A study was performed to compare the preferences of eight

Chapter 15, Problem 41E

(choose chapter or problem)

A study was performed to compare the preferences of eight “expert listeners” regarding 15

models (with approximately equal list prices) of a particular component in a stereo system.

Every effort was made to ensure that differences perceived by the listeners were due to the

component of interest and no other cause (all other components in the system were identical, the same type of music was used, the music was played in the same room, etc.). Thus, the results of the listening test reflect the audio preferences of the judges and not judgments regarding quality, reliability, or other variables. Further, the results pertain only to the models of the components used in the study and not to any other models that may be offered by the various manufacturers. The data in the accompanying table give the results of the listening

tests. The models are depicted simply as models A, B, . . . , O. Under each column heading

are the numbers of judges who ranked each brand of component from 1 (lowest rank) to 15

(highest rank).

a Use the Friedman procedure to test whether the distributions of the preference scores differ

in location for the 15 component models. Give bounds for the attained significance level.

What would you conclude at the \(\alpha=.01\) level of significance? [Hint: The sum of the ranks

associated with the component of model O is \(5+6+8+8+9+10+10+11=67\); other rank sums can be computed in an analogous manner.]

b If, prior to running the experiment, we desired to compare components of models G and H,

this comparison could be made by using the sign test presented in Section 15.3. Using the

information just given, we can determine that model G was preferred to model H by all eight

judges. Explain why. Give the attained significance level if the sign test is used to compare

components of models G and H.

c Explain why there is not enough information given to use the sign test in a comparison of

only models H and M.

Equation transcription:

Text transcription:

5+6+8+8+9+10+10+11=67

alpha=.01