HW 12 - Regression Analysis
Tora Mullings
2024-04-14
df <- read.csv("who.csv", header=TRUE)
head(df)## 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 13046Simple linear regression
LifeExp ~ TotExp
plot(df$TotExp, df$LifeExp, xlab="TotExp", ylab="LifeExp")
df.lm <- lm(LifeExp ~ TotExp, data=df)
summary(df.lm)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = df)
##
## 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-14- The F-statistic is 65.26 on 1 and 188 degrees of freedom. It’s p-value is very small which means it’s highly unlikely that all of the coefficients in the model are zero. It compares the model with all of the predictors to the intercept only model. There is only one predictor for this model so it doesn’t matter much.
- \(R^2=0.2577\), which means the variation in TotExp explains only about 25% of the variability in LifeExp.
- The Standard Error for the cofficient of the only predictor in this model, TotExp, is 7.7795e-06. The ratio of the TotExp coefficient and its standard error is 6.297e-05/7.795e-06=8.08, which means that the standard error for TotExp is about 8 times smaller than the TotExp coefficient. This means that there is large variability in the coefficient estimate.
- The p-value for the coefficient of TotExp is 7.71e-14, which is also very small. This means that we cannot reject the hypothesis which states that the coefficient of TotExp is not zero.
plot(LifeExp ~ TotExp, data=df)
abline(df.lm)
We can see that the relationship between LifeExp and TotExp is not
linear.
par(mfrow=c(2,2))
plot(df.lm)
The assumptions for linear regression are not met. The relationship between LifeExp and TotExp is not linear, the distribution of the residuals is not normal and don’t have equal variance for each value of TotExp. The residuals are distributed in some pattern, which means that there is likely another model that would fit the relationship better.
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).
df1 <- df
df1$`LifeExp^4.6` <- df1$LifeExp^4.6
df1$`TotExp^0.06` <- df1$TotExp^0.06
plot(df1$`LifeExp^4.6`, df1$`TotExp^0.06`, xlab="TotExp^0.06", ylab="LifeExp^4.6")
df1.lm <- lm(`LifeExp^4.6` ~ `TotExp^0.06`, data=df1)
summary(df1.lm)##
## Call:
## lm(formula = `LifeExp^4.6` ~ `TotExp^0.06`, data = df1)
##
## 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-16- The F-statistic p-value is very small which means it’s highly unlikely that all of the coefficients in the model are zero. there is still only one predictor in this model so it doesn’t matter much.
- \(R^2=0.7298\), which means the variation in the (transformed) total expenditures explains about 73% of the variability in the (transformed) life expenditure.
- The Standard Error for the cofficient of TotExp, is 27518940. The ratio of the TotExp coefficient and its standard error is 620060216/27518940=22.53, which means that the standard error for TotExp is about 23 times smaller than the TotExp coefficient. This means that there is little variability in the coefficient estimate.
- The p-value for the coefficient of TotExp is <2e-16, which is very small. We cannot reject the hypothesis which states that the coefficient of TotExp is not zero.
par(mfrow=c(2,2))
plot(df1.lm)
This model looks better. The relationship between the transformed variables is more linear than the variables in the previous model. The residuals look more random.
Forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
pred1 <- predict(df1.lm, newdata=list(`TotExp^0.06`=1.5))
pred1## 1
## 193562414pred2 <- predict(df1.lm, newdata=list(`TotExp^0.06`=2.5))
pred2## 1
## 813622630LifeExp = b0+b1 x PropMd + b2 x TotExp +b3 x PropMD x TotExp
df.lm1 <- lm(LifeExp ~ PropMD + TotExp + PropMD*TotExp, data=df)
summary(df.lm1)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = df)
##
## 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-16Build the following multiple regression model and interpret the F Statistics, R^2, standard error, and p-values. How good is the model?
- The F-statistic p-value is very small which means it’s highly unlikely that all of the coefficients in the model are zero.
- The adjusted \(R^2=0.3471\), which means the variation in the predictors explains about 35% of the variability in the life expenditure.
- The ratio of the TotExp coefficient and its standard error is 7.233e-05/8.982e-06=8.05, which means that the standard error for TotExp is about 8 times smaller than the TotExp coefficient. This means that there is large variability in the coefficient estimate. For the other two coefficients, the ratios are also small.
- The p-values for the coefficients of each of the predictors are very small. We cannot reject the hypothesis which states that the coefficient of TotExp is not zero.
par(mfrow=c(2,2))
plot(df.lm1)## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
## Warning in sqrt(crit * p * (1 - hh)/hh): NaNs produced
The model is not good. The residuals have a clear pattern. The R-squared is small.
Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
pred3 <- predict(df.lm1, newdata=list(PropMD=1.5, TotExp=14))
pred3## 1
## 2308.888This forecast does not seem realistic for the context. This number is supposed to be within the range of a human’s life expectancy.