Three replicate water samples were taken at each of four locations in a river to determine whether the quantity of dissolved oxygen€”a measure of water pollution€” varied from one location to another (the higher the level of pollution, the lower the dissolved oxygen reading). Location 1 was adjacent to the wastewater discharge point for a certain industrial plant, and locations 2, 3, and 4 were selected at points 10, 20, and 30 miles downstream from this discharge point. The resulting data appear in the accompanying table. The quantity Yij denotes the value of the dissolved oxygen content for the jth replicate at location i (j = 1, 2, 3 and i = 1, 2, 3, 4), and i denotes the mean of the three replicates taken at location i.

a. Do the data provide sufficient evidence to suggest that values for mean dissolved oxygen content differ significantly among the four locations? (Use α = .05.) Make sure to construct the appropriate ANOVA table.
b. Given that µi represents the true mean level of dissolved oxygen at location i (i = 1, 2, 3, 4), test the null hypothesis
H0: -3µ1 - µ2 + µ3 + 3µ4 = 0
versus
HA: -3µ1 - µ2 + µ3 + 3µ4 ‰  0
at the 2% level. The quantity (-3µ1 - µ2 + µ3 + 3µ4) is a contrasr based on orthogonal polynomials, which can be shown to be a measure of the linear relationship between "location" (the four equally spaced distances 0, 10, 20, and 30 miles downstream from the plant) and "dissolved oxygen content."
c. Another way to quantify the strength of this linear relationship is to fit by least squares the model
Y = β0 + β1X1 + E
where

Fitting such a regression model to the n = 12 data points yields the accompanying ANOVA table. Use this table to perform a test of H0: β1 = 0 at the 2% significance level.

d. The regression model in part (c) amounts to saying that µi = β0 + β1Xi. Show that the hypothesis tested in part (b) is equivalent to the hypothesis tested in part (c).
e. Why do the two test statistics calculated in parts (b) and (c) not have the same numerical value? What reasonable modification of the test in part (c) would yield the same F-value as that obtained in part (b)?
f. Given the results of part (b), a test for a nonlinear trend in mean response can be obtained by subtracting the sum of squares for the contrast

from the sum of squares for treatments and then dividing this difference by the appropriate degrees of freedom to yield an F statistic of the form

Carry out this test based on the results obtained in parts (a) and (b). (Use α = .05.)
g. Carry out the usual regression lack-of-fit test for adequacy of the straight-line model fit in part (c), using α = .05. Does the value of the F statistic equal the value obtained in part (f)?

Transcribed Image Text:

Dissolved Oxygen Content (V Location Mean (%) 4.5,6 6, 6,6 7,8,9 8, 9, 10 6 8 0 for location 1 10 or location 2 20 for location 3 30 for location 4 Source d.f Regression 29.40 Residual Lack of fit 3 0.60 6.00 10 6.60 Pure error F(Nonlinear trend) = [SST-SSZ)]/df MSE