Question

Toughness and fibrousness of asparagus are major determinants of quality. This was the focus of a study reported in “Post-Harvest Glyphosphate Application Reduces Toughening, Fiber Content, and

Lignification of Stored Asparagus Spears” (J. of the Amer. Soc. Of Horticultural Science, 1988: 569-572).

The article reported the accompanying data on x = shear force (kg) and y = percent fiber dry weight.

X =,46 48, 55, 57, 60, 72, 81, 85, 94

Y,= 2.18, 2.10, 2.13, 2.28, 2.34, 2.53, 2.28, 2.62, 2.63

X =109, 121, 132, 137, 148, 149, 184, 185, 187

Y=2.50, 2.66, 2.79, 2.80, 3.01, 2.98, 3.34, 3.49, 3.26

Summary Statistics:

?=18,Σ??=1950,Σ??2=251,970,Σ??=47.92,Σ??2=130.6075,Σ????=5530.92

(a) Calculate the value ?, the sample correlation coefficient. Based on this value, how would you

describe the nature of the relationship between the two variables?

(b) If a first specimen has a larger value of shear force than does a second specimen, what tends to be

true of percent dry fiber weight for the two specimens?

(c) If shear force is expressed in pounds, what happens to the value of ?? Why?

(d) Compute the least squares regression line, ?̂, for predicting percent fiber dry weight from shear

force.

(e) If two specimens differ in shear force by 2, by how much would you predict their percent fiber dry

weight to differ?

(f) Predict the percent dry weight of a specimen with a shear force of 150 kg.

(g) If a specimen has a shear force of 20 kg, should we use our model to predict its percent fiber dry

weight? If yes, compute your prediction. If not, explain why it should not be used.

(h) What percentage of variation in percent fiber dry weight can be explained by its linear association

with shear strength?

Answer 1

We have the following summary statistics

We can calculate the sum of squares as

a) The sample correlation coefficient is

ans: The sample correlation coefficient is 0.9662. The positive value indicates that there is a strong positive linear relationship between the variables, shear force and percent fiber dry weight.

b) Since the relationship is positive, an increase in the value of shear force, indicates an increase in the value of percent dry weight.

ans:  If a first specimen has a larger value of shear force than does a second specimen, then the percent dry fiber weight of first specimen is larger than the second specimen

c) r is a coefficient having values between -1 to +1 and hence does not have any units.

ans: If shear force is expressed in pounds, r would remain the same.

d) The model that we want to estimate is

where y=percent dry weight

x=shear force (kg)

is the intercept, is the slope of regression line and is a random error

The estimate of slope is

The estimate of intercept is

ans: the least squares regression line, ?̂, for predicting percent fiber dry weight from shear force is

(e) The slope of 0.0083 indicates that if the shear force changes by 1 kg, the percent fiber dry fiber weight would change by 0.0083 percent.

ans: If two specimens differ in shear force by 2, we would predict their percent fiber dry fiber weight differ by 0.0083*2=0.0167

f) Given the shear force x=150, we get

ans: The Predicted percent dry weight of a specimen with a shear force of 150 kg is 3.01

g) The value of the shear force in the sample data varies between 46kg to 187 kg. Hence this linear model is valid for predicting the percent fiber dry weight for shear force in the range 46 to 187.

ans: If a specimen has a shear force of 20 kg, we should not use our model to predict its percent fiber dry weight as we are not sure if the linear model holds good when the shear force is beyond the range 46 to 187.

h) The coefficient of determination is

ans: 93.35 percentage of variation in percent fiber dry weight can be explained by its linear association with shear strength.