The Analysis of Biological Data
2nd Edition
ISBN: 9781936221486
Author: Michael C. Whitlock, Dolph Schluter
Publisher: W. H. Freeman
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Chapter 3, Problem 1PP
(a)
To determine
To find: The
(a)
Expert Solution
Explanation of Solution
Given:
The data set is:
| 112 |
| 128 |
| 108 |
| 129 |
| 125 |
| 153 |
| 155 |
| 132 |
| 137 |
The provided data represents that there are total 9 observations. Thus, the sample size is 9.
(b)
To determine
To find: The sum of the provided observations.
(b)
Expert Solution
Answer to Problem 1PP
The sum is 1179.
Explanation of Solution
The sum of the provided observations can be calculated as:
Thus, the required sum is 1179.
(c)
To determine
To fin: The
(c)
Expert Solution
Answer to Problem 1PP
The mean is 131 mm Hg
Explanation of Solution
The mean for the provided data set can be computed as:
Thus, the mean is 131 mm Hg.
(d)
To determine
To find: The sum of square.
(d)
Expert Solution
Answer to Problem 1PP
The value of the sum of square is 2036.
Explanation of Solution
The sum of square is computed as:
| 112 | -19 | 361 |
| 128 | -3 | 9 |
| 108 | -23 | 529 |
| 129 | -2 | 4 |
| 125 | -6 | 36 |
| 153 | 22 | 484 |
| 155 | 24 | 576 |
| 132 | 1 | 1 |
| 137 | 6 | 36 |
|
|
Thus, the sum of square is 2036.
(e)
To determine
To find: The variance for the provided data set.
(e)
Expert Solution
Answer to Problem 1PP
The variance is 254.2
Explanation of Solution
Variance for the provided data set is computed as:
(f)
To determine
To find: The standard deviation for the provided data set.
(f)
Expert Solution
Answer to Problem 1PP
The standard deviation is 15.950
Explanation of Solution
Standard deviation for the provided data set is computed as:
Thus, the standard deviation is 15.950.
(g)
To determine
To find: The coefficient of variation.
(g)
Expert Solution
Answer to Problem 1PP
The coefficient of variation is 12.176%
Explanation of Solution
The coefficient of variation can be computed as:
The coefficient of variation is 12.176%.
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Bob and Teresa each collect their own samples to test the same hypothesis. Bob’s p-value turns out to be 0.05, and Teresa’s turns out to be 0.01.
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Review a classmate's Main Post.
1. State if you agree or disagree with the choices made for additional analysis
that can be done beyond the frequency table.
2. Choose a measure of central tendency (mean, median, mode) that you
would like to compute with the data beyond the frequency table. Complete
either a or b below.
a. Explain how that analysis can help you understand the data better.
b. If you are currently unable to do that analysis, what do you think you
could do to make it possible? If you do not think you can do
anything, explain why it is not possible.
0|0|0|0
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Consider the time series X₁ and Y₁ = (I – B)² (I – B³)Xt. What transformations were
performed on Xt to obtain Yt?
seasonal difference of order 2
simple difference of order 5
seasonal difference of order 1
seasonal difference of order 5
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Chapter 3 Solutions
The Analysis of Biological Data
Ch. 3 - Prob. 1PPCh. 3 - Prob. 2PPCh. 3 - Prob. 3PPCh. 3 - Prob. 4PPCh. 3 - Prob. 5PPCh. 3 - Prob. 6PPCh. 3 - Prob. 7PPCh. 3 - Prob. 8PPCh. 3 - Prob. 9PPCh. 3 - Prob. 10PP
Ch. 3 - Prob. 11PPCh. 3 - Prob. 12PPCh. 3 - Prob. 13PPCh. 3 - Prob. 14APCh. 3 - Prob. 15APCh. 3 - Prob. 16APCh. 3 - Prob. 17APCh. 3 - Prob. 18APCh. 3 - Prob. 19APCh. 3 - Prob. 20APCh. 3 - Prob. 21APCh. 3 - Prob. 22APCh. 3 - Prob. 23APCh. 3 - Prob. 24APCh. 3 - Prob. 25APCh. 3 - Prob. 26APCh. 3 - Prob. 27APCh. 3 - Prob. 28APCh. 3 - Prob. 29AP
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