LEVEL-LEVEL

perf1 <- lm(math10 ~ lnchprg, data = meap93)
summary(perf1)
## 
## Call:
## lm(formula = math10 ~ lnchprg, data = meap93)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -24.386  -5.979  -1.207   4.865  45.845 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept) 32.14271    0.99758  32.221   <2e-16 ***
## lnchprg     -0.31886    0.03484  -9.152   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 9.566 on 406 degrees of freedom
## Multiple R-squared:  0.171,  Adjusted R-squared:  0.169 
## F-statistic: 83.77 on 1 and 406 DF,  p-value: < 2.2e-16


The estimated regression equation is \[ \widehat{math10}= 32.143-0.319lnchprg \]
This means that if the percentage of students who are eligible for the lunch program increases, the percentage of 10th graders receiving a passing score in a standardized math exam decreases by 0.319 holding other factors equal.

\(R^2=0.171\): Thus, this means that the percentage of students who are eligible for the lunch program can explain about 17.1% variation in the percentage of 10th graders receiving a passing score in a standardized math exam.

\(RMSE=9.566\) This is not a desirable value since it is far from 0.

Based on the \(R^2\) and RMSE, the model does not provide a good fit to the data.


LOG-LEVEL

perf2 <- lm(log(math10)~lnchprg, data = meap93)
summary(perf2)
## 
## Call:
## lm(formula = log(math10) ~ lnchprg, data = meap93)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.34067 -0.22219  0.03436  0.27521  1.29532 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  3.497277   0.046338   75.47   <2e-16 ***
## lnchprg     -0.016734   0.001618  -10.34   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4443 on 406 degrees of freedom
## Multiple R-squared:  0.2085, Adjusted R-squared:  0.2065 
## F-statistic: 106.9 on 1 and 406 DF,  p-value: < 2.2e-16


The estimated regression equation is \[\widehat{log(math10)}= 3.497-0.017lnchprg\]
This means that an increase in the percentage of students who are eligible for the lunch program is predicted to decrease the percentage of 10th graders receiving a passing score in a standardized math exam by 1.7% holding other factors equal.

\(R^2=0.2085\): Thus, the percentage of students who are eligible for the lunch program explains about 20.85% of the variation in log percentage of 10th graders receiving a passing score in a standardized math exam..

\(RMSE=0.4443\) This is not really a desirable value since it is far from 0.

Based on the \(R^2\) and RMSE, the model does not provide a good fit to the data.


LEVEL-LOG

perf3 <- lm(math10~log(lnchprg), data = meap93)
summary(perf3)
## 
## Call:
## lm(formula = math10 ~ log(lnchprg), data = meap93)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -23.336  -6.253  -1.417   4.724  46.218 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   45.6269     2.2732  20.072   <2e-16 ***
## log(lnchprg)  -7.0500     0.7287  -9.675   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 9.471 on 406 degrees of freedom
## Multiple R-squared:  0.1874, Adjusted R-squared:  0.1854 
## F-statistic:  93.6 on 1 and 406 DF,  p-value: < 2.2e-16


The estimated regression equation is: \[ \widehat{math10}= 45.627-7.050log(lnchprg)\]
This means that a 1% increase in the percentage of students who are eligible for the lunch program is associated with 0.0705 decrease in percentage of 10th graders receiving a passing score in a standardized math exam holding other factors equal.

\(R^2=0.1874\) Thus the log of the percentage of students who are eligible for the lunch program can explain about 18.74% of the variation in percentage of 10th graders receiving a passing score in a standardized math exam.

\(RMSE=9.471\) This is not a desirable value since it is far from 0.

Based on the \(R^2\) and RMSE, the model does not provide a good fit to the data.


LOG-LOG

perf4 <- lm(log(math10)~log(lnchprg), data = meap93)
summary(perf4)
## 
## Call:
## lm(formula = log(math10) ~ log(lnchprg), data = meap93)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -2.27639 -0.22457  0.03033  0.25315  1.29443 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)   4.08338    0.10842  37.661   <2e-16 ***
## log(lnchprg) -0.33017    0.03476  -9.499   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4518 on 406 degrees of freedom
## Multiple R-squared:  0.1818, Adjusted R-squared:  0.1798 
## F-statistic: 90.24 on 1 and 406 DF,  p-value: < 2.2e-16


The estimated regression equation is: \[\widehat{log(math10)}= 4.083-0.330log(lnchprg)\]



This means that a 1% increase in the percentage of students who are eligible for the lunch program decreases the percentage of 10th graders receiving a passing score in a standardized math exam by 0.33% holding other factors equal.

\(R^2=0.1818\) Thus,the log of the percentage of students who are eligible for the lunch program can explain about 18.18% of the variation in log of the percentage of 10th graders receiving a passing score in a standardized math exam.

\(RMSE=0.4518\) This is not really a desirable value since it is far from 0.

Based on the \(R^2\) and RMSE, the model does not provide a good fit to the data.


plot(meap93$lnchprg,meap93$math10,
     col="skyblue",
     pch=20,
     xlab = "students eligible for lunch program (%)",
     ylab= "10th graders recieving a passing score (%)",
     main = "")

abline(perf1, col="red", lwd=2)
text(x=60, y=20, "level-level", col="red")
abline(perf2, col="yellow", lwd=2)
text(x=40, y=5, "log-level", col="yellow")
abline(perf3, col="green", lwd=2)
text(x=5, y=50, "level-log", col="green")
abline(perf4, col="purple", lwd=2)
text(x=5, y=5, "log-log", col="purple")