Question: Problem 11.16 analyzed the number of weekly hours worked per person at five different plants. An F value of 3.10 was obtained with a probability
Problem 11.16 analyzed the number of weekly hours worked per person at five different plants. An F value of 3.10 was obtained with a probability of .0266. Because the probability is less than .05, the null hypothesis is rejected at α = .05. There is an overall difference in the mean weekly hours worked by plant. Which pairs of plants have significant differences in the means, if any? To answer this question, a Minitab computer analysis was done. The data follow. Study the output in light of problem and discuss the results.
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons
Individual confidence level = 99.32%
Plant 1 subtractedfrom:
.png)
Plant 2 17.910 7.722 2.466 Plant3 Plant 4-21.830-8.665 4.499 Plant 512.7210. Lower Cnter Upper 3.598 14.939 1 10.880 -7.743 (- -15 15 3 0 Plant 2 subtracted from: Lo wer Center Upper -t Plant 3 Plant 4-13.935 -0.944 12.04 8 Plant 5 0.180 11.320 22.46O -4.807 6.801 18.409 Plant 3 subtracted from: -15 15 30 Lower Center Upper --+--+- Plant 4 26.178-12.263 1.651 -) Plant 5 17.151.519 8.113 Plant 4 subtracted from: -15 15 30 Lower Cr Upper 7.745 22.036 Plant 9S -6.547 -15 15 30
Step by Step Solution
3.23 Rating (164 Votes )
There are 3 Steps involved in it
a 05 There were five plants and ten pairwise comparisons The Mi... View full answer
Get step-by-step solutions from verified subject matter experts
Document Format (1 attachment)
243-M-S-S-I (359).docx
120 KBs Word File
