Data605wk12
Charls Joseph
April 21, 2019
R Markdown
who_data <- read.csv(text = getURL("https://raw.githubusercontent.com/charlsjoseph/data605wk12/master/who.csv"), header = T, stringsAsFactors =F)
kable(who_data[sample(nrow(who_data), 10), ], align='l', row.names=FALSE)| Country | LifeExp | InfantSurvival | Under5Survival | TBFree | PropMD | PropRN | PersExp | GovtExp | TotExp |
|---|---|---|---|---|---|---|---|---|---|
| Brazil | 72 | 0.981 | 0.980 | 0.99945 | 0.0010466 | 0.0034814 | 371 | 13940 | 14311 |
| Namibia | 61 | 0.955 | 0.939 | 0.99342 | 0.0002921 | 0.0030020 | 165 | 3888 | 4053 |
| Belarus | 69 | 0.994 | 0.992 | 0.99929 | 0.0047587 | 0.0124571 | 204 | 11315 | 11519 |
| Turkey | 73 | 0.976 | 0.974 | 0.99968 | 0.0015694 | 0.0029448 | 383 | 18632 | 19015 |
| Armenia | 69 | 0.979 | 0.976 | 0.99920 | 0.0036987 | 0.0049189 | 88 | 1856 | 1944 |
| Chad | 46 | 0.876 | 0.791 | 0.99430 | 0.0000330 | 0.0002387 | 22 | 234 | 256 |
| Bulgaria | 73 | 0.990 | 0.988 | 0.99959 | 0.0002535 | 0.0045532 | 272 | 11550 | 11822 |
| Venezuela (Bolivarian Republic of) | 74 | 0.982 | 0.979 | 0.99948 | 0.0017653 | 0.0010298 | 247 | 10528 | 10775 |
| Slovenia | 78 | 0.997 | 0.996 | 0.99985 | 0.0023603 | 0.0078516 | 1495 | 55233 | 56728 |
| South Africa | 51 | 0.944 | 0.931 | 0.99002 | 0.0007214 | 0.0038205 | 437 | 10920 | 11357 |
summary(who_data)## 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
## Min. :0.9870 Min. :0.0000196 Min. :0.0000883
## 1st Qu.:0.9969 1st Qu.:0.0002444 1st Qu.:0.0008455
## Median :0.9992 Median :0.0010474 Median :0.0027584
## Mean :0.9980 Mean :0.0017954 Mean :0.0041336
## 3rd Qu.:0.9998 3rd Qu.:0.0024584 3rd Qu.:0.0057164
## Max. :1.0000 Max. :0.0351290 Max. :0.0708387
## PersExp GovtExp TotExp
## Min. : 3.00 Min. : 10.0 Min. : 13
## 1st Qu.: 36.25 1st Qu.: 559.5 1st Qu.: 584
## Median : 199.50 Median : 5385.0 Median : 5541
## Mean : 742.00 Mean : 40953.5 Mean : 41696
## 3rd Qu.: 515.25 3rd Qu.: 25680.2 3rd Qu.: 26331
## Max. :6350.00 Max. :476420.0 Max. :482750Qn1. 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.
who.lm <- lm(LifeExp ~ TotExp, data = who_data)
plot(who_data$TotExp, who_data$LifeExp, xlab = 'Total Expenditure - $', ylab = 'Average Life Expectancy- in Years', main='Avg. Life Expectancy Vs. Expenditure')
abline(who.lm, col="red")
The scatter plot doesnt fit properly into the linear line. It shows the expenditure is constant until the age 70-75 and slowly increases beyond 70.
Lets find the summary of linear regression and analyze the key parameters.
summary(who.lm)##
## Call:
## lm(formula = LifeExp ~ TotExp, data = who_data)
##
## 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) 64.753374534 0.753536611 85.933 < 2e-16 ***
## TotExp 0.000062970 0.000007795 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-14Regression model
\[ Avg Life Expectancy=64.75337453+0.00006297∗Total Expenditure \] The model explains on every unit of increase in Total expenditure, the life expenditure goes up by 0.00006297.
plot(who.lm, which=c(1,1))
The resduals are not scattererd, but having an unsual pattern which tells that relation doesnt hold the linear behaviour. P-value is very less which explains there is a strong relation between LifeExp and TotExp. Adjusted R^2 and R^2 is very less which explains the totalExp is not only an explanatory variable and model can be possibly more accurate or better.
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 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?”
#raise total expenditure to the power of 0.06
who_data$TotExp_1 <- (who_data$TotExp)^0.06
#raise life expectancy to the power of 4.6
who_data$LifeExp_1 <- (who_data$LifeExp)^4.6
#build new linear model
who_1.lm <- lm(LifeExp_1 ~ TotExp_1, data = who_data)
plot(who_data$TotExp_1, who_data$LifeExp_1, xlab = 'Total Expenditure(in US Dollars) raised to 0.06', ylab = 'Average Life Expectancy(in Years) raised to 4.6', main='Avg. Life Expectancy Vs. Expenditure')
abline(who_1.lm, col="red")
This explains a better linear relation. Lets check the model key paramaters.
summary(who_1.lm)##
## Call:
## lm(formula = LifeExp_1 ~ TotExp_1, data = who_data)
##
## 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_1 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-16Regression model is
\[LifeExp1=−736527909+620060216∗TotExp1 \]
plot(who_1.lm, which=c(1,1))
The residual looks to be scatter and model looks to follow the linear pattern. Adjusted R^2 and R^2 looks to be closer to 1 and better than previous model.
3. Using the results from 2, forecast life expectancy when TotExp^.06 =1.5. Then forecast life expectancy when TotExp^.06=2.5.
result1 <- (-736527909 + (620060216 * 1.5))^(1/4.6)
result2 <- (-736527909 + (620060216 * 2.5))^(1/4.6)
result1## [1] 63.31153result2## [1] 86.50645Estimated life expectancy is 63.31153 years when TotExp(0.06) is 1.5. Estimated life expectancy is 86.50645 years when TotExp(0.06) is 2.5.
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
who_2.lm <- lm(LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = who_data)
summary(who_2.lm)##
## Call:
## lm(formula = LifeExp ~ PropMD + TotExp + PropMD * TotExp, data = who_data)
##
## 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) 62.772703255 0.795605238 78.899 < 2e-16 ***
## PropMD 1497.493952519 278.816879652 5.371 2.32e-07 ***
## TotExp 0.000072333 0.000008982 8.053 9.39e-14 ***
## PropMD:TotExp -0.006025686 0.001472357 -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-16Regression model is \[ Average Life Expectancy=62.77270326+1497.49395252×PropMD+0.00007233×TotExp−0.00602569×PropMD∗TotExp \]
p-Value for PropMD, TotExp and PropMD * TotExp are highly relevant to estimate average life expectancy. R2 and Adj. R2 values decreased compared to earlier
5. Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
TotExp = 14
PropMD = 0.03
forcast.life.exp = 62.77270326 + (1497.49395252 * PropMD) + (0.00007233 * TotExp) - (0.00602569 * PropMD * TotExp)
forcast.life.exp## [1] 107.696The result doesnt looks to be realistic. This could be due to in-efficiency of the model variables.