620060216 * 2.5## [1] 813622630- 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
Answer:
scatter_plot3 <- lm(LifeExp ~ PropMD + TotExp, data = who)
summary(scatter_plot3)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp, data = who)
##
## Residuals:
## Min 1Q Median 3Q Max
## -23.996 -4.880 3.042 6.958 13.415
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 6.397e+01 7.706e-01 83.012 < 2e-16 ***
## PropMD 6.508e+02 1.946e+02 3.344 0.000998 ***
## TotExp 5.378e-05 8.074e-06 6.661 2.95e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 9.127 on 187 degrees of freedom
## Multiple R-squared: 0.2996, Adjusted R-squared: 0.2921
## F-statistic: 39.99 on 2 and 187 DF, p-value: 3.479e-15F statistics : 39.99
R^2 : 0.2996 which means the model describes 29.96% of the whole data.
p-value : 3.479 * 10^-15 means the TotExp is not relevant.
standard error: 5.378 * 10^-5 for TotExp and 6.508 * 10^2 for PropMD
This model is not that bad as the first model.
- Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
Answer:
Life expectancy is forecasted to be 83.49475 when PropMD = 0.03 and TotExp = 14. It seems relalistic, because averagely people live that long.
#lets see the calculation
life <- 63.97 + (650.8*0.03) + (5.378 * 10^(-5) * 14)
life## [1] 83.49475