Question

Evaluation of flexography printing plates. Refer to the Journal of Graphic Engineering and Design (Vol. 3, 2012) study of the quality of flexography printing, Exercise 9.24 (p. 544). Recall that four different exposure times were studied—8, 10, 12, and 14 minutes—and that the measure of print quality used was dot area (hundreds of dots per square millimeter). Tukey’s multiple comparisons procedure (at an experimentwise error rate of .05) was used to rank the mean dot areas of the four exposure times. The results are summarized below. Which exposure time yields the highest mean dot area? Lowest?

Short answer

Exposure times 8 min and 14 min yields the highest mean dot area. Exposure times 12 min yields the lowest mean dot area.

Step-by-step solution

 Given Information

The four different exposure times are 8, 10, 12 and 14 minutes.

Tukey’s multiple comparison method is used to rank at an experiment wise error rate of .05.

Tukey’s Method

Tukey’s method requires an equal sample size of treatments. Tukey's procedure is assigned when pairwise comparisons are of interest. The formulas for constructing confidence intervals for differences between treatment means using Tukey’s is given by

Where

k= number of treatments

v = number of degrees of freedom

n_t= Number of observations in each k

Explanations

The mean dot areas are .571, .582, .588, and .594.

From the above information, we can see that there is no significant difference between the mean dot area of exposures times 8 min and 14 min as both times are joined by line. There is less significance for the exposure time 12 min.

Therefore, the exposure times 8 min is equal to .588.

And the times 14 min is equal to .594

Therefore, exposure times 8 min and 14 min yields the highest mean dot area. Exposure times 12 min yields the lowest mean dot area.