Data 605 HW 12
Bryan Persaud
4/26/2020
who <- read.csv("https://raw.githubusercontent.com/bpersaud104/Data605/master/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 130461
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.
Solution:
plot(LifeExp ~ TotExp, data = who)
model1 <- lm(LifeExp ~ TotExp, data = who)
summary(model1)##
## 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-14The F statistics is 65.26 on 1 and 188 DF. Multiple R^2 is 0.2577 and Adjusted R^2 is 0.2537. The standard error is 9.371 on 188 degrees of freedom. The p-values are 7.714e-14.
plot(fitted(model1), resid(model1))
qqnorm(resid(model1))
qqline(resid(model1))
Based on the plots, assumptions of simple linear regression is not met since the plots show that the residuals are not normally distributed.
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 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?”
Solution:
plot((LifeExp ^ 4.6) ~ I(TotExp ^ 0.06), data = who)
model2 <- lm((LifeExp ^ 4.6) ~ I(TotExp ^ 0.06), data = who)
summary(model2)##
## Call:
## lm(formula = (LifeExp^4.6) ~ I(TotExp^0.06), data = who)
##
## 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 ***
## I(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-16The F statistics is 507.7 on 1 and 188 DF. Multiple R^2 is 0.7298 and Adjusted R^2 is 0.7283. The standard error is 90490000 on 188 degrees of freedom. The p-values are < 2.2e-16.
plot(fitted(model2), resid(model2))
qqnorm(resid(model2))
qqline(resid(model2))
The model that is “better” would be the second model.
3
Using the results from 3, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
Solution:
Using the model we made in 2, we can use the coefficients to make the formula y = 620060216x - 736527910. Here we are looking at two values for x, 1.5 and 2.5. Then we raise y to (1 / 4.6) to get our forecast life expectancy.
y_1.5 <- (620060216 * 1.5) - 736527910
life_expectancy_1.5 <- y_1.5 ^ (1 / 4.6)
round(life_expectancy_1.5)## [1] 63y_2.5 <- (620060216 * 2.5) - 736527910
life_expectancy_2.5 <- y_2.5 ^ (1 / 4.6)
round(life_expectancy_2.5)## [1] 87The life expectancy when TotExp ^ .06 = 1.5 is 63 years. The life expectancy when TotExp ^ .06 = 2.5 is 87 years.
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:
model3 <- lm(LifeExp ~ PropMD + TotExp + (PropMD * TotExp), data = who)
summary(model3)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + (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 ***
## 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-16The F statistics is 34.49 on 3 and 186 DF. Multiple R^2 is 0.3574 and Adjusted R^2 is 0.3471. The standard error is 8.765 on 186 degrees of freedom. The p-values are < 2.2e-16.
plot(fitted(model3), resid(model3))
qqnorm(resid(model3))
qqline(resid(model3))
This model is good since the p-values are less than 0.05. However looking at the residual analysis we see that it is similar to the first model we made as the distribution does not look normal.
5
Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
life_expectancy_5 <- (6.277 * (10 ^ 1)) + (1.497 * (10 ^ 3) * 0.03) + (7.233 * (10 ^ (-5)) * 14) - (6.026 * (10 ^ (-3)) * 0.03 * 14)
round(life_expectancy_5)## [1] 108max(who$LifeExp)## [1] 83LifeExp when PropMD = 0.03 and TotExp = 14 is 108. This forecast does not seem realistic because the max LifeExp in the dataset is 83.