Practice of Statistics in the Life Sciences
4th Edition
ISBN: 9781319013370
Author: Brigitte Baldi, David S. Moore
Publisher: W. H. Freeman
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Question
Chapter 2, Problem 2.6AYK
(a)
To determine
To find: the five-number summary of the dataset.
(a)
Expert Solution
Answer to Problem 2.6AYK
| Statistics | Group Instruction | Individual instruction |
| Minimum: | 78 | 128 |
| First | 117.50 | 159.75 |
| 147.50 | 191.50 | |
| Third quartile: | 172.00 | 221.50 |
| Maximum: | 359 | 283 |
Explanation of Solution
Given:
| Group Instruction | 141 | 158 | 112 | 153 | 134 | 95 | 96 | 78 | 148 |
| 172 | 200 | 271 | 103 | 172 | 359 | 145 | 147 | 255 | |
| Individual instruction | 128 | 195 | 188 | 158 | 227 | 198 | 163 | 164 | |
| 159 | 128 | 283 | 226 | 223 | 221 | 220 | 160 |
Calculation:
The five-point statistics calculated using the Excel is given below:
|
| Group Instruction | Individual instruction |
| count | 18 | 16 |
| 163.28 | 190.06 | |
| sample variance | 4,992.21 | 1,755.66 |
| sample standard deviation | 70.66 | 41.90 |
| minimum | 78 | 128 |
| maximum | 359 | 283 |
| 281 | 155 | |
| 1st quartile | 117.50 | 159.75 |
| median | 147.50 | 191.50 |
| 3rd quartile | 172.00 | 221.50 |
| 54.50 | 61.75 | |
| 172.00 | 128.00 | |
The five-point summery is:
| Statistics | Group Instruction | Individual instruction |
| Minimum: | 78 | 128 |
| First quartile: | 117.50 | 159.75 |
| Median: | 147.50 | 191.50 |
| Third quartile: | 172.00 | 221.50 |
| Maximum: | 359 | 283 |
(b)
To determine
To construct: the box plot for the given datasets.
(b)
Expert Solution
Explanation of Solution
Calculation:
The summery shown by the box plots is given below:
| Statistics | Group Instruction | Individual instruction |
| Minimum: | 78 | 128 |
| First quartile: | 117.50 | 159.75 |
| Median: | 147.50 | 191.50 |
| Third quartile: | 172.00 | 221.50 |
| Maximum: | 359 | 283 |
| low extremes | 0 | 0 |
| low outliers | 0 | 0 |
| high outliers | 2 | 0 |
| high extremes | 1 | 0 |
(c)
To determine
To find: the mean and standard deviation and explain whether it defines anything about shape or not.
(c)
Expert Solution
Answer to Problem 2.6AYK
Group Instruction:
Mean = 163.28
Standard Deviation = 70.66
Individual instruction:
Mean =190.66
Standard Deviation = 41.90
Explanation of Solution
Formula used:
Mean = Average(“Array”)
Standard Deviation = stdev.s(“array”)
Calculation:
| Statistics | Group Instruction | Individual instruction |
| Mean | 163.28 | 190.06 |
| Standard Deviation | 70.66 | 41.90 |
No, mean and standard are the values that tells about the central value and the standard deviation tell about the spread of the data but not the shape of the data.
(d)
To determine
To represent the mean, one standard deviation above and one standard deviation below the mean on the dot plot of the given two datasets.
(d)
Expert Solution
Explanation of Solution
The dot plots showing the mean and one standard deviation below and one standard deviation above than then mean is given below.
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Chapter 2 Solutions
Practice of Statistics in the Life Sciences
Ch. 2 - Prob. 2.1AYKCh. 2 - Prob. 2.2AYKCh. 2 - Prob. 2.3AYKCh. 2 - Prob. 2.4AYKCh. 2 - Prob. 2.5AYKCh. 2 - Prob. 2.6AYKCh. 2 - Prob. 2.7AYKCh. 2 - Prob. 2.8AYKCh. 2 - Prob. 2.9AYKCh. 2 - Prob. 2.10AYK
Ch. 2 - Prob. 2.11AYKCh. 2 - Prob. 2.12AYKCh. 2 - Prob. 2.13AYKCh. 2 - Prob. 2.14CYSCh. 2 - Prob. 2.15CYSCh. 2 - Prob. 2.16CYSCh. 2 - Prob. 2.17CYSCh. 2 - Prob. 2.18CYSCh. 2 - Prob. 2.19CYSCh. 2 - Prob. 2.20CYSCh. 2 - Prob. 2.21CYSCh. 2 - Prob. 2.22CYSCh. 2 - Prob. 2.23CYSCh. 2 - Prob. 2.24ECh. 2 - Prob. 2.26ECh. 2 - Prob. 2.27ECh. 2 - Prob. 2.28ECh. 2 - Prob. 2.29ECh. 2 - Prob. 2.30ECh. 2 - Prob. 2.31ECh. 2 - Prob. 2.32ECh. 2 - Prob. 2.33ECh. 2 - Prob. 2.34ECh. 2 - Prob. 2.35ECh. 2 - Prob. 2.36ECh. 2 - Prob. 2.37ECh. 2 - Prob. 2.38ECh. 2 - Prob. 2.39ECh. 2 - Prob. 2.40ECh. 2 - Prob. 2.41E
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