Answers Curve Fitting

 

The data is also provided in Q8-1.rda (the data frame is called Q8). To call the data in the console:

Q8
  x    y
1 3 1.20
2 4 1.40
3 5 1.60
4 6 1.75
5 7 1.85

  1. Plot the data, draw the regression line and estimate the equation of the line:

ggplot() + geom_point(aes(x = x,y = y),data=Q8) + geom_smooth(aes(x = x,y = y),data=Q8,method = ‘lm’) + theme_bw()

Q8-1fit<-lm(y~x,data=Q8)
fit

Call:
lm(formula = y ~ x, data = Q8)

Coefficients:
(Intercept)            x  
      0.735        0.165 

The equation of the regression line therefore is:

y = 0.735 + 0.165 × x

      1. What is the correlation coefficient?

cor(Q8$x,Q8$y,method=’pearson’)
[1] 0.9913889

The correlation coefficient therefore is 99%.

  1. Interpolate the y-value for x = 5.5

y(x = 5.5) = 0.735 + 0.165 × 5.5 = 1.6425

  1. Extrapolate the y-values for x = 0.1 and x = 15

y(x = 0.1) = 0.735 + 0.165 × 0.1 = 0.7515

y(x = 15) = 0.735 + 0.165 × 15 = 3.21

The data is also provided in Q8-2.rda (the data frame is called Q8Extended). To call the data in the console:

Q8Extended
      x     y
1   0.1 -1.25
2   0.2 -0.70
3   1.0  0.40
4   2.0  0.90
5   3.0  1.20
6   4.0  1.40
7   5.0  1.60
8   6.0  1.75
9   7.0  1.85
10  8.0  1.95
11  9.0  2.05
12 10.0  2.10
13 12.0  2.25
14 14.0  2.35
15 15.0  2.40 

      1. Plot these data in a graph.

To create a scatterplot to evaluate the relation between the two variables (without a regression line):

ggplot() + geom_point(aes(x = x,y = y),data=Q8Extended)+ theme_bw()

Q8-2

  1. What is the relation between x and y and what is the value of the correlation coefficient?

A linear regression is clearly hopeless:

ggplot() + geom_point(aes(x = x,y = y),data=Q8Extended) + geom_smooth(aes(x = x,y = y),data=Q8Extended,method = ‘lm’) + theme_bw()

Q8-3x and y appear to have a logarithmic relation and the general equation of the regression line is:

y = b + a × log(x)

or:

ggplot() + geom_point(aes(x = log(x),y = y),data=Q8Extended) + geom_smooth(aes(x = log(x),y = y),data=Q8Extended,method = ‘lm’) + theme_bw()

Q8-4The equation of the regression line is:

fit<-lm(y~log(x),data=Q8Extended)
fit

Call:
lm(formula = y ~ log(x), data = Q8Extended)

Coefficients:
(Intercept)       log(x)
0.4271       0.7276 

Or:

y = 0.4271 + 0.7276 × log(x)

To find the value of the correlation coefficient:

cor(log(Q8Extended$x),Q8Extended$y,method=’pearson’)
[1] 0.9997933

  1. What are the y-values for x = 0.1 and x = 15?

As described under 6., the equation of the regression line is:

y = 0.4271 + 0.7276 × log(x)

Therefore,

y(x = 0.1) = 0.4271 + 0.7276 × log(0.1)-1.25

y(x = 15) = 0.4271 + 0.7276 × log(15)2.40

This question illustrates again the danger of extrapolating data!

    1. The scatterplot and line x=y can be created in the Deducer GUI with plot builder, or the following command in the R / JGR console:

ggplot() + geom_point(aes(x = Lafayette,y = iPhone),data=goniometer) + theme_bw() + geom_abline(data=goniometer,intercept = 0.0,slope = 1.0,colour = ‘#339900′,linetype = 2)

iccquestion

    1. The Deducer GUI or the console can be used to calculate the Pearson correlation coefficient:

cor(goniometer$Lafayette,goniometer$iPhone,method=”pearson”)
[1] 0.9473263

Therefore, Pearson’s correlation coefficient is 95%.

    1. The Deducer GUI or the console can be used to calculate the ICC:

library(irr)
icc(goniometer,model=”twoway”,type=”agreement”)
 Single Score Intraclass Correlation

   Model: twoway
   Type : agreement

   Subjects = 60
     Raters = 2
   ICC(A,1) = 0.948

 F-Test, H0: r0 = 0 ; H1: r0 > 0
 F(59,59.8) = 37 , p = 1.6e-31

 95%-Confidence Interval for ICC Population Values:
  0.914 < ICC < 0.968

Therefore, the ICC is 95%.