BirdnestS22.pptx
Nest predation: Developing a question and purpose:
Example of a scientific question: Does nest height matter for influencing nest predation?
Remember developing a question(hypothesis) for scientific research must meet the following requirements:
Testable
Repeatable
Measurable; through statistical analysis
Should be influenced by the scientific literature
Should have a reason for testing; i.e., Why do we care, or why is it important to society, the environment, the public, etc.
Tentative Location: (M,W, F classes)
Flat Rock State Natural Area
2381 Factory Road,
Murfreesboro, TN 37130
860 17’ 43”W, 350 51’ 30”N
Analyzing Ecology:
Independent variables: factors that are presumed to cause other variables to change.
Dependent variables: factors that are being changed.
Example:
We hypothesize that variation in the nest condition has led to different predation rates.
Independent variable = Nest condition
Dependent variable = Predation
* Experimental unit = nest
5.5ft
Use transects and marking them
Map it or sketch it
Assign treatment/control conditions along transects
Random table generator or random # table
Pattern
Random
Combination
Layout experiment
Brown Egg Nest
Speckled Egg Nest
POSITIVE SIGNS OF PREDATION
For this class we focus on Chi Squared Statistical Method:
“Goodness of Fit” Test
2X2 Contingency Table analysis
It is intended to test how likely it is that an observed distribution is due to chance. It is also called a "goodness of fit" statistic, because it measures how well the observed distribution of data fits with the distribution that is expected if the variables are independent.
| Data | High Nest | Low nest | Total |
| Predation | 5 | 6 | 11 |
| No Predation | 11 | 10 | 21 |
| Total | 16 | 16 | 32 |
Data Collection:
Dependent Variable
Independent Variable
| Pred/No | Trt. | Obs. | Calc. | Calc. | Expected |
| Pred | high | 5 | Col1 x row1/Total | (16 x 11)/32 | 5.5 |
| Pred | low | 6 | Col2 x row1/total | (16 x 11)/32 | 5.5 |
| No Pred | high | 11 | Col1 x row2/total | (16 x 21)/32 | 10.5 |
| No Pred | low | 10 | Col2 x row2/total | (16 x 21)/32 | 10.5 |
How to calculate expected frequencies:
| Data | High Nest | Low nest | Total |
| Predation | 5 | 6 | 11 |
| No Predation | 11 | 10 | 21 |
| Total | 16 | 16 | 32 |
| Pred/No | Trt. | Obs. | Exp. | (Obs – Exp)2 /Exp | |
| Pred | High | 5 | 5.5 | (5 – 5.5)2/ 5.5 | 0.045 |
| Pred | Low | 6 | 5.5 | (6 – 5.5)2/5.5 | 0.045 |
| No Pred | High | 11 | 10.5 | (11 – 10.5)2/10.5 | 0.024 |
| No Pred | Low | 10 | 10.5 | (10 – 10.5)2/10.5 | 0.024 |
| Sum of Chi-square | 0.139 |
Calculating Chi-square:
From the observed and expected frequencies
x2 = Σ(observed – expected)2
expected
x2 = (5-5.5) 2+ (11-10.5) 2+ (6-5.5) 2+ (10-10.5) 2
5.5 10.5 5.5 10.5
x2 = 0.139
Hypothesis – Additional info and X2(Chi square):
H0: The height of the nest has no effect on predation of eggs
HA: The height of the nest does have an effect on predation of eggs
Each group will have a X2(chi square critical value of 3.841), so if the calculated test statistic is…..
≥ critical value means that it supports the hypothesis of a difference (reject Ho)
≤ critical value means that it does NOT support the hypothesis of a difference (accept Ho))
Interpretation
Because our x2 is less than the critical value for p = 0.05, we fail to reject the null hypothesis
There is no significant difference in the predation of nests placed high versus nests that low
X2 = 0.139
critical value (from table, p = 0.05) = 3.84
0.139 < 3.84
13
Your Project:
Model birds find at Flat Rock
Lists on D2L
Control & treatment
Manipulate Nest or egg
Shape, size, color, materials, or decoy
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