Exercise

Exercise 6.1 (Correlation)

Assume Proteins A, B and C are members of a cellular pathway that is related to a certain disease. The concentration of these protein was measured in \(n=5000\) patients.

Read the corresponding data study2.csv into the R-environment.

X = read.table("data/study2.csv", sep=";", header=TRUE)
str(X)
## 'data.frame':    500 obs. of  3 variables:
##  $ Protein.A: num  10.7 11.7 10.3 11.9 10 ...
##  $ Protein.B: num  13.5 14.8 12.8 14.4 13.2 ...
##  $ Protein.C: num  7.06 7.34 8.73 6.53 7.68 ...

Make scatter plots of each pair of proteins (plot).

plot(X)

Determine Pearson’s correlation between each pair of proteins (cor.test).

cor.test(X$Protein.A, X$Protein.B)
cor.test(X$Protein.A, X$Protein.C)
cor.test(X$Protein.B, X$Protein.C)
## 
##  Pearson's product-moment correlation
## 
## data:  X$Protein.A and X$Protein.B
## t = 50.271, df = 498, p-value < 2.2e-16
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  0.8982970 0.9273551
## sample estimates:
##       cor 
## 0.9139905 
## 
## 
##  Pearson's product-moment correlation
## 
## data:  X$Protein.A and X$Protein.C
## t = 1.9296, df = 498, p-value = 0.05423
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.00155775  0.17253187
## sample estimates:
##        cor 
## 0.08614461 
## 
## 
##  Pearson's product-moment correlation
## 
## data:  X$Protein.B and X$Protein.C
## t = 0.28971, df = 498, p-value = 0.7722
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
##  -0.0747947  0.1005572
## sample estimates:
##        cor 
## 0.01298103

Exercise 6.2 (Linear Regression)

From the correlation analysis, protein A appears to be related to protein B. From earlier research, it is known that protein B depends on protein A. Model the relation between the two proteins using a simple linear regression (lm). 
L1 = lm(X$Protein.B ~ X$Protein.A)

Evaluate the summary over the linear model to check the goodness of fit.

summary(L1)
## 
## Call:
## lm(formula = X$Protein.B ~ X$Protein.A)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -1.41408 -0.30749  0.00183  0.28727  1.33353 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  4.03927    0.17893   22.57   <2e-16 ***
## X$Protein.A  0.89633    0.01783   50.27   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.4258 on 498 degrees of freedom
## Multiple R-squared:  0.8354, Adjusted R-squared:  0.835 
## F-statistic:  2527 on 1 and 498 DF,  p-value: < 2.2e-16

Do you know the meaning behind each section of summary?

T-statistic

The t-Test checks the nullity of the coefficient.

Multiple R-squared

Percentage of variability explained by the model.

F-statistic

The F-Statistic checks if at least one of the coefficients is nonzero, namely the null hypothesis is that all of the regression coefficients are equal to zero.

The output of the lm is a list which contains the estimates of the linear regression model. Use the latter to draw the regression line into the scatter plot (abline).

par(mfrow=c(1,1))
plot(X$Protein.A, X$Protein.B, cex.axis=1.5, cex.lab=1.5, 
     xlab="Concentration of Protein A",
     ylab="Concentration of Protein B")
grid()
#The coefficients of the linear regression model
#L1$coefficients
abline(a=L1$coefficients[1], b=L1$coefficients[2], col=2, lwd=3)