Homework 9.2
Jett Facey
September 29, 2016
Problem 25
Much stronger, since human life expectance has high leverage, and influencial
Yes, they do not fit in the data becuase humans have devloped new ways to artifically live longer in comparison to other animals, skewing the data.
Problem 27
Hippos, since they’re a relative outlier. Elephants increas the association since the cofirm the apparent trend.
slope would increase
no, it’s important to keep the model real. If we remove too many data points becuase we want our model to be stronger we will be manipulating the data.
No, they were not influencial since they confirmed the previous trend.
Problem 33
##enter the data, 2 variable quatitative data
YearSince1900 = c(14,18,22,26,30,34,38,42,46,50,54,58,62,66,70,74,78,82,86,90,94,98,102,106)
CPI = c(10.0,15.1,16.8,17.7,16.7,13.4,14.1,16.3,19.5,24.1,26.9,28.9,30.2,32.4,38.8,49.3,65.2,96.5,109.6,130.7,148.2,163.0,179.9,201.6)
## make the scatterplot
plot(YearSince1900, CPI, col = "purple", type ='p', pch = 16)
## calculate the linear regression model
lm.r = lm(CPI~YearSince1900)
## add the regression line to the scatterplot
abline(lm.r, col = "dark green")
## state the correlation coeficient
cor(YearSince1900, CPI)## [1] 0.8832971## summary provides lots of information
summary(lm.r)##
## Call:
## lm(formula = CPI ~ YearSince1900)
##
## Residuals:
## Min 1Q Median 3Q Max
## -41.228 -24.142 -2.986 23.928 53.208
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -52.9032 14.1999 -3.726 0.00117 **
## YearSince1900 1.8990 0.2149 8.837 1.09e-08 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 29.15 on 22 degrees of freedom
## Multiple R-squared: 0.7802, Adjusted R-squared: 0.7702
## F-statistic: 78.1 on 1 and 22 DF, p-value: 1.088e-08##look at the residuals
resid(lm.r)## 1 2 3 4 5 6
## 36.3170000 33.8209565 27.9249130 21.2288696 12.6328261 1.7367826
## 7 8 9 10 11 12
## -5.1592609 -10.5553043 -14.9513478 -17.9473913 -22.7434348 -28.3394783
## 13 14 15 16 17 18
## -34.6355217 -40.0315652 -41.2276087 -38.3236522 -30.0196957 -6.3157391
## 19 20 21 22 23 24
## -0.8117826 12.6921739 22.5961304 29.8000870 39.1040435 53.2080000plot(CPI,resid(lm.r), col = "red", type ='p', pch = 16, main = "Residual Plot")
##enter the data, 2 variable quatitative data
YearSince1900 = c(70,74,78,82,86,90,94,98,102,106)
CPI = c(38.8,49.3,65.2,96.5,109.6,130.7,148.2,163.0,179.9,201.6)
## make the scatterplot
plot(YearSince1900, CPI, col = "purple", type ='p', pch = 16)
## calculate the linear regression model
lm.r = lm(CPI~YearSince1900)
## add the regression line to the scatterplot
abline(lm.r, col = "dark green")
## state the correlation coeficient
cor(YearSince1900, CPI)## [1] 0.997492## summary provides lots of information
summary(lm.r)##
## Call:
## lm(formula = CPI ~ YearSince1900)
##
## Residuals:
## Min 1Q Median 3Q Max
## -6.9497 -2.5744 0.4158 2.9559 5.8982
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -287.6667 10.2706 -28.01 2.85e-09 ***
## YearSince1900 4.6130 0.1157 39.86 1.73e-10 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.205 on 8 degrees of freedom
## Multiple R-squared: 0.995, Adjusted R-squared: 0.9944
## F-statistic: 1589 on 1 and 8 DF, p-value: 1.726e-10##look at the residuals
resid(lm.r)## 1 2 3 4 5 6
## 3.5545455 -4.3975758 -6.9496970 5.8981818 0.5460606 3.1939394
## 7 8 9 10
## 2.2418182 -1.4103030 -2.9624242 0.2854545plot(CPI,resid(lm.r), col = "red", type ='p', pch = 16, main = "Residual Plot")
lm.r$coefficients[1]+lm.r$coefficients[2]*116## (Intercept)
## 247.4448-287.6667+4.61303*116## [1] 247.4448