Moose 1 2 3 4 5 6 xa: WOLVES ABSENT Fat (x) 43 49 14 57 31 19 35.5 x-xa (x-x₂)² Moose 7.5 56.25 13.5 -21.5 21.5 -4.5 182.25 462.25 o a: 462.25 20.25 -16.5 272.25 Sum of squared differences: 1455.5 1 2 3 4 5 6 xp: WOLVES PRESENT Fat (x) 76 68 52 38 62 81 62.8
Moose 1 2 3 4 5 6 xa: WOLVES ABSENT Fat (x) 43 49 14 57 31 19 35.5 x-xa (x-x₂)² Moose 7.5 56.25 13.5 -21.5 21.5 -4.5 182.25 462.25 o a: 462.25 20.25 -16.5 272.25 Sum of squared differences: 1455.5 1 2 3 4 5 6 xp: WOLVES PRESENT Fat (x) 76 68 52 38 62 81 62.8
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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Transcribed Image Text:Moose
1
2
3
4
5
6
xa:
WOLVES ABSENT
Fat (x)
43
49
14
57
31
19
35.5
x-xa (x-x₂)² Moose
7.5
56.25
13.5
-21.5
21.5
-4.5
182.25
462.25
o a:
462.25
20.25
-16.5 272.25
Sum of squared differences: 1455.5
1
2
3
4
5
6
WOLVES PRESENT
Fat (x)
76
68
52
38
62
81
¯p: 62.8

Transcribed Image Text:M
3
0²/12
Here, na
(sum of squared differences),
na - 1
is the number of moose sampled from the
population with wolves absent.
Estimate the variance in moose fat stores (WOLVES
ABSENT). To do this, divide the Sum of squared
differences by the sample size minus one (na-1).
Enter this value in the bottom row of the left half of the
table.
Q4.10. Make sure you're on track. What is your
estimated variance for fat stores with wolves
absent?
Check Answer
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