Question: Tukey's Quick Test. In this exercise, we examine an alternative method, conceived by the late Professor John Tukey, for performing a two-tailed hypothesis test for
Step 1 Count the number of observations in the high group that are greater than or equal to the largest observation in the low group. Count ties as 1/2.
Step 2 Count the number of observations in the low group that are less than or equal to the smallest observation in the high group. Count ties as 1/2.
Step 3 Add the two counts obtained in Steps 1 and 2, and denote the sum c.
Step 4 Reject the null hypothesis at the 5% significance level if and only if c ≥ 7; reject it at the 1% significance level if and only if c ≥ 10; and reject it at the 0.1% significance level if and only if c ≥ 13.
a. Can Tukey's quick test be applied to Exercise 10.42 on page 407? Explain your answer.
b. If your answer to part (a) was yes, apply Tukey's quick test and compare your result to that found in Exercise 10.42, where a t-test was used.
c. Can Tukey's quick test be applied to Exercise 10.76? Explain your answer.
d. If your answer to part (c) was yes, apply Tukey's quick test and compare your result to that found in Exercise 10.76, where a t-test was used.
For more details about Tukey's quick test, see J. Tukey, "A Quick, Compact, Two-Sample Test to Duckworth's Specifications" (Technometrics, Vol. 1, No. 1, pp. 31-48).
Step by Step Solution
3.47 Rating (167 Votes )
There are 3 Steps involved in it
a No The highest value 246 and the lowest value 93 are bo... View full answer
Get step-by-step solutions from verified subject matter experts
Document Format (1 attachment)
503-M-S-S-I (1546).docx
120 KBs Word File
