Essentials of Statistics (5th Edition)

Published by Pearson
ISBN 10: 0-32192-459-2
ISBN 13: 978-0-32192-459-9

Chapter 3 - Statistics for Describing, Exploring, and Comparing Data - 3-2 Measures of Center - Page 93: 22

Answer

Mean:Brinks as collection contractor:1.55, not Brinks as collection contractor:1.73. Median:Brinks as collection contractor:1.55, not Brinks as collection contractor:1.65. It does not show evidence because the mean of Brinks employees is less than the mean of not Brink employees.

Work Step by Step

The mean can be counted by summing all the data and dividing it by the number of data: Brinks as collection contractor: $\frac{1.3+1.5+1.3+1.5+1.4+1.7+1.8+1.7+1.7+1.6}{10}=1.55$, not Brinks as collection contractor: $\frac{2.2+1.9+1.5+1.6+1.5+1.7+1.9+1.6+1.6+1.8}{10}=1.73$. The median is the average of the 1 or 2 (here 2) middle data: Brinks as collection contractor:(1.5+1.6)/2=1.55, not Brinks as collection contractor:(1.6+1.7)/2=1.65. It does not show evidence because the mean of Brinks employees is less than the mean of not Brink employees.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.