Urban Runoff

  1. The equation of the best-fit line is IBI=37.8242-0.5136\(\times\)imp.
  2. If the percent of impervious surface increases by 1%, then the IBI score decreases by 0.51, on average.
  3. If the percent of impervious surface is 0, then the IBI score is 37.8, on average.
  4. The IBI score is predicted to be 22.4 when the percent of impervious surface is 30%.
  5. This question is an extrapolation and should not be answered (see plot below).
  6. The residual for an observed IBI score of 30 and a percent of impervious surface of 10% is 30-32.7=-2.7.
  7. The IBI will change by -25 slopes or 12.8 if the percentage of impervious surface is decreased by 20%. In other words the IBI will increase by 12.8 units.
  8. The correlation coefficient between IBI score and percent of impervious surface is -0.874.
  9. The proportion of variability in IBI scores that is explained by knowing the percent of impervious surface is 0.763.
  10. There appears to be homoscedasticity, but there is slight evidence of a curve suggesting that the relationship is nonlinear (see plot below).

R Code and Results

> d <- read.csv("IBI.csv")
> ( lm1 <- lm(IBI~imp,data=d) )
Coefficients:
(Intercept)          imp  
    37.8242      -0.5136  
> rSquared(lm1)
[1] 0.7632712
> ggplot(data=d,mapping=aes(x=imp,y=IBI)) +
    geom_point(pch=21,color="black",fill="lightgray") +
    labs(x="Percent Impervious Land",y="Benthic Index of Biotic Integrity") +
    geom_smooth(method="lm",se=FALSE) +
    theme_NCStats()
`geom_smooth()` using formula 'y ~ x'


Red-Imported Fire Ants and Deer Fawns

  1. The response variable is the index of fawn recruitment. [Note that the index of fawn recruitment could depend on the RIFA index and the RIFA index is what is predicted or explained in the ensuing questions.]
  2. The explanatory variable is RIFA index.
  3. The best-fit line is fawnrec=-0.000209\(\times\)RIFA+0.854.
  4. The slope indicates that for every increase of one for the RIFA index the index of fawn recruitment will decrease by 0.000209, on average.
  5. The intercept indicates that if the RIFA index was zero, then the index of fawn recruitment would be 0.854, on average.
  6. If the RIFA index increased by 500 then the predicted index of fawn recruitment would decrease by 500 slopes or 0.105.
  7. This prediction should not be made as a RIFA index of 500 is outside the observed results for this variable (see plot below) and is, thus, an extrapolation.
  8. The predicted index of fawn recruitment if the RIFA index is 1700 is 0.498.
  9. The residual for an individual with a RIFA index of 2200 and a fawn recruitment index of 0.3 is 0.3-0.393=-0.093. Thus, this individual would have a lower fawn recruitment than the average for locations with a RIFA index of 2200.
  10. The correlation coefficient between RIFA and fawn recruitment indices is -0.700.
  11. The proportion of variability in the index of fawn recruitment that is explained by knowing the RIFA index is 0.49.
  12. I don’t have any strong concerns as the data appear linear (i.e., there is no obvious curve) and homoscedastic (i.e., there is no funnel or cone shape evident).

R Code and Results

> d <- read.csv("RIFA.csv")
> ( lm1 <- lm(fawnrec~rifa,data=d) )
Coefficients:
(Intercept)         rifa  
  0.8536285   -0.0002093  
> rSquared(lm1)
[1] 0.4893269
> ggplot(data=d,mapping=aes(x=rifa,y=fawnrec)) +
    geom_point(pch=21,color="black",fill="lightgray") +
    labs(x="RIFA Index",y="Fawn Recruitment Index") +
    geom_smooth(method="lm",se=FALSE) +
    theme_NCStats()
`geom_smooth()` using formula 'y ~ x'