Question
In: Math
5. In the table below are simulated data of net photosynthesis rate for five replicate leaves...
5. In the table below are simulated data of net photosynthesis rate for five replicate leaves from two plant species, sawgrass (Cladium jamaiscence) and willow (Salix caroliniana). Using the data below and the ‘Analyzing Ecology’ box in Chapter 2 of the text (pg 42), calculate the mean (x), variance (s¬2), standard deviation (s), and standard error (SE) for the two pH conditions. Construct a graph using the means and standard deviations to visually compare photosynthetic capacity between the two species. Make sure to label your axes appropriately and indicate which is the dependent and the independent variable.
Table 1. Photosynthesis rate (µmol CO2 / m2 / sec) for sawgrass and willow leaves.
|
Replicate |
Sawgrass |
Willow |
|
1 |
6.4 |
8.0 |
|
2 |
6.1 |
8.4 |
|
3 |
6.8 |
7.7 |
|
4 |
6.3 |
8.1 |
|
5 |
6.3 |
7.9 |
.
Solutions
milcah answered 1 year agoRelated Solutions
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The daily net cash flows in the table below represent a 5-day
sample. Use the Data Analysis tool in Excel to calculate the 95%
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Day
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15,000
2
17,000
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13,000
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20,000
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14,000
The daily net cash flows in the table below represent a 5-day sample. Use the Data...
The daily net cash flows in the table below represent a 5-day
sample. Use the Data Analysis tool in Excel to calculate the 95%
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with your calculation, interpret the meaning behind your confidence
interval estimates and discuss why the 99% confidence interval is
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Date
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Unit Cost
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25
$40
May/06
Sale
20
$50
May/08
Purchase
20
$41
May/16
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10
$55
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20
$42
May/23
Sale
25
$55
May/30
Purchase
15
43
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NPV
Year
1 Year
2
Year 3
1
12.8
4.40
3.30
3.50
2
10.5
4.30
2.50
2.10
3
9.4
4.10
3.40
1.50
4
11.7
4.70
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3.10
...
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COMPUTE ONLY THE SUM of SQUARES
FOR TREATMNET
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10, 10, 13
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6
11
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SS Treatment =
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Year 2
Year 3
Year 4
Year 5
Year 6
January
5550
4782
5283
4760
4314
February
5398
4897
4544
4930
4476
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4236
4693
4370
4930
4476
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2977
3747
2419
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2345
2235
2463
2546
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1198
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The table below will be used for the following five questions. The table represents the mean...
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Age
Mean Graduation Rate
Standard Deviation
16
76.6
1.8
17
74.2
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18
79.5
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19
79.6
2.1
20
81.2
2.1
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84.2
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B1
B2
n
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0
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1
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