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= 46,48,55,57,60,72,81,85,94,109,121,132,137,148,149,184,185,187

Y= 2.18,2.10,2.13,2.28,2.34,2.53,2.28,2.62,2.63,2.5,2.66,2.79,2.8,3.01,2.98,3.34,3.49,3.26

Summary Statistics:

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

(a) Distributional assumptions in regression are based on the assumption that the model error terms

satisfy ??~?(0,?^2), ?=1,2,⋯,?, and are independent. We estimate ? with ? = sqrt(???/?−2). Compute ?

using the given data.

(b) Compute a 99% confidence interval for ?1.

(c) Test the hypothesis ?0: ?1≥.01 versus ?1: ?1<.01 at the ?=.05 level. State your conclusion.

(d) Suppose x=150. Compute a 95% confidence interval for ??|?=150, the population mean percent fiber

dry weight for asparagus with shear force 150 kg. Write an interpretation for this interval.

(e) Suppose x=150. Compute a 95% prediction interval for ?|? = 150, a future value of percent dry

weight for asparagus with shear force 150 kg. Write an interpretation for this interval.

Answer 1

a) Using the given data s=0.1123

b) 99% CI for slope parameter is (0.006714386, 0.009964723)

c) Here the test statistic is T= (estimate of  ?1-.01)/SE, which under the null has a t distribution with 16 DF.

The p value is P(t₁₆< observed T)

Now observed T=( 0.0083396 -.01)/ 0.0005564=-2.984184
Then p value=P(t₁₆<-2.984184)=0.004381909, which is less than 5% . Hence, evidence against the null is enough to reject the null and we reject the null.

d) 95% CI is (2.935118., 3.084289)

If x is fixed at 150, the above interval, in the long run, contains the true ??|?=150 in 95% cases .

e) 95% prediction interval is (2.760268, 3.259139)

A future observation for x=150 is expected to lie within the interval (2.760268, 3.259139) in 95% cases in repeated sampling.

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