Scheruku_Assign12
SantoshCheruku
4/25/2020
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_data <- read.csv("who.csv")
str(who_data)## 'data.frame': 190 obs. of 10 variables:
## $ Country : Factor w/ 190 levels "Afghanistan",..: 1 2 3 4 5 6 7 8 9 10 ...
## $ LifeExp : int 42 71 71 82 41 73 75 69 82 80 ...
## $ InfantSurvival: num 0.835 0.985 0.967 0.997 0.846 0.99 0.986 0.979 0.995 0.996 ...
## $ Under5Survival: num 0.743 0.983 0.962 0.996 0.74 0.989 0.983 0.976 0.994 0.996 ...
## $ TBFree : num 0.998 1 0.999 1 0.997 ...
## $ PropMD : num 2.29e-04 1.14e-03 1.06e-03 3.30e-03 7.04e-05 ...
## $ PropRN : num 0.000572 0.004614 0.002091 0.0035 0.001146 ...
## $ PersExp : int 20 169 108 2589 36 503 484 88 3181 3788 ...
## $ GovtExp : int 92 3128 5184 169725 1620 12543 19170 1856 187616 189354 ...
## $ TotExp : int 112 3297 5292 172314 1656 13046 19654 1944 190797 193142 ...summary(who_data)## Country LifeExp InfantSurvival Under5Survival
## Afghanistan : 1 Min. :40.00 Min. :0.8350 Min. :0.7310
## Albania : 1 1st Qu.:61.25 1st Qu.:0.9433 1st Qu.:0.9253
## Algeria : 1 Median :70.00 Median :0.9785 Median :0.9745
## Andorra : 1 Mean :67.38 Mean :0.9624 Mean :0.9459
## Angola : 1 3rd Qu.:75.00 3rd Qu.:0.9910 3rd Qu.:0.9900
## Antigua and Barbuda: 1 Max. :83.00 Max. :0.9980 Max. :0.9970
## (Other) :184
## 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
## plot(who_data$LifeExp, who_data$TotExp, xlab = 'Life exp', ylab='Total exp')
linearReg <- lm(LifeExp~TotExp, data = who_data)
summary(linearReg)##
## 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) 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-14qqnorm(linearReg$residuals)
qqline(linearReg$residuals)
- Going through the summary of model, we can observe that
- R square value is quite less than 1 indicating the variability is low and hence the model is not strong enough.
- The P-value is quite less than .05 which indicates the rejection of null hypothesis.
- The F-statistic number if high enough to indicate a co-relation between the parameters.
- Standard Error is very low indicating the model could be closely representing the response variable
Also, the residuals are not normally distributed suggesting the model may not be properly fit.
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?”
who_data$LifeExp4.6 = who_data$LifeExp^4.6
who_data$TotExp.06 = who_data$TotExp^0.06
linearReg_updated <- lm(LifeExp4.6~TotExp.06, data = who_data)
summary(linearReg_updated)##
## Call:
## lm(formula = LifeExp4.6 ~ TotExp.06, 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.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-16qqnorm(linearReg_updated$residuals)
qqline(linearReg_updated$residuals)
- Going through the summary of model, we can observe that
- R square value is much close to 1 usually indicating a variability close to the mean, indicating high strength in predicting the response variable
- The P-value is quite less than .05 which indicates the rejection of null hypothesis.
- The F-statistic number if high enough to indicate a co-relation between the parameters.
Also, Model 2 residuals are normally distributed when compared to Model 1 residuals. Therefore, Model 2 is better when compared to Model 1.
The model gives the below relation between LifeExp4.6 and TotExp.06
LifeExp4.6 = -736527910 + 620060216*TotExp.06
3. Using the results from 3, forecast life expectancy when TotExp^.06 =1.5.
TotExp.06 <- 1.5
LifeExp4.6 <- -736527910 + 620060216*TotExp.06
LifeExp4.6^(1/4.6)## [1] 63.311533. Then forecast life expectancy when TotExp^.06=2.5.
TotExp.06 <- 2.5
LifeExp4.6 <- -736527910 + 620060216*TotExp.06
LifeExp4.6^(1/4.6)## [1] 86.506454. 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
linearReg_newFormula <- lm(LifeExp ~ PropMD + TotExp + PropMD*TotExp, who_data)
summary(linearReg_newFormula)##
## 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) 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- Observations from the model
- P-value is less than .05 representing that there is a direct co-relation between the parameters
- R square value is less than expected, suggesting that model is not a great fit
- F statistic is not as high as observed in previous models indicating model is not a strong fit
qqnorm(linearReg_newFormula$residuals)
qqline(linearReg_newFormula$residuals)
LifeExp <- 6.277 + 1.497e+03 X PropMD + 7.233e-05 X TotExp - 6.026e-03 X PropMD X TotExp
5. Forecast LifeExp when PropMD=.03 and TotExp = 14. Does this forecast seem realistic? Why or why not?
PropMD <- .03
TotExp <- 14
LifeExp <- 6.277e+01 + 1.497e+03 * PropMD + 7.233e-05 * TotExp - 6.026e-03 * PropMD * TotExp
LifeExp## [1] 107.6785As per the summary of the data structure ‘who’ the median of The value of LifeExp is 70 and max is 83. 107 seems much unrealistic when compared.