Homework 12
Julia Ferris
2024-04-12
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.
countryData <- read.csv("~/who.csv")
summary(countryData)## 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. :482750- 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 are met.
The scatterplot is below titled Total Expenditures cs. Life Expectancy. The F statistic was 65.26 with a significant p-value of less than 0.01. This means the model is a better fit than just using the mean. The R squared values show a weak correlation since they are much closer to 0 than 1. The standard errors are very low in comparison to the coefficients, so they are not of concern. The p-values for the coefficients are all significant. Based on these results, the regression appears to be useful for some values, but it is not a strong model for this data. Also, the assumptions of a linear regression are not met. For example, the data is not normally distributed as seen in the Q-Q plot, and the residuals do not have constant variance as seen in the Residuals vs Fitted plot.
myLM <- lm(LifeExp ~ TotExp, data = countryData)
plot(countryData[,"TotExp"], countryData[,"LifeExp"], main = "Total Expenditures vs. Life Expectancy", xlab="Total Expenditures", ylab="Life Expectancy")
abline(myLM)
summary(myLM)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = countryData)
##
## 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))
plot(myLM)
- 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?”
The F statistic was 507.7 with a significant p-value of less than 0.01. This means the model is a much better fit than just using the mean. The R squared values show a stronger correlation since they are much closer to 1 than 0. The standard errors are very low in comparison to the coefficients, so they are not of concern. The p-values for the coefficients are all significant. Based on these results, the regression appears to be useful for many more values than the original linear model. It appears to be a stronger model in a few ways. Also, the assumptions of a linear regression are more closely met. For example, the data is almost normally distributed as seen in the Q-Q plot, and the residuals seem to have constant variance as seen in the Residual vs Fitted plot. The model is not perfect, but it is a much better fit compared to the original linear model without transformations. The line also appears to visually fit the data better.
#Models
first <- lm(LifeExp ~ TotExp, data = countryData)
countryData['transformedTot'] <- countryData['TotExp']^0.06
countryData['transformedLife'] <- countryData['LifeExp']^4.6
transformedLM <- lm(transformedLife ~ transformedTot, data = countryData)
summary(transformedLM)##
## Call:
## lm(formula = transformedLife ~ transformedTot, data = countryData)
##
## 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 ***
## transformedTot 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(countryData[,"TotExp"]^0.06, countryData[,"LifeExp"]^4.6, main = "Total Expenditures cs. Life Expectancy - Transformed", xlab="Total Expenditures", ylab="Life Expectancy")
abline(transformedLM)
par(mfrow=c(2,2))
plot(transformedLM)
- Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
The answers below are rounded down to the nearest age.
Life expectancy when Total Expenditures ^ 0.06 = 1.5:
63 years
Life expectancy when Total Expenditures ^ 0.06 = 2.5:
86 years
(620060216*1.5 - 736527910)^(1/4.6)## [1] 63.31153(620060216*2.5 - 736527910)^(1/4.6)## [1] 86.50645- 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
LifeExp = 62.77 + 149.7(PropMD) + 0.00007233(TotExp) - 0.006026(PropMD)(TotExp)
The F Statistic of 34.49 seems a bit low. The R squared values show relationships with low strength. The standard errors are on the border of being concerning. They are within the range of being 5 to 10 times smaller than the coefficients except for the interaction term. The error is large compared to the slope of the interaction term. Also, the standard error for PropMD is a little more than 5 times smaller than the slope. The p-values all show high levels of signficance.
Therefore, since the R squared values showed such a weak relationship and the standard errors were overall not too low, it seems that the model is weak. Although the p-values showed significance, the model overall does not appear to be a good fit for the data.
lifeExp <- lm(LifeExp ~ PropMD * TotExp, data = countryData)
summary(lifeExp)##
## Call:
## lm(formula = LifeExp ~ PropMD * TotExp, data = countryData)
##
## 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-16- Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
This forecast could be realistic. The proportion of people who were doctors in the population is very high at 0.03, despite the low sum of expenditures. In other countries with a high proportion of doctors, life expectancy was high. In other countries with a very low sum of expenditures, life expectancy was low. The countries with the highest proportion of doctors had life expectancies around 80. The countries with the lowest sum of expenditures had life expectancies between 40 and 60. In this example, we have a combination of the two, so it stands to reason that the life expectancy would fall in the middle of these ranges. 67 is much less than 80 when thinking about the proportion of doctors, but it is above the typical range for countries with low expenditure. Since it falls in the middle, it seems like it could make sense. I graphed the correlations of the three variables from the data to compare them, and it seems that the proportion of doctors to the population is correlated stronger with life expectancy than total expenditures, but not by too much. It is probably a little high, but the avtual value, if it existed, might be within 10 years of the prediction. Therefore, it seems like this forecast could be realistic, but I would not say that definitively, since the data does not contain one example where both variables had similar values to the ones in the problem.
estimate <- 62.77 + (149.7 * 0.03) + (0.00007233 * 14) - (0.006026 * 0.03 * 14)
estimate## [1] 67.25948plot(countryData$PropMD, countryData$TotExp)
plot(countryData$PropMD, countryData$LifeExp)
plot(countryData$TotExp, countryData$LifeExp)