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- StudyBlue
- Washington
- University of Washington - Seattle Campus
- Statistics
- Statistics 390
- Marzban
- hw 3 solutions

Maziar R.

Problem 3.11 (a) SSxy = 5530.92 - (1950)(47.92)/18 = 339.586667, SSxx = 251,970 - (1950)2/18 = 40,720,and SSyy = 130.6074 - (47.92)2/18 = 3.033711, so r = EMBED Equation.3 = .9662 There is a very strong positive correlation between the two variables. Because the association between the variables is positive, the specimen with the larger shear force will tend to have a larger percent dry fiber weight. (c) Changing the units of measurement on either (or both) variables will have no effect on the calculated value of r, because any change in units will affect both the numerator and denominator of r by exactly the same multiplicative constant. Problem 3.12 (a) Using R: ## Read in the data file dat=read.csv(file.choose(),header=T) ## Assign variable names x=dat[,1] y=dat[,2] ## Calculate correlation coefficient cor(x,y) A value of r = .9233 means that there is a strong positive linear relationship between TOST time and RBOT time. The value of r does not depend on which of the two variables is labeled as the x variable. Thus, had we let x = RBOT time and y = TOST time, the value of r would have remained the same. The value of r does not depend on the unit of measure for either variable. Thus, had we expressed RBOT time in hours instead of minutes, the value of r would have remained the same. Problem 3.16 The value of the sample correlation coefficient using the squared y values would not necessarily be approximately 1. If the y-values are greater than 1, then the squared y-values would differ from each other by more than the y-values differ from one another. Hence, the relationship between x and y2 would be less like a straight line, and the resulting value of the correlation coefficient would decrease.

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